Important Questions

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 7 Equilibrium Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 7 Equilibrium

Very Short Answer Type Questions

Question 1.
A tank is full of water. Water is coming in as well as going out at same rate. What will happen to level of water in a tank? What is name given to such state?
Answer:
It will remain the same because rate of inflow is equal to rate of outflow. This state is called state of ‘equilibrium’.

Question 2.
The ionization of hydrogen chloride in water is given t
HCl(aq) + H₂O(l) ⇌ H₃O⁺+(aq) + Cl^–(aq)
Label two conjugate acid-base pairs in this ionization.
Answer:

Question 3.
Why solution of sugar in water does not conduct electricity whereas that of common salt in water does?
Answer:
Common salt (NaCl) is an electrolyte which gives Na⁺ and Cl^– ions in the aqueous solution. Hence, it conducts electricity. Sugar is sucrose (C₁₂H₂₂O₁₁) which is a non-electrolyte and does not give ions in the solution. Hence, it does not conduct electricity.

Question 4.
Why is ammonia termed as a base though it does not contain OH^– ions?
Answer:
Ammonia is termed as a base due to its tendency to donate electron pair. Therefore it is a Lewis base.

Question 5.
K_b for NH₄O, H is 1.8 x 10^-5 and for CH₃NH₂ is 44 x 10^-4. Which of them is strongest base and why?
Answer:
CH₃NH₂ is strongest base because it has high value of base dissociation constant.

Question 6.
pK_a value of acids A, B, C, D are 1.5, 3.5, 2.0 and 5.0. Which of them is strongest acid?
Answer:
Acid A with pK_a = 1.5 is strongest acid, lower the value of pK_a stronger will be the acid.

Question 7.
What will be the pH of 1M Na₂SO₄ solution?
Answer:
Na₂SO₄ is salt of strong acid and strong base, thus its aqueous solution will be neutral. Therefore, its pH will be 7.

Question 8.
Is it possible to get precipitate of Fe(OH)₃ at pH = 2? Give reason.
Answer:
No, because Fe(OH)₃ will dissolve in strongly acidic medium.

Question 9.
What happens to ionic product of water if some acid is added to it?
Answer:
Ionic product will remain unchanged.

Question 10.
How does common ion affect the solubility of electrolyte?
Answer:
Solubility of electrolyte decreases due to common ion effect.

Short Answer Type Questions

Question 1.
A certain buffer is made by mixing sodium form ate and formic acid in water. With the help of equations explain how this buffer neutralises addition of a small amount of an acid or a base?
Answer:
HCOONa → HCOO^– + Na⁺
HCOOH ⇌ HCOO^– + H⁺

HCOO^– is common ion in the above acidic buffer. When small amount of H⁺ ions is added, these H⁺ ions combine with HCOO^– which are in excess to form HCOOH back and [H⁺] remains practically same, so pH remains constant. When small amount of OH^– ions are added, OH^– ions will take up H⁺ and association of HCOOH will increase so as to maintain concentration of H⁺ ions. So, pH would not be affected.

Question 2.
How much volume of 0.1 M CH₃COOH should he added to 50 ml of 0.2 M CH₃COONa solution to prepare a buffer solution of pH 4.91. (pA_a of AcH is 4.76).
According to Henderson’s equation

Required volume of 0.1 M acetic acid = 70.92 mL

Question 3.
Some processes are given below. What happens to the process if it is subjected to a change given in the brackets?

(ii) Dissolution of NaOH in water (Temperature is increased)
(iii) N₂(g) + O₂(g) ⇌ 2NO(g) -180.7 kJ (Pressure is increased and temperature is decreased.)
Answer:
(i) Equilibrium will shift in the forward direction, i.e., more ice will melt.
(ii) Solubility will decrease because it is an exothermic process.
(iii) Pressure has no effect. Decrease of temperature will shift the equilibrium in the backward direction.

Question 4.
50.0 g of CaCO₃ are heated to 1073 K in a 5 L vessel. What percent of the CaCO₃ would decompose at equilibrium? K_p for the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g) is 1.15 atm at 1073 K.
Answer:
The reaction is : CaCO₃(s) ⇌ CaO(s) + CO₂(g)
K_p = P_Co2 = 1.15 atm, pV = nRT
\(\mathrm{n}_{\mathrm{CO}_{2}}=\frac{p_{\mathrm{CO}_{2}} \mathrm{~V}}{R T}=\frac{1.15 \times 5}{0.082 \times 1073}\) = 0.065 mol

1 mole of CO₂ is obtained by decomposition of 1 mole CaCO₃. Therefore, moles of CaCO₃ decomposed is equal to the moles of CO₂ = 0.065 mol.
Mole of CaCO₃ initially present = \(\frac{50}{100}\) = 0.5 mol
[Molecular mass of CaCO₃ = 100]
Per cent of CaCO₃ decomposed = \(\frac{0.065}{0.5}\) x 100 = 13%

Question 5.
Arrange the following in increasing order of pH.
KNO₃(aqr), CH₃COONa(aq), NH₄Cl(aq), C₆H₅COONH₄(aq)
Answer:
(i) KNO₃ is a salt of strong acid-strong base, hence its aqueous solution is neutral; pH = 7
(ii) CH₃COONa is a salt of weak acid and strong base, hence, its aqueous solution is basic; pH < 7.
(iii) NH₄Cl is a salt of strong acid and weak base, hence its aqueous solution is acidic; pH < 7.
(iv) C₆H₅COONH₄ is a salt of weak acid, C₆H₅COOH and weak base, NH4OH. ButNH4OH is slightly stronger than C6H5COOH. Hence, pH is slightly greater than 7.
Therefore, increasing order of pH of the given salts is,
NH₄Cl < KNO₃ < C₆H₅COONH₄ < CH₃COONa

Long Answer Type Questions

Question 1.
Calculate the pH of a buffer which is 0.1 M in acetic acid and 0.15 M in sodium acetate. Given that the ionisation constants of acetic acid is 1.75 x 10^-5. Also calculate the change in pH of the buffer if the following adds in 1 L of the buffer (i) 1 cc of 1 M NaOH. (ii) 1 cc of 1 M HC1. Assume that the charge in volume is negligible, (iii) What will be the buffer index of the above buffer?
Answer:
pH = pK_a + log\(\frac{Salt}{Acid}\) = – log(1.75 x 10^-5) + log
\(\frac{0.15}{0.10}\)
= (5 – 0.2430) + 0.1761 = 4.757 + 0.1761 = 4.933.

(i) 1 cc of 1M NaOH contains NaOH = 10^-3 mol. This will convert 10^-3 mol of acetic acid into the salt so that salt formed = 10^-3 mol.
[Acid] = 0.10 – 0.001 = 0.099 M
[Salt] = 0.15 + 0.001 = 0.151 M
pH =. 4.757 + log \(\frac{0.151}{0.099}\)
= 4.757 + 0.183 = 4.940
∴ Increase in pH = 4.940 – 4.933 = 0.007 which is negligible.

(ii) 1 cc of 1 M HC1 contains HCl = 1CF₃ mol. This will convert 10^-3 mol CH₃COONa into CH₃COOH.
Now, [Acid] = 0.10 + 0.001 = 0.101 M
[Salt] = 0.15 – 0.001 = 0.149 M 0.149
∴ pH = 4.757 + log\(\frac{0.149}{0.101}\) = 4.757 + 0.169 = 4.925
∴ Decrease in pH = 4.933 = 0.007 which is again negligible.

(iii) Calculation of buffer index No. of moles of HC1 or NaOH added = 0.001 mol
Change in pH = 0.007
Hence, buffer index = \(\frac{\Delta n}{\Delta \mathrm{pH}}=\frac{0.001}{0.007}=\frac{1}{7}\)= 0.143

Question 2.
On the basis of Le-Chatelier’s principle, explain how temperature and pressure can be adjusted to increase the yield of ammonia in the following reaction?
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
What will be the effect of addition of argon to the above reaction mixture at constant volume?
Answer:
N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = -92.38 kJ mol^-1

It is an exothermic process. According to Le-Chatelier’s principle, low temperature is favourable for high yield of ammonia, but practically very low temperatures slow down the reaction. So, optimum temperature, 700 K is favourable in attainment of equilibrium.

Similarly, high pressure about 200 atm is favourable for high yield of ammonia. On increasing pressure, reaction goes in the forward direction because the number of moles decreases in the forward direction.

At constant volume, addition of argon does not affect the equilibrium because it does not change the partial pressures of the reactants or products involved in the reaction and the equilibrium remains undisturbed.

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 6 Thermodynamics Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 6 Thermodynamics

Very Short Answer Type Questions

Question 1.
Identify the state functions and path functions out of the following.
Enthalpy, entropy, heat, temperature, work, free energy.
Answer:
State function Enthalpy, entropy, temperature, free energy.
Path function Heat, work

Question 2.
At 1 atm will the Δ_fH⁰ be zero for Cl₂(g) and Br₂(g)? Explain.
Answer:
Δ_fH⁰ for Cl₂(g) will be zero but Δ_fH⁰ for Br₂(g) will not be zero because liquid bromine state is elementary state not gaseous.

Question 3.
Why for predicting the spontaneity of a reaction, free energy criteria is better than the entropy criteria?
Answer:
Criteria of free energy change is better because it requires free energy change of the system only whereas the entropy change requires the total entropy change of the system and the surroundings.

Question 4.
Water can be lifted into the water tank at the top of the house with the help of a pump. Then why is it not considered to be spontaneous?
Answer:
A spontaneous process should occur continuously by itself after initiation. But this is not so in the given case because water will go up so long as the pump is working.

Question 5.
Given that ΔH = 0 for mixing of two gases. Explain whether the diffusion of these gases into each other in a closed container is a spontaneous process or not?
Answer:
It is a spontaneous process because although ΔH = 0, i.e., energy factor has no role to play but randomness increases, i.e., randomness factor favours the process.

Question 6.
Under what condition, the heat evolved or absorbed in a reaction is equal to its free energy change?
Answer:
As ΔG = ΔH – TΔS. Thus, ΔG = ΔH only when either the reaction is carried out at 0 K or the reaction is not accompanied by any entropy change, i.e., ΔS = 0.

Question 7.
In the equation, N₂(g) + 3H₂(g) ⇌ 2NH₃(g) what would be the sign of work done?
Answer:
The sign of work done will be positive, i.e., work will be done on the system due to decrease in volume.

Question 8.
The molar enthalpy of vaporisation of acetone is less than that of water. Why?
Answer:
Enthalpy of vaporisation of water is more than that of acetone because there is strong hydrogen bonding in H₂O molecules.

Question 9.
One mole of acetone requires less heat to vaporise than 1 mole of water. Which of the two liquids has higher enthalpy of vaporisation?
Answer:
Less the heat required to vaporise 1 mole of a liquid, less is its enthalpy of vaporisation. Hence, water has higher enthalpy of vaporisation.

Question 10.
Which quantity out of Δ_rG and Δ_rG° will be zero at equilibrium?
Answer:
Δ_rG = Δ_rG° + RT In K
At equilibrium, 0 (zero) = Δ_rG° + RT In K
(v Δ_rG = 0)
or Δ_rG° = -RT In It;
Δ_rG° = 0 when it = 1
For all other values of K, Δ_rG° will be non-zero.

Short Answer Type Questions

Question 1.
Define the following :
(i) First law of thermodynamics.
(ii) Standard enthalpy of formation.
Answer:
(i) First law of thermodynamics : It states that energy can neither be created nor be destroyed. The energy of an isolated system is constant. ΔU = q + w
(ii) Standard Enthalpy of Formation : It is defined as the amount of heat evolved or absorbed when one mole of the compound is formed from its constituent elements in their standard states.

Question 2.
Give reason for the following:
(i) Neither q nor w is a state function but q + w is a state function.
(ii) A real crystal has more entropy than an ideal crystal.
Answer:
(a) q + w = ΔU
As ΔU is a state function hence, q + w is a state function.
(b) A real crystal has some disorder due to the presence of defects in its structural arrangement whereas ideal crystal does not have any disorder. Hence, a real crystal has more entropy than an ideal crystal.

Question 3.
Represent the potential energy/enthalpy change in the following processes graphically
(i) Throwing a stone from the ground to roof.
(ii) \(\frac{1}{2}\)H₂(g) + \(\frac{1}{2}\)Cl₂(g) ⇌ HCl(g); Δ_rH^s = – 92.32kJ mol^-1
In which of the processes potential energy/enthalpy change is contributing factor to the spontaneity?
Answer:

Energy increases in (a) and it decreases in (b). Hence, in process (b), enthalpy change is the contributing factor to the spontaneity.

Question 4.
A man takes a diet equivalent to 10000 kJ per day and does work, by expending his energy in all forms equivalent to 12500 kJ per day. What is change in internal energy per day? If the energy lost was stored as sucrose (1632 kJ per 100 g), how many days should it take to lose 2 kg of his weight? (Ignore water loss)
Answer:
Energy taken by a man = 10000 kJ
Change in internal energy per day = 12500 -10000 = 2500 kJ
The energy is lost by the man as he expends more energy than he takes.
Now 100 g of sugar corresponds to energy = 1632 kJ loss in energy.
2000 g of sugar corresponds to energy = \(\frac{1632 \times 2000}{100}\) = 32640 kJ
∴ Number of days required to lose 2000 g of weight or 32640 kJ of energy = \(\frac{32640}{2500}\) = 13 days

Question 5.
Give the appropriate reason :
(i) It is preferable to determine the change in enthalpy rather than the change in internal energy.
(ii) It is necessary to define the ‘standard state’.
(iii) It is necessary to specify the phases of the reactants and products in a thermochemical equation.
Answer:
(i) Because it is easier to make measurement under constant pressure than under constant volume conditions.
(ii) Enthalpy change depends upon the conditions in which a reaction is carried out. For making the comparison of results obtained by different people meaningful, the reaction conditions must be well-defined.
(iii) Because enthalpy depends upon the phase of reactants and products.

Long Answer Type Questions

Question 1.
(i) A cylinder of gas supplied by a company is assumed to contain 14 kg of butane. If a normal family requires 20000 kJ of energy per day for cooking, how long will the cylinder last?
(ii) If the air supplied to the burner is insufficient, a portion of gas escapes without combustion. Assuming that 25% of the gas is wasted due to this inefficiency, how long will the cylinder last (Heat of combustion of butane = 2658kJ!mol.)?
Answer:
(i) Molecular formula of butane = C₄H₁₀
Molecular mass of butane = 4 x 12 +10 x 1 = 58
Heat of combustion of butane 2658 kJ mol^-1
1 mole.or 58 g of butane on complete combustion gives heat = 2658 kJ
∴ 14 x 10³ g of butane on complete combustion will give heat
= \(\frac{2658 \times 14 \times 10^{3}}{58}\) = 641586 kJ
The family needs 20000 kJ of heat per day.
∴ 20000 kJ of heat is used for cooking by a family in 1 day.
∴ 641586 kJ of heat will be used for cooking by a family in
= \(\frac{641586}{20000}\) = 32days
The cylinder will last for 32 days

(ii) 25 per cent of the gas is wasted due to inefficiency. This means that only 75% of butane gets combusted. Therefore, the energy produced by
75% combustion of butane = \(\frac{641586 \times 75}{100}\) = 481189.5 kJ
∴ The number of days the cylinder will last = \(\frac{481189.5}{20000}\) = 24 days.

Question 2.
10 moles of an ideal gas expand isothermally and reversibly from a pressure of 5 atm to 1 atm at 300 K. What is the largest mass that can be lifted through a height of 1 m by this expansion?
Answer:
W_exp = -2.303 nRT log \(\frac{p_{1}}{p_{2}}\)
= -2.303(10) x (8.314)(300) log \(\frac{5}{1}\) = – 40.15 x 10³ J
If M is the mass that can be lifted by this work through a height of 1 m, then work done = Mgh
40.15 x 10³ J = M x 9.81 ms^-1 x 1 m
or M = \(\frac{40.15 \times 10^{3} \mathrm{~kg} \mathrm{~m}^{2} \mathrm{~s}^{-2}}{9.81 \mathrm{~m} \mathrm{~s}^{-2} \times 1 \mathrm{~m}}\) [∵ J = kg m²s^-2]
= 4092.76 kg

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 5 States of Matter Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 5 States of Matter

Very Short Answer Type Questions

Question 1.
Name two intermolecular forces that exist between HF molecules in a liquid state.
Answer:
HF are polar covalent molecules. In the liquid state, there are dipole-dipole interactions and H-bonding.

Question 2.
Explain why Boyle’s law cannot be used to calculate the volume of a real gas when it is converted from its initial state to the final state by an adiabatic expansion.
Answer:
During adiabatic expansion, the temperature is lowered, and therefore, Boyle’s law cannot be applied.

Question 3.
Boyle’s law states that at constant temperature, if pressure is increased on a gas, volume decreases and vice-versa. But when we fill air in a balloon, volume as well as pressure increase. Why?
Answer:
The law is applicable only for a definite mass of the gas. As we fill air into the balloon, we are introducing more and more air into the balloon.
Thus, we are increasing the mass of air inside. Hence, the law is not applicable.

Question 4.
What will be the molar volume of nitrogen and argon at 273.15 K and 1 atm?
Answer:
Every gas has 22.4 L molar volume at 273.15 K and 1 atm pressure (STP).

Question 5.
A gas that follows Boyle’s law, Charles’ law and Avogadro’s law is called an ideal gas. Under what conditions a real gas would behave ideally?
Answer:
At low pressure and high temperature, a real gas behaves as an ideal gas.

Question 6.
Explain why temperature of a boiling liquid remains constant?
Answer:
This is because at the boiling point, the heat supplied is used up in breaking off the intermolecular forces of attraction of the liquid to change it into vapour and not for raising the temperature of the liquid.

Question 7.
Assuming C02 to be van der Waals’ gas, calculate its Boyle temperature.
Given a = 3.59 L² atm mol^-2 and b = 0.0427 L mol^-1.
\(T_{b}=\frac{a}{R b}=\frac{3.59 \mathrm{~L}^{2} \mathrm{~atm} \mathrm{} \mathrm{mol}^{-2}}{\left(0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)\left(0.0427 \mathrm{~L} \mathrm{~mol}^{-1}\right)}\) = 1025.3 K

Question 8.
Name two phenomena that can be explained on the basis of surface tension.
Answer:
Surface tension can explain
(i) capillary action, i.e., rise or fall of a liquid in capillary,
(ii) spherical shape of small liquid drops.

Question 9.
Why are the gases helium and hydrogen not liquefied at room temperature by applying very high pressure?
Answer:
Because their, critical temperature is lower than room temperature. (Gases cannot be liquefied above the critical temperature by applying even very high pressure).

Question 10.
What would have happened to the gas if the molecular collisions were not elastic?
Answer:
On every collision, there would have been loss of energy. As a result, the molecules would have slowed down and ultimately settle down in the vessel. Moreover, the pressure would have gradually reduced to zero.

Short Answer Type Questions

Question 1.
(i) What do you mean by ‘Surface Tension’ of a liquid?
(ii) Explain the factors which can affect the surface tension of a liquid.
Answer:
(i) Surface tension : It is defined as the force acting per unit length perpendicular to the line drawn on the surface. It’s unit is Nm-1.
(ii) Surface tension of a liquid depends upon the following factors :
(a) Temperature : Surface tension decreases with rise in temperature. As the temperature of the liquid increases, the average kinetic energy of the molecules increases. Thus, there is a decrease in intermolecular force of attraction which decrease the surface tension.
(b) Nature of the liquid : Greater the magnitude of
intermolecular forces of attraction in the liquid, greater will be the value of surface tension.

Question 2.
A neon-dioxygen mixture contains 70.6 g dioxygen and 167.5g neon. If pressure of the mixture of gases in the cylinder is 25 bar.
What is the partial pressure of dioxygen and neon in the mixture?
Answer:

Alternatively, mole fraction of neon = 1 – 0.21 = 0.79
Partial pressure of a gas = mole fraction x total pressure
⇒ Partial pressure of oxygen = 0.21 x (25bar) = 5.25bar
Partial pressure of neon = 0.79 x (25bar) = 19.75bar

Question 3.
Give reasons for the following:
(i) The size of weather balloon becomes larger and larger as it ascends into higher altitudes.
(ii) Tyres of automobiles are inflated to lesser pressure in summer than in winter.
Answer:
(i) As we go to higher altitudes, the atmospheric pressure decreases.
Thus, the pressure outside the balloon decreases. To regain equilibrium with the external pressure, the gas inside expands to decrease its pressure, Hence, the size of the balloon increases.
(ii) In summer, due to higher temperature, the average kinetic energy of the air molecules inside the tyre increases, i.e., molecules start moving faster. Hence, the pressure on the walls of the tube increases. If pressure inside is not kept low at the time of inflation, at higher temperature, the pressure may become so high that the tyre may burst.

Question 4.
On the basis of intermolecular forces and thermal energy, explain why .
(i) a solid has rigidity but liquids do not have rigidity?
(ii) gases have high compressibility but liquids and solids have poor compressibility?
Answer:
(i) It is because in solids, the intermolecular forces are very strong and predominate over thermal energy but in liquid, these forces are no longer strong enough.
(ii) Because of very weak intermolecular forces and high thermal energy, molecules of gases are far apart. That is why gases are highly compressible.

Question 5.
A gas is enclosed in room. The temperature, pressure, density and number of moles respectively are t°C,p atm, g cm^-3 and n moles.
(i) What will be the pressure, temperature, density and number of moles in each compartment, if room is partitioned into four equal compartments?
(ii) What will be the value of pressure, temperature, density and number of moles in each compartment if the walls between the two compartments (say 1 and 2) are removed?
(iii) What will be the values of pressure, temperature, density and number of moles, if an equal volume of gas at pressure
(p) and temperature (t) is let inside the same room? .
Answer:
(i) (a) Pressure in each compartment is same, (p atm)
(b) Temperature will remain same (t°C).
(c) Density will remain same (d g cm^-3).
(d) Because of partition, volume of each compartment becomes 1/4 and the number of molecules also become 1/4. The number of moles in each compartment will be n/4.

(ii) (a) Pressure will remain same (p atm).
(b) Temperature will remain same (t°C).
(c) Density will remain same (d g cm^-3).
(d) The number of moles in each compartment will be n/2.

(iii) (a) Pressure will be doubled (2p atm).
(b) Temperature will remain same.
(c) Density will remain same (d g cm^-3) ,
(d) Number of moles will be doubled i.e., 2n.

Long Answer Type Questions

Question 1.
Explain the following:
(i) The boiling point of a liquid rises on increasing pressure. ;
(ii) Drops of liquid assume spherical space.
(iii) The boiling point of water (373 K) is abnormally high when compared to that of H₂S (211.2 K).
(iv) The level of mercury in capillary tube is lower than the s level outside when a capillary tube is inserted in the mercury.
(v) Tea or coffee is sipped from a saucer when it is quite hot.
Answer:
(i) A liquid boils when its vapour pressure becomes equal to the atmospheric pressure. An increase in pressure on liquid, therefore, causes a rise in the boiling temperature of the liquids.
(ii) Liquids have a property, called surface tension, due to which liquids tend to contract (to decrease the surface area). For a given volume of a liquid, since a sphere has the least surface area, hence the liquids tend to form spherical droplet.
(iii) The extensive hydrogen bonding in water gives a polymeric structure. This makes the escape of molecules from the liquid more difficult. Therefore, water requires higher temperature to bring its vapour pressure equal to the atmospheric pressure.
On the other hand, sulphur being less electronegative, does not form hydrogen bonds with H of H2S. As a result, H2S has low boiling point.
(iv) The cohesive forces in mercury are much stronger than the force of adhesion between glass and mercury. Therefore, mercury-glass contract angle is greater than 90°C.
As a result, the vertical component of the surface tension forces acts vertically downward, thereby lowering the level of mercury column in the capillary tube.
(v) Evaporation causes cooling and the rate of evaporation increases with an increase in the surface area. Since, saucer has a large surface area, hence tea/coffee taken in a saucer cools quickly.

Question 2.
Nitrogen molecule (N₂) has radius of about 0.2 nm. Assuming that nitrogen molecule is spherical in shape, calculate
(i) volume of a single molecule of N₂.
(ii) the percentage of empty space in one mole of N₂ gas at STP.
Answer:
(i) The volume of a sphere = \(\frac{4}{3}\)πr³ nr where Volume of a molecule of N₂
= \(\frac{4}{3} \times \frac{22}{7}\) x (2 x 10^-8)³ cm³ – 3.35 x 10^-23 cm³

(ii) To calculate the empty space, let us first find the total volume of 1 mole (6.022 x 10²³ molecules) of N₂.
Volume of 6.022 x 10²³ molecules of N₂
= 3.35 x 10^-23 x 6.022 x 10²³ = 20.17 cm³
Now, volume occupied by 1 mole of gas at STP
= 22.4 litre = 22400 cm³
Empty volume = Total volume of gas – Volume occupied by molecules
= (22400 – 20.17) cm³ – 22379.83 cm³
∴ Percentage of empty space = \(\frac{Empty space}{Total volume}\) x 100
= \(\frac{22379.83}{22400}\) x 100 = 99.9%
Thus, 99.9% of space of 1 mole of N₂ at STP is empty.

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 4 Chemical Bonding and Molecular Structure Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 4 Chemical Bonding and Molecular Structure

Very Short Answer Type Questions

Question 1.
In \(\mathrm{PO}_{4}^{3-}\) ion formal charge on the oxygen atom of P—O bond is
Answer:
In \(\mathrm{PO}_{4}^{3-}\) ion, formal charge on each O-atom of P—O bond
= \(\frac{\text { total charge }}{\text { Number of O-atoms }}=-\frac{3}{4}\) = -0.75

Question 2.
Which of the following molecules show super octet?
CO₂, CIF₃, SO₂, IF₅
Answer:
ClF₃ and IF₅ are super octet molecules.

Question 3.
Which of the following has highest lattice energy and why?
CsF, CsCl, CsBr, Csl
Answer:
CsF has highest lattice energy because ‘F’ is smallest in size and is more electronegative, therefore, it has maximum ionic character and maximum force of attraction, hence, highest lattice energy.

Question 4.
Account for the following:
The experimentally determined N—F bond length in NF₃ is greater than the sum of the single covalent radii of N and F.
Answer:
This is because both N and F are small and hence, have high electron density. So, they repel the bond pairs thereby making the N—F bond length larger.

Question 5.
What is valence bond approach for the formation of covalent * bond?
Answer:
A covalent bond is formed by the overlapping of half-filled atomic orbitals.

Question 6.
Why axial bonds of PCI5 are longer than equatorial bonds?
Answer:
This is due to greater repulsion on the axial bond pairs by the equatorial bond pairs of electrons.

Question 7.
Which type of atomic orbitals can overlap to form molecular orbitals?
Answer:
Atomic orbitals with comparable energies and proper orientation can overlap to form molecular orbitals.

Question 8.
Why KHF₂ exists but KHCl₂ does not?
Answer:
Due to H-bonding in HF, we have

This can dissociate to give \(\mathrm{HF}_{2}^{-}\) ion and hence, KHF₂ exists but there is no H-bonding in H-Cl. So, \(\mathrm{HCl}_{2}^{-}\) ion does not exist and hence, KHCl₂ also does not exist.

Question 9.
How many nodal planes are present in n(2px) and n(2px) molecular orbitals?
Answer:
One and two respectively.

Question 10.
What is the magnetic character of the anion of K02?
Answer:
Anion of KO₂ is \(\mathrm{O}_{2}^{-}\) (superoxide ion) which has one unpaired electron and hence is paramagnetic.

Short Answer Type Questions

Question 1.
Describe the change in hybridisation (if any) of the Al-atom in
the following reaction:
AlCl₃ + Cl^– → \(\mathrm{AlCl}_{\mathbf{4}}^{-}\)
Answer:
Electronic configuration of Al in ground state,
₁₃Al = 1s², 2s²,2p⁶,3s²,3p¹_x
In excited state = 1s², 2s², 2p⁶, 3s¹, 3p¹_x, 3p¹_y
In the formation of AlCl₃, Al undergoes sp² hybridisation and it is trigonal
planar in shape. While in the formation of AlCl₄^–, Al undergoes sp3
hybridisation.
It means empty 3P_z orbital also involved in hybridisation.
Thus, the shape of AlCl₄^– ion is tetrahedral.

Question 2.
Arrange the following in order of decreasing bond angle, with appropriate reason
\(\mathrm{NO}_{2}, \mathrm{NO}_{2}^{+}, \mathrm{NO}_{2}^{-}\)
Answer:
\(\mathrm{NO}_{2}, \mathrm{NO}_{2}^{+}, \mathrm{NO}_{2}^{-}\). This is because \(\mathrm{NO}_{2}^{+}\) has no lone pair of electrons
(i.e., has only bond pairs on two sides) and hence it is linear.

NO₂ has one unshared electron while \(\mathrm{NO}_{2}^{-}\) has one unshared electron pair. There are greater repulsion on N—O bonds in case of \(\mathrm{NO}_{2}^{-}\) than in case of NO₂

Question 3.
Among the molecules, \(\mathbf{O}_{2}^{-}, \mathbf{N}_{2}^{+}, \mathrm{CN}^{-}\) and \(\mathbf{O}_{2}^{+}\) identify the species which is isoelectronic with CO.
Answer:
Isoelectronics species are those species which have the same number of electrons. CO in total has 14 electrons (6 from carbon and 8 from oxygen). Out of the given ions CN is the ion which has 14 electrons (6 from carbon 7 from Nitrogen and 1 from the negative charge). Thus CN^– ion is isoelectronic with CO.

Question 4.
Which is more polar : COa or N20? Give reason.
Answer:
N₂Ois more polar than CO₂. This is because CO₂ is linear and symmetrical.
Its net dipole moment is zero im
on the other hand, is linear but unsymmetrical. It is considered as a resonance hybrid of the following two structures

It.has a net dipole moment of 0.116 D.

Question 5.
Aluminium forms the ion Al3+, but not Al4+ why?
Answer:
Aluminium [Ne]3s² 3p¹ can achieve the electronic configuration of the nearest noble gas (Ne) by losing only three electrons. : Al³⁺ = 1s² 2s² 2p⁶.
Aluminium will not form Al⁴⁺ ion because an extremely high amount of energy would be required to remove an electron from the stable noble gas configuration.

Long Answer Type Questions

Question 1.
On the basis of VSEPR theory, predict the shapes of the following
(i) \(\mathbf{N H}_{2}^{-}\) (ii) O₃
Answer:
(i) Shape of \(\mathbf{N H}_{2}^{-}\)
Number of valence electrons on central N atom = 5 + 1 (due to one unit negative charge) = 6
Number of atoms linked to it = 2
∴ Total number of electron pairs around N
= \(\frac{6+2}{2}\) = 4 and number of bond pairs = 2
∴ Number of lone pairs = 4 —2 = 2. Thus, the ion is of the type AB₂E₂
Hence, it has bent shape (V-shape).

(ii) Shape of O₃
While predicting geometry of molecules containing the double (or multiple) bond is considered as one electron pair, e.g., in case of ozone, its two resonating structures are

Thus, the central O-atom is considered to have two bond pairs and one lone pair, i.e., it is of the type AB₂E. Hence, it is a bent molecule. Thus, the two resonating structures will be

Question 2.
In each of the following pairs of compounds, which one is more covalent and why?
(i) AgCl, Agl
(ii) BeCl₂,MgCl₂
(iii) SnCl₂, SnCl₄
(iv) CuO, CuS
Answer:
Applying Fajans’ rules, the result can be obtained in each case as follows :
(i) Agl is more covalent than AgCl. This is because I^– ion is larger in size than Cl^– ion and hence is more polarised than Cl^– ion.
(ii) BeCl₂ is more covalent thanMgCl₂. This is because Be²⁺ ion is smaller in size than Mg₂ ion and hence has the greater polarising power.
(iii) SnCl₄ is more covalent than SnCl₂. This is because Sn⁴⁺ ion has greater charge and smaller size than Sn²⁺ ion and hence has greater polarising power.
(iv) CuS is more covalent than CuO. This is because S^2- ion has larger size
than O^2- ion and hence is more polarised than O^2- ion.

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties

Very Short Answer Type Questions

Question 1.
Which properties of the elements depend on’ the electronic configuration of the atoms and which do not?
Answer:
Chemical and many physical properties of the elements depends on the electronic configuration of the atoms, whereas the nuclear properties do not.

Question 2.
Write the number designation of a group that has 2 electrons beyond a noble gas configuration.
Answer:
The number designation of a group that has 2 electrons beyond a noble gas configuration will be 2 which means it will belong to group 2 of periodic table.

Question 3.
Why is it more logical to call the atomic radius as the effective atomic radius?
Answer:
This is because the size of atom is very small and it has no sharp boundaries.

Question 4.
A boy has reported the radii of Cu, Cu⁺ and Cu²⁺ as 0.096 nm, 0.122 nm and Question072 nm respectively. However, it has been noticed that he interchanged the values by mistake. Assign the correct values to different species.
Answer:
Cu [0.122 nm], Cu⁺ [0.096 nm], Cu²⁺ [0.072 nm].
∵ Size ∝ \(\frac{1}{\text { positive charge }}\)

Question 5.
Atomic radii of fluorine is 72 pm where as atomic radii of neon is 160 pm. Why? [NCERT Exemplar]
Answer:
Atomic radius of F is expressed in terms of covalent radius while, atomic radius of neon is usually expressed in terms of van der Waals’ radius, van der Waals’ radius of an element is always larger than its covalent radius.
Therefore, atomic radius of F is smaller than atomic radius of Ne (F = 72 pm, Ne = 160 pm)

Question 6.
Arrange the following elements in order of decreasing electron gain enthalpy : B, C, N, O.
Answer:
N has positive electron gain enthalpy while all others have negative electron gain enthalpies. Since size decreases on moving from B → C → O, therefore, electron gain enthalpies become more and more negative from B → C → O. Thus, the overall decreasing order of electron gain enthalpies is N, B, C, O.

Question 7.
Which of the following atoms would most likely form an anion (i) Be, (ii) Al, (iii) Ga, (iv) I ?
Answer:
I, because of high electron gain enthalpy, it can accept an electron readily to form an anion F < Cl < Br > I.

Question 8.
Explain why chlorine can be converted into chloride ion more easily as compared to fluoride ion from fluorine.
Answer:
Electron gain enthalpy of Cl is more negative than that of F.

Question 9.
Among alkali metals which element do you expect to be least electronegative and why? [NCERT Exemplar]
Answer:
On moving down the group, electronegativity decreases because atomic size increases. Fr has the largest size, therefore it is least electronegative.

Question 10.
Arrange the following elements in the increasing order of non-metallic character. B, C, Si, N, F
Answer:
The given non-metals are arranged in the increasing order of non-metallic character as follows:

Short Answer Type Questions

Question 1.
What would be IUPAC names and symbols for elements with atomic numbers 122, 127, 135, 149 and 150? .
Answer:
The roots 2, 7, 5, 9 and 0 are referred as bi, sept, pent, enn and nil respectively. Therefore, their names and symbol are

Z (Atomic number)	Name	Symbol
122	Unbibium	Ubb
127	Unbiseptium	Ubs
135	Untripentium	Dtp
149	Unquadennium	Uqe
150	Unpentnilium	Upn

Question2.
All transition elements are d-block elements, but all d-block elements are not transition elements. Explain.
Answer:
Elements in which the last electron enters in the d-orbitals, are called d-block elements or transition elements. These elements have the general outer electronic configuration (n – 1)d^1-10ns^0-2 Zn, Cd and Hg having the electronic configuration, (n – l1)d¹⁰ns² do not show most of the properties of transition elements. The d-orbitals in these elements are completely filled in the ground state as well as in their common oxidation states. Therefore, they are not regarded as transition elements. Thus, on the basis of properties, all transition elements are d-block elements but on the basis of electronic configuration, all d-block elements are not transition elements.

Question 3.
Arrange the elements N, P, O and S in the order of
(i) increasing first ionisation enthalpy.
(ii) increasing non-metallic character.
Give reason for the arrangement assigned.
Answer:

	Group 15	Group 16
2nd period	N	0
3rd period	P	S

(i) Ionisation enthalpy of nitrogen (₇N = 1s², 2s², 2p³) is greater than oxygen (₈O = 1s² , 2s² , 2p⁴ ) due to extra stable half-filled 2p-orbitals. Similarly, ionisation enthalpy of phosphorus (₁₅P = 1s², 2s², 2p⁶, 3s², 3p³) is greater than sulphur (₁₆S = 1s², 2x², 2p⁶, 3s², 3p⁴).
On moving down the group, ionisation enthalpy decreases with increasing atomic size. So, the increasing order of first ionisation enthalpy is S < P < O < N.

(ii) Non-metallic character across a period (left to right) increases but on moving down the group it decreases. So, the increasing order of non-metallic character is P < S < N < 0.

Question 4.
What do you understand by exothermic reaction and endothermic reaction? Give one example of each type.
Answer:
Exothermic reactions : Reactions which are accompanied by evolution of heat are called exothermic reactions. The quantity of heat produced is shown either along with the products with a ‘+’ sign or in terms if ΔH with a sign, e.g.,

C(s) + O₂(g) → CO₂(g) + 393.5 kJ
H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l) ΔH = -285.8 kJ mol^-1

Endothermic reactions : Reactions which proceed with absorption of heat are called endothermic reactions. The quantity of heat absorbed is shown either alongwith the products with a sign or in terms of ΔH with a ‘-‘ sign, e.g.,

C(s) + H₂O(g) → CO(g) + H₂(g) -131.4 kJ
N₂(g) + 3H₂(g) → 2NH₃(g); ΔH = +92.4 kJ mol^-1

Question5.
How does the metallic and non-metallic character vary on moving from left to right in a period?
Answer:
As we move from left to right in a period, the number of valence electrons increases by one at each succeeding element but the number of shells remains same. Due to this, effective nuclear charge increases.

More is the effective nuclear charge, more is the attraction between nuclei and electron.
Hence, the tendency of the element to lose electrons decreases, this results in decrease in metallic character.
Furthermore, the tendency of an element to gain electrons increases with increase in effective nuclear charge, so non-metallic character increases on moving from left to right in a period.

Long Answer Type Questions

Question 1.
Write the drawbacks in Mendeleev’s Periodic Table that led to its modification.
Answer:
The main drawbacks of Mendeleev’s Periodic Table are:
(i) Some elements having similar properties were placed in different groups whereas some elements having dissimilar properties were placed in the same group. For example alkali metals such as Li, Na, K, etc., (I A group) are grouped together with coinage metals such as Cu, Ag, Au (I B group) though their properties are quite different. Chemically similar elements such as Cu (I B group) and Hg (II B group) have been placed in different groups.

(ii) Some elements with higher atomic weights are placed before the elements with lower atomic weights in order to maintain the similar chemical nature of elements. For example,

(iii) Isotopes did not find any place in the Periodic Table. However, according to Mendeleev’s classification, these should be placed at different places in the Periodic Table.
(All the above three defects were however removed when modern periodic law based on atomic number was given.)

(iv) Position of hydrogen in the Periodic Table is not fixed but is
controversial. ,
(v) Position of elements of group VIII could not be made clear which have been arranged in three triads without any justification.
(vi) It could not explain the even and odd series in IV, V and VI long periods.
(vii) Lanthanides and actinides which were discovered later on, have not been given proper positions in the main frame of Periodic Table.

Question 2.
p-block elements form acidic, basic and amphoteric oxides. Explain each property by giving two examples and also write the reactions of these oxides with water.
Answer:
In p-block, when we move from left to right in a period, the acidic character of the oxides increases due to increase in electronegativity, e.g.,

(i) 2nd period
B₂O₃ < CO₂ < N₂O₃ acidic character increases.

(ii) 3rd period
Al₂O₃ < SiO₂ < P₄O₁₀ < SO₃ < Cl₂O₇ acidic character increases.
On moving down the group, acidic character decreases and basic character increases, e.g.,

(a) Nature of oxides of 13 group elements

(b) Nature of oxides of 15 group elements

Among the oxides of same element, higher the oxidation state of the element, stronger is the acid. e.g., SO₃ is a stronger acid than SO₂. B₂O₃ is weakly acidic and on dissolution in water, it forms orthoboric acid. Orthoboric acid does not act as a protonic acid (it does not ionise) but acts as a weak Lewis acid.

Al₂O₃ is amphoteric in nature. It is insoluble in water but dissolves in alkalies and react with acids.

Tl₂O is as basic as NaOH due to its lower oxidation state (+1).
Tl₂O + 2HCl → 2TlCl + H₂O

P₄O₁₀ on reaction with water gives orthophosporic acid.

Cl₂O₇ is strongly acidic in nature and on dissolution in water, it gives perchloric acid.

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 2 Structure of Atom Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 2 Structure of Atom

Very Short Answer Type Questions

Question 1.
Which of the following will not show deflection from the path on passing through an electric field? Proton, cathode rays, electron, neutron
Answer:
Neutron is a neutral particle. Hence, it will not be deflected on passing through an electric field.

Question 2.
What is the nuclear radius of an atom whose mass number is 125?
Answer:
Nuclear radius, r = R₀A^1/3 where, R₀ = 1.4 x 10^-15 m,
∴ r = (1.4 x 10^-15 m) x (125)^1/3 = 7.0 x 10^-15 m.

Question 3.
The magnitude of charge on the electron is 4.8 x 10^-10 esu. What is the charge on the nucleus of a helium atom?
Answer:
Helium nucleus contains 2 protons and charge of a proton is same as that of an electron.
Therefore, the charge on the nucleus of a helium atom is (+2) x 4.8 x 10^-10 = + 9.6 x 10^-10 esu.

Question 4.
What is the difference in the origin of cathode rays and anode rays?
Answer:
Cathode rays originate from the cathode whereas anode rays do not originate from the anode. They are produced from the gaseous atoms by knock out of the electrons with high speed cathode rays.

Question 5.
What is the difference between atomic mass and mass number?
Answer:
Mass number is a whole number because it is the sum of number of protons and number of neutrons whereas atomic mass is fractional because it is the average relative mass of its atom as compared with mass of an atom of C-12 isotope taken as 12.

Question 6.
What is the difference between a quantum and a photon?
Answer:
The smallest packet of energy of any radiation is called a quantum whereas that of light is called photon.

Question 7.
Arrange s, p and rf-subshells of a shell in the increasing order of , effective nuclear charge (Zeff) experienced by the electron present in them. [NCERT Exemplar]
Ans. s-orbital is spherical in shape, it shields the electrons from the nucleus more effectively than p-orbital which in turn shields more effectively than d-orbital. Therefore, the effective nuclear charge (Zeff) experienced by electrons present in them is d < p < s.

Question 8.
Show the distribution of electrons in oxygen atom (atomic number 8) using orbital diagram.
Answer:

Question 9.
Nickel atom can lose two electrons to form Ni ion. The atomic number of Ni is 28. From which orbital will nickel lose two electrons? [NCERT Exemplar]
Answer:
₂₈Ni = 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁸, 4s²; Nickel will lose 2 electrons from 4s (outer most shell) to form Ni²⁺ ion.

Question 10.
Which of the following orbitals are degenerate?
3d_xy, 4d_xy, \(3 d_{z^{2}}\), 3d_yz, \(4 d_{z^{2}}\)
Answer:
The orbitals which belongs to same subshell and same shell are called degenerate orbitals. (3d_xy, \(3 d_{z} 2\), 3d_yz) and (4d_xy, 4dyz, 4d 2) are the two sets of degenerate orbitals.

Short Answer Type Questions

Question 1.
The Balmer series in the hydrogen spectrum corresponds to the transition from n₁ = 2 to n₂ = 3, 4 … . This series lies in the visible region. Calculate the wave number of line associated with the transition in Balmer series when the electron moves to n = 4 orbit. (R_H = 109677 cm1)
Answer:
From Rydberg formula,

Question 2.
Out of electron and proton which one will have, a higher velocity to produce matter waves of the same wavelength? Explain it.
Answer:
From de Broglie equation, wavelength, \(\lambda=\frac{h}{m v}\)
For same wavelength for two different particles, i.e., electron and proton, m₁v₁ = m₂v₂ (h is constant). Lesser the mass of the particle, greater will be the velocity. Hence, electron will have higher velocity.

Question 3.
Wavelengths of different radiations are given helow.
λ(A) = 300 nm, λ(B) = 300 pm, λ(C) = 3 nm, λ(D) = 30Å Arrange these radiations in the increasing order of their energies.
Answer:
(A) λ=3OOnm=3OO x 10^-9m
(B) λ =300µm=300 x 10^-6m
(C) λ =3nm = 3 x 10^-9 m
(D) λ = 30 = 30 x 10^-10m= 3 x 10^-9m
Energy, E = \(\frac{h c}{\lambda}\)
Therefore, E ∝ \(\frac{1}{\lambda}\)
Increasing order of energy is B

Question 4.
The electronic configuration of valence shell of Cu is 3d¹⁰4s¹ and not 3d⁹4s². How is this configuration explained?
Answer:
Configurations with completely filled and half-filled orbitals have extra stability. In 3d¹⁰4s¹, d-orbitals are completely filled and s-orbital is half-filled. Hence, it is a more stable configuration for Cu as compare to 3d⁹4s².

Question 5.
In each of the following pairs of salts, which one is more stable? (i) Ferrous and ferric salts (ii) Cuprous and cupric salts
Answer:
(i) Ferrous and ferric salts : In ferrous salts Fe²⁺, the configuration is 1s²,2s²,2p⁶,3s²,3p⁶,3d⁶. In ferric salts Fe³⁺, the configuration is 1s², 2s², 2p⁶, 3s², 3p⁶, 3d⁵. As half-filled 3d⁵ configuration is more stable therefore ferric salts are more stable than ferrous salts.
(ii) Cuprous and cupric salts : In cuprous salts, the configuration of Cu⁺ is 1s²,2s², 2p⁶, 3s², 3p⁶, 3d¹⁰. In cupric salts the configuration of Cu²⁺, is, 1s²,2s²,2p⁶,3s²,3p⁶,3d⁹. Although Cu+ has completely filled d-orbital, yet cuprous salts are less stable. This is because the nuclear charge is not sufficient enough to hold 18 electrons of Cu⁺ ion present in the outermost shell.

Long Answer Type Questions

Question 1.
When an electric discharge is passed through hydrogen gas, the hydrogen molecules dissociate to produce excited hydrogen atoms. These excited atoms emit electromagnetic radiation of discrete frequencies which can be given by the general formula \(\bar{v}=109677\left[\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}\right]\)
What points of Bohr’s model of an atom can he used to arrive at this formula? Based on these points derive the above formula giving description of each step and each term. [NCERT Exemplar]
Answer:
The two important points of Bohr’s model that can be used to derive the given formula are as follows :
(i) Electrons revolve around the nucleus in a circular path of fixed radius and energy. These paths are called orbits, stationary states or allowed energy states.
(ii) Energy is emitted or absorbed when an electron moves from higher stationary state to lower stationary state or from lower stationary state to higher stationary state respectively.
Derivation : The energy of the electron in the nth stationary state is given by the expression,
\(E_{n}=-R_{\mathrm{H}}\left(\frac{1}{n^{2}}\right)\)
n = 1,2,3 …. ……(i)
where R_H is called Rydberg constant and its value is 2.18 x 10^-18 J.
The Energy of the lowest state, also called the ground state, is
E₁ = -2.18 x 10^-18\(\left(\frac{1}{1^{2}}\right)\) = 2.18 x 10^-18J ……. (ii)
The energy gap between the two orbits is given by the equation,
ΔE = E_f – E_i … (iii)
On combining equations (i) and (iii)

Question 2.
Calculate the energy and frequency of the radiation emitted when an electron jumps from n = 3 to n = 2 in a hydrogen atom.
Answer:

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids

Very short answer type questions

Question 1.
Write the product in the following reaction:

Answer:
CH₃– CH = CH-CH₂– CHO Pent-3-en-1-al

Question 2.
Complete the following reaction sequence: (NCERT Exemplar]

Answer:

Question 3.
Illustrate the following name reaction: Clemmensen reduction.
Answer:
Clemmensen reduction: The carbonyl group of aldehydes and ketones is reduced to CH₂ group on treatment with zinc amalgam and concentrated hydrochloric acid.

Question 4.
Name the electrophile produced in the reaction of benzene with benzoyl chloride in the presence of anhydrous AlCl₃.
Name the reaction also. [NCERT Exemplar]
Answer:

Question 5.
Which acid of each pair shown here would you expect to be stronger?
(i) F-CH₂ -COOH or Cl-CH₂ -COOH

Answer:
(i) F-CH₂-COOH > Cl-CH₂-COOH, because of stronger -ve-I effect of F than Cl.

Question 6.
Identify the compounds A, B, and C in the following reaction. [NCERT Exemplar]

Answer:

Question 7.
Write the product(s) in the following reaction:

Answer:

Question 8.
Name the aldehyde which does not give Fehling’s solution test.
Answer:
Benzaldehyde.

Question 9.
Give the name of the reagent that brings the following transformation: But-2-ene to ethanal.
Answer:
O₃/H₂O-Zn dust

Question 10.
Write TUPAC names of the following structures: (NCERT Exemplar)

Answer:

Question 11.
(CH₃) ₃C-CHO does not undergo aldol condensation comment.
Answer:
Because no α-H is present.

Short answer type questions

Question 1.
Write the reactions involved in the following:
(i) Etard reaction
(ii) Stephen’s reduction
Answer:
(i) Etard reaction: Chromyl chloride oxidizes toluene to chromium complex which on hydrolysis gives benzaldehyde.

(ii) Stephen reductIon: Nitriles are reduced to corresponding imines with SnCl₂ in the presence of HCl, which on hydrolysis give the corresponding aldehyde.
SnCl₂ + 2HCl → SnCl₄ + 2[H]

Question 2.
Oxidation of ketones involves carbon-carbon bond cleavage. Name the products formed on oxidation of 2,5-dimethyl hexane-3-one. [NCERT Exemplar]

Answer:

Question 3.
Carboxylic acids contain carbonyl group but do not show the nucleophilic addition reaction like aldehydes or ketones. Why? [NCERT Exemplar]
Answer:
Carboxylic acids contain, carbonyl group but do not show nucleophilic addition reaction like aldehyde and ketone. Due to resonance as shown below the partial positive charge on the carbonyl carbon atom is reduced.

Question 4.
Give reasons:
(i) The α-hydrogen atoms of aldehydes and ketones are acidic in nature.
(ii) Propanone is less reactive than ethanal toward addition of HCN
(iii) Benzoic acid does not give Friedel Crafts reaction.
Answer:
(i) The acidity of a-hydrogen atom of carbonyl carbon is due to the strong withdrawing effect of the carbonyl group and resonance stabilization of the conjugate base.

(ii) This is due to steric and electronic reasons. Sterically, the presence of two methyl groups in propanone hinders more the approach of nucleophile to carbonyl carbon than in ethanal having one methyl group. Electronically two methyl groups reduce the positivity of the carbonyl carbon more effectively in propanone than in ethanal.

(iii) Benzoic acid does not give Friedel Craft reaction because:

• the carboxyl group is strongly deactivating.
• the catalyst AlCl₃ which is a lewis acid gets bonded to the carboxyl group strongly.

Question 5.
An organic compound T having molecular formula C₄H₈O gives orange-red ppt. with a 2,4-DNP reagent.
It does not reduce Tollen’s reagent but gives yellow ppt. of iodoform on heating with NaOI.
Compound X on reduction with LiAlH₄ gives compound T’ which undergoes dehydration reaction on heating with -cone. H₂SO₄ to form but-2-ene. Identify the compounds X and Y.
Answer:

Reaction involved:

Long answer type questions

Question 1.
(a) Account for the following:
(i) Cl – CH₂COOH is a stronger acid than CH₃COOH.
(ii) Carboxylic acids do not give reactions of carbonyl group.
(b) Out of CH₃CH₂ – CO – CH₂ – CH₃ and CH₃CH₂ – CH₂ –
CO – CH₃, which gives iodoform test?
Answer:
(a) (i) Because of -I effect of Cl atom in ClCH₂COOH and +I effect of CH₃ group in CH₃COOH the electron density in the O-H bond in ClCH₂COOH is much lower than CH₃COOH.
As a result O-H bond in ClCH₂COOH is much weaker than in CH₃COOH therefore loses a proton more easily than CH₃COOH.
Hence ClCH₂COOH acid is stronger acid than CH₃COOH.

(ii) Carboxylic acids are resonance hybrid of the following structures:

Similarly, a carbonyl group of aldehydes and ketones may regarded as resonance hybrid of following structures :

Because of contribution of structure (IV), the carbonyl carbon in aldehydes and ketones is electrophilic. On the other hand, electrophilic character of carboxyl carbon is reduced due to contribution of structure (II). As carbonyl carbon of carboxyl group is less electropositive than carbonyl carbon in aldehydes and ketones, therefore, carboxylic acids do not give nucleophilic addition reactions of aldehydes and ketones.
(b) CH₃ – CH₂ – CH₂ – COCH₃.

Question 2.
Write the products formed when ethanal reacts with the following reagents:
(i) CH₃MgBr and then H₃O⁺
(ii) Zn-Hgyconc.HCl
(iii) C₆H₅CHO in the presence of dilute NaOH
Answer:

Question 3.
(a) Write the products of the following reactions :

(b) Give simple chemical tests to distinguish between the following pairs of compounds:
(i) Benzaldehyde and Benzoic acid
(ii) Propanal and Propanone
Answer:

(iii) Cl-CH₂-COOH

(b)
(i) NaHCO₃ test
(ii) Iodoform test or Fehling test or Tollen’s test.

Question 4.
(a) Account for the following:
(i) CH₃CHO is more reactive than CH₃COCH₃ towards reaction with HCN.
(ii) 2-Fluorobutanoic acid is a stronger acid than 3-Fluorobutanoic acid.

(b) Write the chemical equations to illustrate the following name reactions:
(i) Etard reaction.
(ii) Rosenmund’s reaction.
(c) Give the mechanism of cyanohydrin formation when carbonyl compounds react with HCN in the presence of alkali.
Answer:
(a)

• CH₃CHO is more reactive than CH₃COCH₃ towards reaction with HCN due to steric and electronic factors.
• Because the inductive effect decreases with distance and hence the conjugate base of 2-Fluorobutanoic acid is more stable.

(b)

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 3 Electrochemistry

Very short answer type questions

Question 1.
Why on dilution the ∧_m of CH₃COOH increases drastically while that of CH₃COONa increases gradually? [NCERT Exemplar]
Answer:
In the case of CH₃COOH, which is a weak electrolyte, the number of ions increase on dilution due to an increase in degree of dissociation.
CH₃COOH + H₂o → CH₃COO^– + H₃O⁺

Question 2.
Why is alternating current used for measuring resistance of an electrolytic solution? [NCERT Exemplar]
Answer:
Alternating current is used to prevent electrolysis so that concentration of ions in the solution remains constant.

Question 3.
Define electrochemical series.
Answer:
The arrangement of elements in the increasing or decreasing order of their standard reduction potentials is called electrochemical series.

Question 4.
What are secondary cells?
Answer:
Secondary cells are those cells which are rechargeable, i.e., the products can be changed back to reactants.

Question 5.
What is the necessity to use a salt bridge in a Galvanic cell?
Answer:
To complete the inner circuit and to maintain the electrical neutrality of the electrolytic solutions of the half-cells we use a salt bridge in a Galvanic cell.

Question 6.
Why does a dry cell become dead after a long time even if it has not been used?
Answer:
Even though not in use, a dry cell becomes dead after some time because the acidic NH4C1 corrodes the zinc container.

Question 7.
What does the negative sign in the expression \(\boldsymbol{E}_{\mathbf{Z n}^{2+} / \mathbf{Z n}}[latex] = -0.76 V mean?[NCERT Exemplar]
Answer:
It means that Zn is more reactive than hydrogen. When zinc electrode will be connected to SHE, Zn will get oxidised and H⁺ will get reduced.

Question 8.
Write the Nemst equation for the cell reaction in the Daniel cell. How will the £cell be affected when concentration of Zn2+ ions is increased? [NCERT Exemplar]
Answer:
Zn + Cu²⁺ > Zn²⁺ + Cu
E_cell = [latex]E_{\text {cell }}^{\ominus}\) – \(\frac{0.059}{2}\) log \(\frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Cu}^{2+}\right]}\)
E_cellcell decreases when concentration of Zn²⁺ ions, [Zn²⁺ ] increases.

Question 9.
What does the negative value of \(E_{\text {cell }}^{\ominus}\) indicate?
Answer:
Negative \(E_{\text {cell }}^{\ominus}\) value means Δ_rG^Θ will be +ve, and the cell will not work.

Question 10.
Can \(E_{\text {cell }}^{\ominus}\) or Δ_rG^Θ for cell reaction ever be equal to zero? [NCERT Exemplar]
Answer:
No.

Short answer type questions

Question 1.
State Kohlrausch law of independent migration of ions. Why does the conductivity of a solution decrease with dilution?
Answer:
Kohlrausch law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of cation and anion of the electrolyte.

In general, if an electrolyte on dissociation gives v₊ cations and v_– anions then its limiting molar conductivity is given by
\(\Lambda_{m}^{\ominus}=v_{+} \lambda_{+}^{0}+v_{-} \lambda_{-}^{0}\)

Where, \(\lambda_{+}^{0}\) and \(\lambda_{-}^{0}\) are the limiting molar conductivities of cations and anions respectively.

Conductivity of a solution decreases with dilution. This is due to the fact that the number of ions per unit volume that carry the current in a solution decreases with dilution.

Question 2.
Define the following terms:
(i) Fuel cell
(ii) Limiting molar conductivity (\(\))
Answer:
(i) A fuel cell is a device which converts the energy produced during the combustion of fuels like hydrogen, methanol, methane etc. directly into electrical energy. One of the most successful fuel cell is H₂ – O₂ fuel cell.

(ii) When concentration approaches zero, the molar conductivity is known
as limiting molar conductivity. It is represented by \(\Lambda_{m}^{\ominus}\).
\(\Lambda_{m}^{\ominus}\) (∧b)when → c

Question 3.
(i) Write two advantages of H₂ – O₂ fuel cell over ordinary cell.
(ii) Equilibrium constant (K_c) for the given cell reaction is 10. Calculate \(\boldsymbol{E}_{\text {cell }}^{\ominus}\).

Solution:
(i) The two main advantages of H₂ – O₂ fuel cell over ordinary cell are as follows:

• It has high efficiency of 60%-70%.
• It does not cause any pollution.

Question 4.
Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution:
Ag⁺(aq) + e^– → Ag(s) E^Θ = + 0.80 V
H⁺ (aq)+ e^– → \(\frac{1}{2}\)H₂(g) E^Θ = 0.00 V
On the basis of their standard reduction electrode potential (E^Θ) values, which reaction is feasible at the cathode and why?
Answer:
The reaction, Ag⁺ (aq) + e^– → Ag(s) is feasible at cathode as
cathodic reaction is one which has higher standard reduction electrode potential (\(E_{\text {red }}^{\ominus}\)).

Question 5.
Calculate ∆G and log K_c for the following reaction at 298 K:
2Cr(8) + 3Fe²⁺(aq) > 2Cr³⁺ (aq) + 3Fe(s)
Given: \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) = 0.30 V
Solution:

Question 6.
Calculate the emf of the following cell at 298 K:
Cr(s)/Cr³⁺ (0.1M)//Fe<>2+ (0.01M)/(Fe(s) [Given: \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) = + 0.30 V]
Answer:
The cell reaction is as follows :
2Cr(s) + 3Fe²⁺(aq) > 3Fe(s) + 2Cr³⁺(aq)
For this reaction, n = 6
Now,
E_cell = \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) – \(\frac{2.303 R T}{n F}\) log \(\frac{\left[\mathrm{Cr}^{3+}\right]^{2}}{\left[\mathrm{Fe}^{2+}\right]^{3}}\)
E_cell = 0.30 – \(\frac{0.059}{6}\) log \(\frac{\left[10^{-1}\right]^{2}}{\left[10^{-2}\right]^{3}}\)
E_cell = 0.26V

Question 7.
The conductivity of 10^-3 mol/L acetic acid at 25°C is 4.1 x 10^-5 S cm^-1. Calculate its degree of dissociation, if \(\Lambda_{m}^{0}\) for acetic acid at 25°C is 390.5 S cm² mol^-1.
Answer:
We know that ∧_m = \(\frac{1000 \mathrm{~K}}{\mathrm{C}}\)
∧_m = \(\frac{1000 \times 4.1 \times 10^{-5}}{10^{-3}}\)
= 41 S cm² mol^-1
Now, α = \(\frac{\Lambda_{m}^{c}}{\Lambda_{m}^{0}}\)
= \(\frac{41}{390.5}\) = 0.105 390.5

Question 8.
(i) Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases 1.5 times while that of ‘A’ increases 25 times. Which of the two is a strong electrolyte? Justify your answer.
(ii) The products of electrolysis of aqueous NaCl at the respective electrodes are:
Cathode: H₂
Anode: Cl₂ and not 0₂. Explain.
Answer:
(i) ‘B’ is a strong electrolyte.
Because a strong electrolyte is already dissociated into ions, but on dilution inter ionic forces are overcome, ions are free to move. So, there is slight increase in molar conductivity on dilution.

(ii) On anode water should get oxidised in preference to Cl^– but due to overvoltage/over potential Cl” is oxidised in preference to water.

Long answer type questions

Question 1.
(a) The conductivity of 0.20 mol L^-1 solution of KC1 is 2.48 x 10^-2 S cm^-1. Calculate its molar conductivity and degree of dissociation (α). Given λ⁰ (K⁺) = 73.5 S cm² mol^-1 and λ⁰ (Cl^–) = 76.5 S cm² mol^-1.

(b) What type of battery is mercury cell? Why is it more advantageous than dry cell?
Answer:
(a) Substituting K =2.48 x 10^-2 S cm^-1, M = 0.20 molL^-1 in the expression ∧_m = \(\frac{K \times 1000}{M}[latex] , we get

Substituting ∧_m = 124 S cm² mol^-1, [latex]\Lambda_{m}^{\ominus}\) = 150 S cm² mol^-1 in the expression α = \(\frac{\Lambda_{m}}{\Lambda_{m}^{\ominus}}\), we get
Degree of dissociation, α = \(\frac{124 \mathrm{~s} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}{150 \mathrm{~s} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}}\) = 82666
α = 82.67%

(b) Primary cell. Mercury cell is more advantageous than dry cell because its cell potential remains constant during its life as the overall reaction does not involve any ion in solution whose concentration can change during its life period.

Question 2.
(i) The conductivity of 0.001 mol L^-1 solution of CH₃COOH is 3.905 x 10^-5 S cm^-1. Calculate its molar conductivity and degree of dissociation (α).
Given λ⁰ (H⁺) = 349.6 S cm² mol^-1 and λ⁰ (CH₃COO ) = 40.9 S cm² mol^-1

(ii) Define electrochemical cell. What happens if external potential applied becomes greater than \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) of electrochemical cell?
Solution:

(b) A device which is used to convert chemical energy produced in a redox reaction into electrical energy is called an electrochemical cell.
If external potential applied becomes greater than \(\boldsymbol{E}_{\text {cell }}^{\ominus}\) of
electrochemical cell, the reaction gets reversed and the electrochemical cell function as an electrolytic cell.

Question 3.
Calculate e.m.f and ΔG for the following cell at 298 K:
Mg(s) | Mg²⁺ (0.01 M) || Ag⁺ (0.0001 M) | Ag(s)
Given: \(\underset{\left(\mathrm{Mg}^{2+} / \mathrm{Mg}\right)}{\boldsymbol{E}^{\circ}}\) = -2.37V, \(\boldsymbol{E}^{\ominus}{\left(\mathbf{A g}^{+} / \mathbf{A g}\right)}\) = + 0.80V
Solution:

E_cell = 3.17 – 0.0295 log 10⁶
E_cell = 3.17 – 0.177 V = 2.993 V
E_cell = 2.993 V
Substituting n = 2, F = 96500 C mol^-1, E_cell = 2.993 V in the
expression, ΔG = – nFE_cell, we get
ΔG = nFE_cell = -2 x 96500 C mol^-1 x 2.993V
ΔG = – 577649 J mol^-1 = – 577.649 kJ mol^-1

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 13 Amines Important Questions and Answers.

PSEB 12th Class Physics Important Questions Chapter 13 Amines

Very short answer type questions

Question 1.
CH₃NH₂ is more basic than C₆H₅NH₂ why?
Answer:
Aliphatic amines (CH₃NH₂) are stronger bases than aromatic amines (C₆H₅NH₂) because due to resonance in aromatic amines, the lone pair of electrons on the nitrogen atom gets delocalized over the benzene ring and thus is less easily available for protonation.

Question 2.
What is the role of pyridine in the acylation reaction of amines? [NCERT Exemplar]
Answer:
Pyridine and other bases are used to remove the side product, L e., HCl from the reaction mixture.

Question 3.
A primary amine, RNH₂ can be reacted with CH₃ – X to get secondary amine, R-NHCH₃ but the only disadvantage is that 3° amine and quaternary ammonium salts are also obtained as side products. Can you suggest a method where RNH₂ forms only 2° mine? [NCERT Exemplar]
Answer:

Primary amines show carbylamine reaction in which two H-atoms attached to N-atoms of NH₂ are replaced by one C-atom. On catalytic reduction, isocyanide (formed) produces secondary amine and not tertiary or quaternary salts.

Question 4.
What is Hinsberg reagent? (NCERT Exemplar)
Answer:
Benzene sulphonyl chloride (C₆H₅SO₂Cl) is known as Hinsberg reagent. It is used to distinguish between primary, secondary and tertiary amine.

Question 5.
The conversion of primary aromatic amines into diazonium salts is known as……
Answer:
Diazotisation.

Question 6.
Rearrange the following in an increasing order of their basic strengths:
C₆H₅NH₂,C₆H₅N(CH₃)₂,(C₆H₅)₂NH and CH₃NH₂
Answer:
(C₆H₅)₂NH < C₆H₅NH₂ <C₆H₅N(CH₃)₂ <CH₃NH₂

Question 7.
What is the best reagent to convert nitrile to primary amine? [NCERT Exemplar]
Answer:
Reduction of nitriles with sodium/alcohol or LiAlH₄ gives primary amine.

Question 8.
Suggest a route by which the following conversion can be accomplished: [NCERT Exemplar]

Answer:

Question 9.
What is the role of HNO₃ in the nitrating mixture used for the nitration of benzene? [NCERT Exemplar]
Answer:
HNO₃ acts as a base in the nitrating mixture and provides the electrophile NO₂.

Question 10.
Why is benzene diazonium chloride not stored and is used immediately after its preparation? [NCERT Exemplar]
Answer:
Benzene diazonium chloride is very unstable.

Short answer type questions

Question 1.
Write the structures of A, B and C in the following:

Answer:

Question 2.
Give a chemical test to distinguish between C₆H₅CH₂NH₂ and C₆H₅NH₂.
Answer:
C₆H₅CH₂NH₂ reacts with HNO₂ at 273-278 K to give diazonium salt, which being unstable decomposes with brisk evolution of N₂ gas.

whereas, C₆H₅NH₂ reacts with HNO₂ at 273-278 K to form stable benzene diazonium chloride, which upon treatment with an alkaline solution of f3-naphthol, gives an orange dye.

Question 3.
Complete the following reaction: [NCERT Exemplar]
Answer:
The reaction exhibits an azo-coupling reaction of phenols. Benzene diazonium chloride reacts with phenol in such a manner that the para position of phenol is coupled with diazonium salt to form p-hydroxy azobenzene.

Question 4.
A solution contains 1 g mol. each of p-toluene diazonium chloride and p-nitrophenyl diazonium chloride. To this lg mol. of alkaline solution of phenol is added. Predict the major product. Explain your answer. [NCERT Exemplar]
Answer:
The above-stated reaction is an example of electrophilic aromatic substitution. In alkaline medium, phenol generates phenoxide ion which is more electron-rich than phenol and more reactive for electrophilic attack. The electrophile in this reaction is aryldiazonium cation. As we know, stronger the electrophile faster is the reaction. p Nitrophenyldiazonium cation is a stronger electrophile than p-toluene diazonium cation. So, nitrophenyl diazonium chloride couples preferentially with phenol

Question 5.
Under what reaction conditions (acidic, basic) the coupling reaction of aryl diazonium chloride with aniline is carried out? [NCERT Exemplar]
Answer:
In strongly basic conditions, benzene diazonium chloride is converted, into diazohydroxide and diazoate as both of which are not electrophilic and do not couple with aniline.

Similarly, in highly acidic conditions, aniline gets converted into an anilinium ion. From this, result aniline is no longer nucleophilic acid and hence will not couple with diazonium chloride. Hence, the reaction is carried out under mild conditions, i.e., pH 4-5

Question 6.
An organic aromatic compound ‘A’ with the molecular formula C₆H₇N is sparingly soluble in water. ‘A’ on treatment with dil. HCl gives a water-soluble compound ‘B’ ‘A’ also reacts with chloroform in presence of alcoholic KOH to form an obnoxious smelling compound ‘C’. ‘A’ reacts with benzene sulphonyl chloride to form an alkali-soluble compound ‘D’ ‘A’ reacts with NaNO₂ and HCl to form a compound ‘E’ which on reaction with phenol forms an orange-red dye ‘F’ Elucidate the structures of the organic compounds from ‘A’ to ‘F’
Answer:

Long answer type questions

Question 1.
Predict the reagent or the product in the following reaction sequence : [NCERT Exemplar]

Answer:
A correct sequence can be represented as follows including all reagents:

Hence,

(v) 5 = H₃PO₂/H₂O

Question 2.
A hydrocarbon ‘A’ (C₄H₈) on reaction with HC1 gives a compound ‘B’, (C₄H₉Cl), which on reaction with 1 mol of NH₃ gives compound ‘C’, (C₄H₁₁N). On reacting with NaNO₂ and HCl followed by treatment with water, compound ‘C’ yields an optically active alcohol, ‘D’ Ozonolysis of ‘A’ gives 2 mols of acetaldehyde. Identify compounds ‘A’ to ‘D\ Explain the reactions involved. [NCERT Exemplar]
Answer:
(i)Addition of HCl to compound A shows that compound A is alkene.
Compound ‘B’ is C₄H₉Cl.
(ii) Compound‘B’reacts with NH₂. It forms amine‘C’.

(iii) ‘C’ gives diazonium salt with NaNO₂/HCl, which yields an optically active alcohol.
So, ‘C’ is aliphatic amine.
(iv) ‘A on ozonolysis produces 2 moles of CH₃CHO. So, A is CH₃– CH =CH-CH₃ (But-2-ene).
Reactions

Punjab State Board PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions Important Questions and Answers.

PSEB 12th Class Chemistry Important Questions Chapter 2 Solutions

Very short answer type questions

Question 1.
Define mole fraction.
Answer:
Mole fraction of a component in a solution may be defined as the ratio of moles of that compdnent to the total number of moles of all the components present in the solution.

Question 2.
What is the similarity between Raoult’s law and Henry’s law?
Answer:
The similarity between Raoult’s law and Henry’s law is that both state that the partial vapour pressure of the volatile component or gas is directly proportional to its mole fraction in the solution.

Question 3.
Why is the vapour pressure of a solution of glucose in water lower than that of water? (NCERT Exemplar)
Answer:
This is due to decrease in the escaping tendency of the water molecules from the surface of solution as some of the surface area is occupied by non-volatile solute, glucose particles.

Question 4.
What type of deviation is shown by a mixture of ethanol and acetone? What type of azeotrope is formed by mixing ethanol and acetone?
Answer:
Positive deviation.
Minimum boiling azeotropes.

Question 5.
State how does osmotic pressure vary with temperature.
Answer:
Osmotic pressure increases with increase in temperature.

Question 6.
What are isotonic solutions?
Answer:
The solutions of the same osmotic pressure at a given temperature are called isotonic solutions.

Question 7.
Define van’t Hoff factor.
Answer:
van’t Hoff factor may be defined as the ratio of normal molecular mass to observed molecular mass or the ratio of observed colligative property to normal colligative property.

Question 8.
Why are aquatic species more comfortable in cold water in comparison to warm water? (NCERT Exemplar)
Answer:
Solubility of oxygen in water increases with decrease in temperature. Presence of more oxygen at lower temperature makes the aquatic species more comfortable in cold water.

Question 9.
What is semipermeable membrane? (NCERT Exemplar)
Answer:
Continuous sheets or films (natural or synthetic) which contain a network of submicroscopic holes or pores through which small solvent molecules (water etc.) can pass, but solute molecules of bigger size cannot pass are called semipermeable membrane.

Question 10.
Give an example of a material used for making semipermeable membrane for carrying out reverse osmosis. (NCERT Exemplar)
Answer:
Cellulose acetate.

Short answer type questions

Question 1.
State Henry’s law. Write its one application. What is the effect of temperature on solubility of gases in liquid?
Answer:
It states that the partial pressure of a gas in vapour phase (p) is proportional to the mole fraction of the gas (χ) in the solution.
p ∝ χ or p = K_Hχ where K_H is the
Henry’s constant.
Application of Henry’s law:
To increase the solubility of CO₂ in soft drinks and soda water, the bottle is sealed under high pressure.

Effect of temperature on solubility:
As dissolution is an exothermic process, therefore, according to Le Chatelier’s principle solubility should decrease with rise in temperature.

Question 2.
Why does a solution containing non-volatile solute have higher boiling point than the pure solvent? Why is elevation of boiling point a colligative property?
Answer:
The addition of a non-volatile solute to a volatile solvent lowers its vapour pressure. In order to boil the solution, i.e., to make its vapour pressure equal to atmospheric pressure, the solution has to be heated at a higher temperature. In other words, the boiling point of solution becomes higher than solvent.

As elevation in boiling point depends on the number of moles of solute particles and independent of their nature, therefore, it is a colligative property.

Question 3.
Will the elevation in boiling point be same if 0.1 mol of sodium chloride or 0.1 mol of sugar is dissolved in 1L of water? Explain.
Answer:
No, the elevation in boiling point is not the same. NaCl, being an electrolyte, dissociates almost completely to give Na⁺ and Cl^– ions whereas glucose, being non-electrolyte does not dissociate. Hence, the number of particles in 0.1 M NaCl solution is nearly double than 0.1 M glucose solution. Elevation in boiling point being a colligative property, is therefore, nearly twice for 0.1 M NaCl solution than for 0.1 M glucose solution.

Question 4.
Explain the solubility rule ‘like dissolves like’ in terms of intermolecular forces that exist in solutions. (NCERT Exemplar)
Answer:
If the intermolecular interactions are similar in both constituents, i.e., solute and solvent then solute dissolves in the solvent. e.g., polar solutes dissolve in polar solvents and non-polar solutes in non-polar solvents. Thus, the statement ‘like dissolved like’ proves to be true.

Question 5.
Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of temperature. However, molarity is a function of temperature. Explain.
(NCERT Exemplar)
Answer:
Molarity is defined as the number of moles of solute dissolved per litre of a solution. Since, volume depends on temperature and changes with change in temperature, the molarity will also change with change in temperature. On the other hand, mass does not change with change in temperature, so other concentration terms given in the question also do not do so. According to the definition of all these terms, mass of solvent used for making the solution is related to the mass of solute.

Question 6.
Which of the following solutions has higher freezing point?
0.05 M Al₂(SO₄)₃, 0.1 M K₃ [Fe(CN)₆] Justify.
Answer:
0.05 M Al₂(SO₄)₃ has higher freezing point.

Justification:
For 0.05 MAl₂(SO₄) ₃,i = 5
Number of particles = i × concentration
= 5 × 0.05
= 0.25 moles of ions
For 0.1MK₃ [Fe(CN)₆], i = 4
Number of particles = i × concentration
= 4 × 0.1
= 0.4 moles of ions
We know that, ΔT_f. Number of particles
Hence, 0.05 M Al₂(SO₄)₃ has higher freezing point because it has lower number of particles than 0.1 M K₃ [Fe (CN)₆].

Question 7.
The freezing point of benzene decreases by 2.12 K when 2.5 g of benzoic acid (C₆H₅COOH) is dissolved in 25 g of benzene. If benzoic acid forms a dimer in benzene, calculate the van’t Hoff factor and the percentage association of benzoic acid. (K_f for benzene = 5.12 K kg mol^-1)
Answer:
Given, ΔT_f = 2.12 K, K_f = 5.12 K kg mol^-1
We know that, ΔT_f = i K_f m
2.12 = \(\frac{i \times 5.12 \times 2.5 \times 1000}{122 \times 25}\)
or i = 0.505
For association,
i = 1 – \(\frac{\alpha}{2}\)
0.505 = 1 – \(\frac{\alpha}{2}\)
or α =0.99
Hence, percentage association of benzoic acid is 99%.

Long answer type questions

Question 1.
(a) What type of deviation is shown by a mixture of ethanol and acetone? Give reason.
(b) A solution of glucose (molar mass = 180 g mol^-1) in water is labelled as 10% (by mass). What would be the molality and molarity of the solution?
(Density of solution = 1.2 g mL^-1)
Answer
(a) A mixture of ethanol and acetone shows positive deviation from Raoult’s law.
In pure ethanol hydrogen bond exist between the molecules. On adding acetone to ethanol, acetone molecules get in between the molecules of . ethanol thus breaking some of the hydrogen bonds and weakening molecular interactions considerably. Weakening of molecular interactions leads to increase in vapour pressure resulting in positive deviation from Raoult’s law.

(b) Let the mass of solution = 100 g
∴ Mass of glucose = 10 g
Number of moles of glucose \(=\frac{\text { Mass of glucose }}{\text { Molar mass }}\)
= \(\frac{10 \mathrm{~g}}{180 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 0.056 mol

Question 2.
Discuss biological arid industrial importance of osmosis.
(NCERT Exemplar)
Answer:
The process of osmosis is of great biological and industrial importance as is evident from the following examples:

• Movement of water from soil into plant roots and subsequently into upper portion of the plant occurs partly due to osmosis.
• Preservation of meat against bacterial action by adding salt.
• Preservation of fruits against bacterial action by adding sugar.
  Bacterium in canned fruit loses water through the process of osmosis, shrivels and dies.
• Reverse osmosis is used for desalination of water.