NCERT solutions- chapter1 Class 11th Chemistry
1.1 Calculate the molar mass of the following:
(i) H₂O (ii) CO₂ (iii) CH₄
Answer.
(i) 1u x 2 + 16u = 18u = 18 g/mol
(ii) 12u + 16u x 2 = 44u = 44 g/mol
(iii) 12u + 1u x 4 = 16u = 16 g/mol
1.2 Calculate the mass per cent of different elements present in sodium sulphate (Na₂SO₄).
Answer. Different elements present in Na₂SO₄ are- sodium, sulphur, and oxygen.
Mass % of an element = mass of that element in the compound ➗ molar mass of the compound ❌ 100
Molar mass of Na₂SO₄ = 23 x 2 + 32 + 16 x 4 = 142 g/mol
∴ Mass % of sodium = (23 x 2) ÷ 142 x 100 = 32.4%
Mass % of sulphur = 32 ÷ 142 x 100 = 22.5%
Mass % of oxygen = (16 x 4) ÷ 142 x 100 = 45%
1.3 Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.
Answer.
|
Elements |
Atomic
mass |
% composition or mass of each element |
Number
of moles of each element |
Divide
each with the smallest value |
Convert
to a simple whole number |
|
Fe |
56 |
69.9%
=69.9g
|
69.9 ÷
56 = 1.25 |
1.25 ÷
1.25 = 1 |
1 x
2 =2 |
|
O |
16 |
30.1%
=30.1g
|
30.1 ÷
16 = 1.88 |
1.88 ÷ 1.25 =1.5 |
1.5 x
2 =3 |
The empirical formula is - Fe₂O₃
1.4 Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Answer. (i) C + O₂ ⟶ CO₂
It can be seen from the above reaction that 1mole carbon reacts with 1mole of oxygen to form 1mole carbon-di-oxide.
Number of moles of CO₂ (n) = Mass of CO₂ ➗ Molar mass of CO₂
n=1mole, molar mass of CO₂ = 44g/mol, mass of CO₂ we have to calculate = ❓
1mole = Mass of CO₂ produced ÷ 44g/mol
Mass or amount of CO₂ produced = 44g
(ii) C + O₂ (16g) ⟶ CO₂
number of moles of oxygen = Mass of O₂ ➗ Molar mass of O₂
n = 16g ÷ 32g/mol
n = 0.5mol
So, this means that only 0.5mole of C will react with 0.5moles of O₂ to form 0.5mole of CO₂. Here, 0.5mole of CO₂ will be wasted as there is not enough oxygen.
0.5mole = Mass of CO₂ produced ÷ 44g/mol
Mass or amount of CO₂ produced = 22g
Extra info- In this case, oxygen is the limiting agent as it limits the amount of CO₂ produced. Despite the presence of enough carbon the amount of CO₂ produced is low.
(iii) C (2mole) + O₂ (16g) ⟶ CO₂
The amount of available oxygen is only 16g (0.5mole) and will behave as a limiting agent. Thus, it does not matter how much the amount of carbon is increased, as only 0.5mole of carbon can react in this case because oxygen is only 0.5mole.
Thus, Mass or amount of CO₂ produced = 22g
1.5 Calculate the mass of sodium acetate (CH₃COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g/mol.
Answer. Molarity = No. of moles of solute ➗ Volume of solution in litres
No. of moles of solute = Mass ÷ Molar mass
Mass we have to calculate = ❓
Given, Molar mass = 82.0245g/mol
Molarity = 0.375M
Volume of solution = 500ml = 0.5L
Thus, 0.375 = Mass/ 82.0245/ 0.5
Mass = 15.379 = 15.38g
1.6 Calculate the concentration of nitric acid in moles per litre in a sample which has a density, 1.41 g/ml and the mass per cent of nitric acid in it being 69%.
Answer. Mass % of nitric acid = 69%
Density of solution = 1.41g/ml
Concentration or Molarity of nitric acid = ?
Molarity = No. of moles of solute ➗ Volume of solution in litres
Mass of nitric acid = 69g
Mass of solution = 100g
Density = Mass of solution ÷ Volume of solution
1.41g/ml = 100 ÷ Volume of solution
Volume of solution = 100g ÷ 1.41g/ml
= 70.92 ml = 0.07092 L
Number of moles of nitric acid = Mass ÷ Molar mass
= 69g ÷ 63g/mol
= 1.095 mol
Molarity = 1.095 mol ÷ 0.07092 L
= 15.55 mol/L
1.7 How much copper can be obtained from 100 g of copper sulphate (CuSO₄)?
Answer. Molar mass of CuSO₄ = 63.5 + 32 + 16x4 = 159.5g
As, 159.5g of CuSO₄ contains = 63.5g of Cu
∴ The amount of copper from 100g CuSO₄ will be
63.5 x 100/ 159.5 = 39.81g
1.8 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. Given that molecular mass is 159.69g.
Answer.
Elements | Atomic mass | % composition or mass of each element | Number of moles of each element | Divide each with the smallest value | Convert to a simple whole number |
Fe | 56 | 69.9% =69.9g | 69.9 ÷ 56 = 1.25 | 1.25 ÷ 1.25 = 1 | 1 x 2 =2 |
O | 16 | 30.1% =30.1g | 30.1 ÷ 16 = 1.88 | 1.88 ÷ 1.25 =1.5 | 1.5 x 2 =3 |
The empirical formula is - Fe₂O₃
n = Molecular formula mass ÷ Emperical formula mass
(Where n can be any integer)
Emperical formula mass = 2 x 55.85 + 3 x 16 = 159.7g
n = 159.69 ÷ 159.7
n = 0.999 = 1
Molecular formula is calculated by multiplying n with Emperical formula.
Molecular formula = Emperical formula x n
∴ Molecular formula = Fe₂O₃ x 1 = Fe₂O₃
1.9 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. (75.77 ÷ 100 x 34.9689) + (24.23 ÷ 100 x 36.9659)
= 35.4527u
1.10 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. 3 moles of ethane = 3 C₂H₆
(i) Number of moles of carbon atoms = 3 x 2 = 6mol
(ii) Number of moles of hydrogen atoms = 3 x 6 = 18mol
(iii) 1 mole = 6.022 x 10²³ molecules
So, total molecules in 3 moles of ethane = 3 x 6.022 x 10²³
= 18.069 x 10²³ molecules
1.11 What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol/L if its 20 g are dissolved in enough water to make a final volume up to 2L?
Answer.
Molarity = No. of moles of solute ➗ Volume of solution in litres
No. of moles of solute = Mass/ Molar mass
= 20/12x12 + 22x1 + 11x16
= 20/342 = 0.0585 mol
Molarity or concentration of sugar = 0.0585 mol/ 2L
= 0.02925 mol/L
1.12 If the density of methanol is 0.793 kg/L, what is its volume needed for making 2.5 L of its 0.25 M solution?
Answer. Density of methanol is given = 0.793 kg/L
That means, 0.793 kg of methanol is dissolved in 1L
Mass of methanol = 0.793 kg = 793g
The molar mass of methanol (CH₃OH) = 32g/mol
No. of moles of methanol = 793/ 32 = 24.78mol
Molarity = No. of moles of solute ➗ Volume of solution in litres
Molarity = 24.78mol/1L = 24.78mol/L
Using, M₁V₁ = M₂V₂
24.78 x V₁ = 0.25 x 2.5
V₁ = 0.0252 L = 25.2 ml
Volume of methanol needed for making 2.5 L of its 0.25 M solution will be = 25.2 ml
1.13 Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:
1Pa = 1N/m²
If mass of air at sea level is 1034 g/cm², calculate the pressure in pascal.
Answer. Pressure = Force/Area = m x g/Area
Pressure = 1034g x 9.8 m/s² / 1cm²
Pressure = 1.034kg x 9.8 m/s² / 0.0001m²
Pressure = 101332 kg/m/s² -----------(eq.1)
We know 1N = 1kgm/s²
1Pa = 1N/m² (Given)
So, 1Pa = 1kgm/s²/m² (Putting 1N = kgm/s²)
1Pa = 1kg/m/s²---------------(eq.2)
Putting eq.2 in eq.1, we get
Pressure = 101332 Pa
1.14 What is the SI unit of mass? How is it defined?
Answer. SI unit of mass- Kg. 1kg is defined as the mass of 1liter of pure water at 4℃.
1.15 Match the following prefixes with their multiples:
Prefixes Multiples
(i) micro 10⁶
(ii) deca 10⁹
(iii) mega 1/10⁶
(iv) giga 1/10¹⁵
(v) femto 10
Answer.
(i) micro 1/10⁶
(ii) deca 10
(iii) mega 10⁶
(iv) giga 10⁹
(v) femto 1/10¹⁵
1.16 What do you mean by significant figures?
Answer. Significant figures are those meaningful digits that are
known with certainty. They indicate uncertainty in an experiment or calculated
value. For example, if 22.6 mL is the result of an experiment, then 22 is certain while 6 is
uncertain, and the total number of significant figures are 3.
Hence, significant figures are defined as the total number
of digits in a number including the last digit that represents the uncertainty of the
result.
1.17 A sample of drinking water was found to be severely contaminated with chloroform, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15
ppm (by mass).
(i) Express this in per cent by mass.
(ii) Determine the molality of chloroform in the water sample
Answer. (i)
1 ppm is equivalent to 1 part out of 1 million parts.
Mass % of 15 ppm chloroform in water = (15/ 106) x(100)
= 1.5 x 10-3 %
(ii) That means 100 g of sample contains 1.5 x 10-3 g of CHCl3.
1000 g of the sample contains 1.5 x 10-2 g of CHCl3.
Molality = No. of moles of solute ➗ Mass of solvent in Kg
Mass of solvent = 1kg
No. of moles of solute = Mass/Molar mass of CHCl3
Molar mass of CHCl3 = 12 + 1 + 3 x 35.3 = 119.5 g/mol
Molality of CHCl3 in water = (1.5 x 10-2)/119.5/1kg
Molality of chloroform in water = 0.0125 x 10-2 m
= 1.25 x 10-4 m
1.18 Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Answer.
(i) 0.0048 = 4.8 x 10-3
(ii) 234,000 = 2.34 x105
(iii) 8008 = 8.008 x 103
(iv) 500.0 = 5.000 x102
(v) 6.0012 = 6.0012
1.19 How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034
Answer.
(i) 2
(ii) 3
(iii) 4
(iv) 3
(v) 4
(vi) 5
(iii) 4
(iv) 3
(v) 4
(vi) 5
1.20 Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810
1.21 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.
Mass of dinitrogen Mass of dioxygen
(i) 14 g x 2= 28g 16 gx 2= 32 g
(ii) 14 g x 2= 28g 32 gx 2= 64 g
(iii) 28 g 32 g
(iv) 28 g 80 g
(a) Law of multiple proportions. If we fix the mass of dinitrogen to 28 g then masses of dioxygen that will combine with the fixed mass of dinitrogen are 32 g, 64 g, 32 g and 80 g.
Thus, when different masses of one element (dioxygen) combines with the fixed mass of another element (dinitrogen), then the masses of dioxygen bear a whole number ratio of 1:2:2:5 within themselves.
(b)
(i) 1 Km = 106 mm = 1015 pm
(ii) 1 mg = 10-6 kg = 106 ng
(iii) 1 mL = 10-3 L = 10-3 dm3
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