PSEB 11th Class Chemistry Solutions Chapter 1 Some Basic Concepts of Chemistry

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 1 Some Basic Concepts of Chemistry Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 1 Some Basic Concepts of Chemistry

PSEB 11th Class Chemistry Guide Some Basic Concepts of Chemistry InText Questions and Answers

Question 1.
Calculate the molecular mass of the following:
(i) H₂O (ii)CO₂ (iii) CH₄
Answer:
(i) H₂O
The molecular mass of water (H20)
= (2 x Atomic mass of hydrogen) + (1 x Atomic mass of oxygen)
= [2(1.008 u) +1(16.00 u)]
= 2.016u +16.00u = 18.016 u=18.02u

(ii) CO₂
The molecular mass of carbon dioxide (CO₂)
= (1 x Atomic mass of carbon) + (2 x Atomic mass of oxygen)
= [1(12.011u) + 2(16.00u)]
= 12.011u + 32.00u = 44.01u

(iii) CH₄
The molecular mass of methane (CH₄)
= (1 x Atomic mass of carbon) + (4 x Atomic mass of hydrogen)
– [1(12.011u) + 4(1.008u)]
= 12.011u + 4.032u = 16.043u

Question 2.
Calculate the mass per cent of different elements present in sodium sulphate (Na₂SO₄).
Answer:
The molecular formula of sodium sulphate is Na₂SO₄
Molar mass of Na₂SO₄ = 2 x Atomic mass of Na + 1 x Atomic mass of S + 4 x Atomic mass of O
= [(2 x 23.0) + (1 x 32.066) + (4 x 16.00)]
= 46.0 + 32.066 + 64.00 = 142.066 g
Mass per cent of sodium

Question 3.
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Answer:

Question 4.
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mol of carbon is burnt in air
(ii) 1 mole of carbon is burnt in 16g of dioxygen
(iii) 2 moles of carbon are burnt in 16g of dioxygen
Answer:

∵ 32.0g of O₂ produce 44.0 g of CO₂
∵ 16.0 g of O₂ produce = \(\frac{44}{32}\) x 16 = 22.0 g of CO₂
Amount of CO₂produced = 22.0g
(iii) Amount of CO₂produced when 2 moles (= 24 g) of C are burnt in 16.0g (limited amount ) of O₂ = 22.0g

Question 5.
Calculate the mass of sodium acetate (CH₃COONa) required to make 500mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0.245 g mol^-1.
Answer:
0.375M aqueous solution of sodium acetate = 1000mL of solution containing 0.375 moles of sodium actate
∴ Number of moles of sodium acetate in 500 mL.
= \(\frac{0.375}{1000}\) x 500 = 0.1875 mole.
Molar mass of sodium acetate = 82.0245 g mol^-1
∴ Required mass of sodium acetate
= 82.0245 g mol^-1 x 0.1875 mole
= 15.38 g

Question 6.
Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 gmLT1 and the mass per cent of nitric acid in it being 69%.
Answer:
Mass percent of nitric acid in the sample = 69%
Thus, 100 g of sample contains 69 g of nitric acid
Molar mass of nitric acid (HNO₃)
= {1 +14 + 3(16)} g mol^-1
= 1 +14 + 48 = 63 g mol^-1
∴ Number of moles in 69 g of HNO₃

Question 7.
How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?
Answer:
1 mole of CuSO₄ contains 1 mole of copper.
Molar mass of CuSO₄
= (63.5) + (32.00) + 4(16.00)
= 63.5 + 32.00 + 64.00 = 159.5 g mol^-1
159.5 g of CuSO₄ contains = 63.5 g of copper
=> 100 g of CuSO₄ contains = \(\frac{63.5 \times 100 \mathrm{~g}}{159.5}\) = 39.81g
∴ Mass of copper that can be obtained from 100 g of CuSO₄ = 39.81 g

Question 8.
Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are 69.9 and 30.1 respectively.
Answer:

Question 9.
Calculate the atomic mass (average) of chlorine using the following data:

	% Natural Abundance	Molar Mass
35cl	75.77	34.9689
37Cl	24.23	36.9659

Answer:
The average atomic mass of chlorine

= 26.4959 + 8.9568 = 35.4527
∴ The average atomic mass of chlorine = 35.4527

Question 10.
In three moles of ethane (C₂H₆), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Answer:
(i) 1 mole of C₂H₆ contains 2 moles of carbon atoms.
∴ 3 moles of C₂H₆ will contain = 2 x 3 = 6 moles of carbon atoms
(ii) 1 mole of C₂H₆ contains 6 moles of hydrogen atoms.
∴ 3 moles of C₂H₆ will contain =3 x 6 = 18 moles of hydrogen atoms
(iii) 1 mole of C₂H₆ contains 6.022 x 10²³ molecules of ethane.
∴ 3 moles of C₂H₆ will contain = 3 x 6.022 x 10²³
= 18.066 x 10²³ molecules of ethane

Question 11.
What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L^-1 if 20 g of sugar are dissolved in enough water to make a final volume up to 2 L?
Answer:
Molar mass of sugar (C₁₂H₂₂O₁₁)
= 12 x 12 + 22 x 1 +11 x 16 = 342 g mol^-1
Molar concentration = \(\frac{\text { Moles of solute }}{\text { Volume of solution in L }}=\frac{0.0585}{2 \mathrm{~L}}\)
= 0.0293 mol L^-1 = 0.0293 M

Question 12.
If the density of methanol is 0.793 kg L^-1, what is its volume needed for making 2.5 L of its 0.25 M solution?
Answer:
Final volume, V₂ = 2.5 L
Final molarity, M₂ = 0.25 M
Density of methanol = 0.793 kg L^-1
Molarity of initial solution M₁ = ?
Initial volume, V₁ = ?
Molar mass of methanol (CH₃OH) = (1 x 12) + (4 x 1) + (1 x 16)
= 32 g mol^-1 = 0.032 kg mol^-1
Molarity of methanol solution, M₁ = \(\frac{0.793 \mathrm{~kg} \mathrm{~L}^{-1}}{0.032 \mathrm{~kg} \mathrm{~mol}^{-1}}\) = 24.78 mol L^-1
Applying,
M₁V₁ = M₂V₂
(24.78 mol L^-1)V₁ = (2.5L) (0.25mol L^-1)
V₁ = 0.02522 L
V₁ = 25.22 mL

Question 13.
Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below :
1 Pa= IN m^-2
If mass of air at sea level is 1034 g cm-2, calculate the pressure in pascal.
Answer:
Pressure is defined as force acting per unit area of the surface.
i.e, P = F/A

= 101332.0 Nm^-2 [1N = 1kg ms^-2]
∴ Pressure = 1.01332 x 10 ⁵

Question 14.
What is the SI unit of mass? How is it defined?
Answer:
The SI unit of mass is kilogram (kg).
1 kilogram is defined as the mass equal to the mass of the international prototype of kilogram.

Question 15.
Match the following prefixes with their multiples :

	Prefixes	Multiples
(i)	micro	106
(ii)	deca	109
(iii)	mega	10-6
(iv)	giga	10-15
(v)	femto	10

Answer:

	Prefixes	Multiples
(i)	micro	10-6
(ii)	deca	10
(iii)	mega	106
(iv)	giga	109
(v)	femto	10-15

Question 16.
What do you mean by significant figures?
Answer:
The number of significant figures in a given data is the number of all certain digits plus one uncretain digit.
For example, if 15.6 mL is the result of an experiment, then 15 is certain while 6 is uncertain, and the total number of significant figures are 3.

Question 17.
A sample of drinking water was found to he severely contaminated with chloroform, CHC13, supposed to he carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in per cent by mass.
(ii) Determine the molarity of chloroform in the water sample.
Answer:
(i) ∵ 10⁶ g of solution contains 15 g of CHCl₃
∴ 1 g of solution contains = \(\frac{15}{10^{6}}\) g of CHCl_{3
100 g of solution contains = \(\frac{15}{10^{6}}\) x 10² = 15 x 10^-4 g of CHCl3
∴ Percent by mass = 0.0015%}

(ii) Molar mass of CHCl₃ = 12 +1 + 3 x 35.5 = 119.5 g mol^-1
0.0015% means 15 x 10^-4 g chloroform is present in 100 g sample.
Molarity = \(\frac{W \times 1000}{m \times \text { Volume of sample }}\)
[For water density = 1 g cm^-3; so mass = volume]
= \(\frac{15 \times 10^{-4} \times 1000}{119.5 \times 100}\)
= 1.25 x 10^-4 M

Question 18.
Express the following in the scientific notation :
(i) 0.0048
(ii) 234000
(iii) 8008
(iv) 500.0
(v) 6.0012
Answer:
(i) 0.0048 = 4.8 x 10^-3
(ii) 234000 = 2.34 x 10⁵
(iii) 8008 = 8.008 x 10³
(iv) 500.0 = 5.00 x 10²
(v) 6.0012 = 6.0012 x 10⁰

Question 19.
How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126000
(v) 500.0
(vi) 2.0034
Answer:
(i) 0.0025
There are 2 significant figures,
(ii) 208
There are 3 significant figures.
(iii) 5005
There are 4 significant figures.
(iv) 126000
There are 3 significant figures.
(v) 500.0
There are 4 significant figures.
(vi) 2.0034
There are 5 significant figures.

Question 20.
Round up the following upto three significant figures :
(i) 34.216 (ii) 10.4107
(iii) 0.04597 (iv) 2808
Answer:
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810

Question 21.
The following data are obtained when dinitrogen and dioxygen react together to form different compounds :

	Mass of dinitrogen	Mass of dioxygen
(i)	14 g	16 g
(ii)	14 g	32 g
(iii)	28 g	32 g
(iv)	28 g	80 g

(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement.
(b) Fill in the blanks in the following conversions :
(i) 1 km = ………mm = ………pm
(ii) 1 mg = ………kg = ………ng
(iii) 1 mL = ………L = ………dm³
Answer:
On fixing the mass of dinitrogen as 14 g, then the masses of dioxygen which combines with the fixed mass (= 14 g) of dinitrogen will be 16, 32, 16, 40 which are in the simple whole number ratio of 1 : 2 : 1 : 2.5 or 2 : 4 : 2 : 5.

(a) Law of multiple proportions : This law was proposed by Dalton in 1803. According to this law, if tvo elements can combine to form two or more than two compounds, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

Question 22.
If the speed of light is 3.0 x 10⁸ms^-1, calculate the distance covered by light in 2.00 ns.
Answer:
According to the question,
Time taken to cover the distance = 2.00 ns
2ns = 2.00 x 10^-9s [∵ 1ns = 10^-9s]
Speed of light = 3.0 x 10⁸ms^-1
Distance travelled by light in 2.00 ns
= Speed of light x Time taken
= (3.0 x 10⁸ms^-1)(2.00 x 10^-9s) = 6.00 x 10^-1 m = 0.6 m

Question 23.
In a reaction A + B₂ → AB₂
Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
Answer:
(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B.
Thus, 300 atoms of A will react with 200 molecules of B thereby, leaving 100 atoms of A unused. Hence, B is the limiting reagent.
(ii) According to the given reaction, 1 mol of A reacts with 1 mol of B. Thus, 2 mol of A will react with 2 mol of B. As a result, 1 mol of B will not be consumed. Hence, A is the limiting reagent.
(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B. Thus, all 100 atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where no limiting reagent is present.
(iv) 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent.
(v) 1 mol of atom A combines with 1 mql of molecule B. Thus, 2.5 mol of A will combine only 2.5 mol of B and the remaining 2.5 mol of B will be left as such. Hence, A is the limiting reagent.

Question 24.
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: N2(g)+H2(g) → 2NH₃(g)
(i) Calculate the mass of ammonia produced if 2.00 x 10³ g dinitrogen reacts with 1.00 x 10³ g of dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
On balancing the given chemical equation,
Answer:

⇒ 28 g N₂ reacts with 6 g H₂; 1 g N₂ reacts with \(\frac{6}{28}\) g H₂;
2.0 x 10³ g of N2 will react with \(\frac{2000 \times 6}{28}\) = 428.6 g of H₂.
Hence, N₂ is the limiting reagent.
∵ 28 g N2 produces 34 g NH3 34
∴ 1 g N2 produces \(\frac{34}{28}\) g NH₃
2.00 x 10³ f N₂ will produce = \(\frac{34 g}{28 g}\) x 2000g = 2428.57 g NH₃
(ii) N₂ is the limiting reagent and H₂ is the excess reagent. Hence, H₂ will remain unreacted.
(iii) Mass of H₂ remain unreacted = 1.00 x 10³ g – 428.6 g = 571.43 g

Question 25.
How are 0.50 mol Na₂CO₃ and 0.50 M Na₂CO₃ different?
Answer:
Molar mass of Na₂CO₃ = (2 x 23) +12.00 + (3 x 16) = 106 g mol^-1
Now, 1 mol Na₂CO₃ means 106 g Na₂CO₃
∴ 0.5 mol Na₂CO₃ = x °-5 mo1 Na₂C0₃ = 53 g Na₂C0₃
⇒ 0.50 M Na₂CO₃ = 0.50 mol/L Na₂CO₃
Hence, 0.50 mol Na₂CO₃ means 53 g Na₂CO₃ is present in 1 L of the solution.

Question 26.
If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Answer:
Reaction of dihydrogen with dioxygen can be written as:

Now, 2 volumes of dihydrogen react with 1 volume of O₂ to produce 2 volumes of water vapour.
Hence, 10 volumes of dihydrogen will react with 5 volumes of dioxygen to produce 10 volumes of water vapour.

Question 27.
Convert the following into basic units :
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg
Answer:
(i) 28.7 pm :
∵ 1 pm = 10^-12 m
∴ 28.7pm = 28.7 x 10^-12 m = 2.87 x 10^-11 m

(ii) 15.15pm
∵ 1 pm =10 12 m
∴ 15.15 pm = 15.15 x 10^-12 m = 1.515 x 10^-11 m

(iii) 25365 mg :
1 mg = 10^-3 g
25365 mg = 2.5365 x 10⁴ x 10^-3 g
= 2.5365 x 10¹ g
Since,
1 g = 10^-3 kg
2.5365 x 10¹ g = 2.5365 x 10¹ x 10^-3 kg.
∴ 25365 mg = 2.5365 x 10^-2 kg

Question 28.
Which one of the following will have largest number of atoms?
(i) 1 g Au(s)
(ii) 1 g Na(s)
(iii) 1g Li (s)
(iv) 1 g of Cl₂(g)
Answer:
(i) 1 g Au (s) = \(\frac{1}{197}\) mol atoms of Au (s)
= \(\frac{6.022 \times 10^{23}}{197}\) mol atoms of Au (s)
= 3.06 x 10²¹ atoms of Au (s)

(ii) 1 g Na (s) = \(\frac{1}{23}\) mol atoms of Na (s)
= \(\frac{6.022 \times 10^{23}}{23}\) atoms of Na (s)
= 0.262 x 10²³ atoms of Na (s)
= 26.2 x 10²¹ atoms of Na (s)

(iii) 1 g Li (s) = \(\frac{1}{7}\) mol atoms of Li (s)
= \(\frac{6.022 \times 10^{23}}{23}\) atoms of Li (s)
= 86.0 x 1021 atoms of Li (s)

(iv) 1 gCl₂ (g) = \(\frac{1}{71}\) mol molecules of Cl₂ (g)
= \(\frac{6.022 \times 10^{23}}{71}\) molecules of Cl? (g)
= 0.0848 x 10²³ molecules of Cl₂ (g)
= 8.48 x 10²¹ molecules of Cl₂ (g)
(one molecule of Cl₂ contains two atoms of Cl)
Number of atoms of Cl = 2 x 8.48 x 10²¹ = 16.96 x 10²¹ atoms of Cl
Hence, 1 g of Li (s) has largest number of atoms.

Question 29.
Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one).
Answer:
Mole fraction of ethanol (C₂H₅OH)

Number of moles present in 1L water;
ⁿH₂O = \(\)
ⁿH₂O = 55.55 moles
Substituting the value of ⁿH₂O in eq (i).
\(\frac{n_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}}{n_{\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}}+55.55}\) = 0.040
ⁿC₂H₅OH – 0.040ⁿC₂H₅OH = 2.222mol
0.96 ⁿC₂H₅OH = 2.222 mol ⁿC₂H₅OH = \(\frac{2.222}{0.96}\) mol.
ⁿC₂H₅OH = 2.314 mol
∴ Molarity of solution = \(\frac{2.314 \mathrm{~mol}}{1 \mathrm{~L}}\) = 2.314M

Question 30.
What will be the mass of one 12C atom in g?
Answer:
1 mol of carbon atoms = 6.022 x 10²³ atoms of carbon = 12 g of carbon
∴ Mass of one ¹²C atom = \(\frac{\text { Atomic mass of } \mathrm{C}}{\text { Avogadro’s number }}=\frac{12 \mathrm{~g}}{6.022 \times 10^{23}}\)
= 1.993 x 10^-23 g

Question 31.
How many significant figures should be present in the answer of the following calculations?
(i) \(\frac{0.02856 \times 298.15 \times 0.112}{0.5785}\)
(ii) 5 x 5.364
(iii) 0.0125 + 0.7864 + 0.0215
Answer:
\(\frac{0.02856 \times 298.15 \times 0.112}{0.5785}\)
Least precise number of calculation = 0.112
Number of significant figures in the answer = Number of significant figures in the least precise number = 3 /*
(ii) 5 x 5.364
Least precise number of calculation = 5.364
Number of significant figures in the answer = Number of significant figures in 5.364 = 4
(iii) 0.0125 + 0.7864 + 0.0215
Since the least number of decimal places in each term is four, the number of significant figures in the answer is also 4.

Question 32.
Use the data given in the following table to calculate the molar mass of naturally occurring argon isotopes :

Isotope	Isotopic molar mass	Abundance
36 Ar	35.96755 g mol-1	0.337%
38 Ar	37.96272 g mol-1	0.063%
40 Ar	39.9624 g mol-1	99.600%

Answer:
Molar mass of argon

= 0.121 + 0.024 + 39.802 g mol^-1 = 39.947 g mol^-1

Question 33.
Calculate the number of atoms in each of the following
(i) 52 moles of Ar (ii) 52 u of He (iii) 52 g of He.
Answer:
(i) ∵ 1 mol of Ar = 6.022 x 10²³ atoms of Ar
∴ 52 moles of Ar = 52 x 6.022 x 10²³ atoms of Ar
= 3.131 x 10²⁵ atoms

(ii) ∵ 1 atom of He = 4 u of He
Or, 4 u of He = 1 atom of He
∴ 1 u of He = 1/4 atom of He
∴ 52 u of He = 52/4 atoms of He = 13 atoms

(iii) ∵ 4 g of He = 6.022 x 10²³ atoms
6022 1023 52
∴ 52 g of He = \(\frac{6.022 \times 10^{23} \times 52}{4}\) = 7.8286 x 10²⁴ atoms

Question 34.
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.
Answer:
(i) 44 g CO₂ = 12 g carbon
3.38 g CO₂ = 12/44 x 3.38g = 0.9218 g carbon
18 g H₂O = 2g hydrogen
0.690 g H₂O = 2/18 x 0.690 g = 0.0767 g hydrogen
Total mass of compound = 0.9218+0.0767 = 0.9985 g (because compound contains only carbon and hydrogen)
% of C in the compound = \(\frac{0.9218}{0.9985}\) x 100 = 92.32
% of H in the compound = \(\frac{0.0767}{0.9985}\) x 100 = 7.68
Calculation for Empirical Formula

Hence, empirical formula = CH
(ii) Calculation for molar mass of the gas
10.0 L of the given gas at STP weigh = 11.6 g
∴ 22.4 L of the given gas at STP will weigh = \(\frac{11.6 \times 22.4}{10}\) = 25.984 g
Molar mass = 25.984 = 26 g mol^-1
(iii) Empirical formula mass (CH) =12 + 1 = 13
∴ n = \(\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{26}{13}\) = 2
Hence, molecular formula = n x CH = 2 x CH = C₂H₂

Question 35.
Calcium carbonate reacts with aqueous HC1 to give CaCl₂ and CO₂ according to the reaction,
CaC0₃(s) + 2 HCl(aq) -» CaCl₂(aq) + C0₂(g) + H₂O(l) What mass of CaC0₃ is required to react completely with 25 mL of 0.75 M HC1?
Answer:
The given reaction is
CaC0₃(s) + 2HCl(aq) → CaCl₂(aq) + C0₂(g) + H₂0(l)
Let us find out the weight of HCl present in 25 mL of 0.75 M HCl
1000 mL of 1.0 M HCl contains = 36.5 g
25 mL of 0.75 M HCl contains = \(\frac{36.5}{1000}\) x 25 x 0.75
= 0.6844 g of HCl
According to the equation;
73 g of HCl [2(1 + 35.5)] reacts with 100.0 g of CaC0₃
0.6844 g of HCl reacts with = \(\frac{100}{73}\) x 0.6844 = 0.94 g of CaC0₃

Question 36.
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction
4HCl(oq) + MnO₂(s) > 2H₂O(l) + MnCl₂(aq) + Cl₂(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Answer:

According to the balanced chemical equation,
∵ 87 g of MnO₂ react with 4 x 36.5 g HCl
∴ 5 g of MnO₂ will react \(\frac{4 \times 36.5 \times 5}{87}\) = 8.39 g HCl