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# NCERT Solutions - Basic Concepts of Chemistry (Part - 1) Class 11 Notes | EduRev

## Chemistry Class 11

Created by: Mohit Rajpoot

## JEE : NCERT Solutions - Basic Concepts of Chemistry (Part - 1) Class 11 Notes | EduRev

The document NCERT Solutions - Basic Concepts of Chemistry (Part - 1) Class 11 Notes | EduRev is a part of the JEE Course Chemistry Class 11.
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Question 1.1: Calculate the molecular mass of the following:
(a) H2O
(b) CO2
(c) CH4
Ans.
(a) H2O:
The molecular mass of water, H2O
= (2 × Atomic mass of hydrogen) + (1 × Atomic mass of oxygen)
= [2(1.0084)  + 1(16.00 u)]
= 2.016 u + 16.00 u
= 18.016
= 18.02 u
(b) CO2:
The molecular mass of carbon dioxide, CO2
= (1 × Atomic mass of carbon) + (2 × Atomic mass of oxygen)
= [1(12.011 u) + 2 (16.00 u)]
= 12.011 u +  32.00 u
= 44.01 u
(c) CH4:
The molecular mass of methane, CH4
= (1 × Atomic mass of carbon) + (4 × Atomic mass of hydrogen)
= [1(12.011 u) + 4 (1.008 u)]
= 12.011 u + 4.032 u
= 16.043 u

Question 1.2: Calculate the mass percent of different elements present in sodium sulfate (Na2SO4).
Ans. The molecular formula of sodium sulphate is Na2SO4
Molar mass of Na2SO4 = [(2 × 23.0) + (32.066) + 4 (16.00)]
= 142.066 g
Mass percent of an element ∴ Mass percent of sodium: = 32.379
=32.4%
Mass percent of sulphur: = 22.57
= 22.6 %
Mass percent of oxygen: = 45.049
= 45.05%

Question 1.3: Determine the empirical formula of an oxide of iron which has 69.9% iron & 30.1% dioxygen by mass.
Ans.
Given:
% of iron by mass = 69.9 %
% of oxygen by mass = 30.1 %
Relative moles of iron in iron oxide:  = 1.25
Relative moles of oxygen in iron oxide:  =1.88
Simplest molar ratio of iron to oxygen:
= 1.25: 1.88
= 1:1.5 = 2:3
∴ The empirical formula of the iron oxide is Fe2O3.

Question 1.4: Calculate the amount of carbon dioxide that could be produced when
(a) 1 mole of carbon is burnt in air.
(b) 1 mole of carbon is burnt in 16 g of dioxygen.
(c) 2 moles of carbon are burnt in 16 g of dioxygen.
Ans.
The balanced reaction of combustion of carbon can be written as: (a) As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen (air) to produce 1 mole of carbon dioxide.
(b) According to the question, only 16 g (0.5 mol) of dioxygen is available. Hence, it will react with 0.5 mole of carbon to give 22 g of carbon dioxide. Hence, it is a limiting reactant
(c) Since oxygen is the limiting reactant here, the 16g (0.5 mol) of O2 will react with 6 g of carbon (0.5 mol) to form 22 g of carbon dioxide. The remaining 18 g of carbon (1.5 mol) will not undergo combustion.

Question 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–1.
Ans.
0.375 M aqueous solution of sodium acetate ≡ 1000 mL of solution containing 0.375 moles of sodium acetate.
∴Number of moles of sodium acetate in 500 mL.
= 0.375/1000 x 500
= 0.1875 mole
Molar mass of sodium acetate = 82.0245 g mol–1 (Given)
∴ Required mass of sodium acetate = (82.0245 g mol–1) (0.1875 mole)
= 15.38 g

Question 1.6: Calculate the concentration of nitric acid in moles per liter in a sample which has a density, 1.41 g mL–1 and the mass percent of nitric acid in it being 69%.
Ans.
Mass percent of nitric acid in the sample = 69 % (Given)
Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.
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 = 1.095 mol
Volume of 100 g of nitric acid solution  = 70.92 mL ≡ 70.92 x 10-3 L
Concentration of nitric acid = 15.44 mol/L
∴ Concentration of nitric acid = 15.44 mol/L

Question 1.7: How much copper can be obtained from 100 g of copper sulphate (CuSO4)?
Ans.
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
159.5 g of CuSO4 contains 63.5 g of copper.
⇒ 100 g of CuSO4 will contain of copper.
∴ Amount of copper that can be obtained from 100 g CuSO = 39.81 g

Question 1.8: Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.69 g mol–1.
Ans.
Mass percent of iron (Fe) = 69.9% (Given)
Mass percent of oxygen (O) = 30.1% (Given)
Number of moles of iron present in the oxide = 69.90/55.85 = 1.25
Number of moles of oxygen present in the oxide = 30.1/16.0 = 1.88
Ratio of iron to oxygen in the oxide,
= 1.25 : 1.88
= = 1 : 1.5
= 2 : 3
The empirical formula of the oxide is Fe2O3.
Empirical formula mass of Fe2O3 = [2(55.85) + 3(16.00)] g
Molar mass of Fe2O3 = 159.69 g Molecular formula of a compound is obtained by multiplying the empirical formula with n.
Thus, the empirical formula of the given oxide is Fe2O3 and n is 1.
Hence, the molecular formula of the oxide is Fe2O3.

Question 1.9: Calculate the atomic mass (average) of chlorine using the
following data: Ans:
The average atomic mass of chlorine = 26.4959 + 8.9568
= 35.4527 u
The average atomic mass of chlorine = 35.4527 u

Question 1.10: In three moles of ethane (C2H6), calculate the following:
(a) Number of moles of carbon atoms.
(b) Number of moles of hydrogen atoms.
(b) Number of molecules of ethane.
Ans.
(a) 1 mole of C2H6 contains 2 moles of carbon atoms.
Number of moles of carbon atoms in 3 moles of C2H
= 2 × 3 = 6
(b) 1 mole of C2Hcontains 6 moles of hydrogen atoms.
Number of moles of carbon atoms in 3 moles of C2H
= 3 × 6 = 18
(c) 1 mole of C2Hcontains 6.023 × 1023 molecules of ethane.
Number of molecules in 3 moles of C2H
= 3 × 6.023 × 1023
= 18.069 × 102

Question 1.11: What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water to make a final volume up to 2 L?
Ans.
Molarity (M) of a solution is given by: = 0.02925 mol L–1
Molar concentration of sugar = 0.02925 mol L–1.

Question 1.12: 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?
Ans. Molar mass of methanol (CH3OH) = (1 × 12) + (4 × 1) + (1 × 16)
= 32 g mol–1
= 0.032 kg mol–1
Molarity of methanol solution = 24.78 mol L–1
(Since density is mass per unit volume)
Applying,
M1V1 = M2V2
(Given solution) = (Solution to be prepared)
(24.78 mol L–1) V= (2.5 L) (0.25 mol L–1)
V= 0.0252 L
V= 25.22 mL

Question 1.13: Pressure is determined as force per unit area of the surface.
an SI unit of pressure, Pascal is as shown below:
1 Pa = 1 N m–2
If mass of air at sea level is 1034 g cm–2, calculate the pressure in Pascals.
Ans.
Pressure is defined as force acting per unit area of the surface.
P = F/A = 1.01332 × 105 kg m-1 s-2
We know,
1 N = 1 kg ms–2
Then,
1 Pa = 1 N m–2 = 1 kg m–2s–2
1 Pa = 1 kg m1s2 Pressure = 1.01332 × 105 Pa.

Question 1.14: What is the SI unit of mass? How is it defined?
Ans. 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 1.15: Match the following prefixes with their multiples Ans. Question 1.16: What do you mean by significant figures?
Ans. Significant figures are those meaningful digits that are known with certainty.They indicate uncertainty in an experiment or calculated value.
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.
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.

Question 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).
(a) Express this in percent by mass.
(b) Determine the molality of chloroform in the water sample.
Ans.
(a) 1 ppm is equivalent to 1 part out of 1 million (106) parts.
Mass percent of 15 ppm chloroform in water
= 15/106 × 100
≈1.5 x 10-3%
(b) 100 g of the sample contains 1.5 × 10–3 g of CHCl3.
⇒ 1000 g of the sample contains 1.5 × 10–2 g of CHCl3.
Molality of chloroform in water Molar mass of CHCl3 = 12.00 + 1.00 + 3(35.5)
= 119.5 g mol–1
Molality of chloroform in water = 0.0125 × 10–2 m
= 1.25 × 10–4 m

Question 1.18: Express the following in the scientific notation:
(a) 0.0048
(b) 234,000
(c) 8008
(d) 500.0
(e) 6.0012
Ans.
(a) 0.0048 = 4.8× 10–3
(b) 234, 000 = 2.34 ×105
(c) 8008 = 8.008 ×103
(d) 500.0 = 5.000 × 102
(e) 6.0012 = 6.0012

Question 1.19: How many significant figures are present in the following?
(a) 0.0025
(b) 208
(c) 5005
(d) 126,000
(e) 500.0
(f) 2.0034
Ans.
(a) 0.0025
There are 2 significant figures.
(b) 208
There are 3 significant figures.
(c) 5005
There are 4 significant figures.
(d) 126,000
There are 3 significant figures.
(e) 500.0
There are 4 significant figures.
(f) 2.0034
There are 5 significant figures.

Question 1.20: Round up the following up to three significant figures:
(a) 34.216
(b) 10.4107
(c) 0.04597
(d) 2808
Ans.
(a) 34.2
(b) 10.4
(c) 0.0460
(d) 2810

Question 1.21: The following data are obtained when dinitrogen and dioxygen react together to form different compounds: (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
Ans.
(a) If we fix the mass of dinitrogen at 28 g, then the masses of dioxygen that will combine with the fixed mass of dinitrogen are 32 g, 64 g, 32 g, and 80 g.
The masses of dioxygen bear a whole number ratio of 1:2:2:5. Hence, the given experimental data obeys the law of multiple proportions.
The law states that if two elements combine to form more than one compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers
(b)
(i) 1 km = 1 km × 1000 m/1 km × 100 cm/1m × 10 mm/1cm
1 km = 106 mm
1 km = 1 km × 1000 m/1 km × 1 pm/10-12 m
1 km = 1015 pm
Hence, 1 km = 106 mm = 1015 pm
(ii) 1 mg = 1 mg × 1 g/1000 mg × 1 kg/1000 g
⇒ 1 mg = 10-6 kg
1 mg = 1 mg × 1 g/ 1000 mg × 1 mg/ 10-9 g 1
⇒ 1 mg = 106 ng
1 mg = 10–6 kg = 106 ng
(iii) 1 mL = 1 mL × 1 L/1000 mL
⇒ 1 mL = 10–3 L
1 mL = 1 cm3 = 1 cm× ⇒ 1 mL = 10–3 dm3
1 mL = 10–3 L = 10–3 dm3

Question 1.22: If the speed of light is 3.0 × 108 m s-1, calculate the distance covered by light in 2.00 ns.
Ans.
According to the question:
Time taken to cover the distance = 2.00 ns = 2.00 × 10–9 s
Speed of light = 3.0 × 108 ms–1
Distance travelled by light in 2.00 ns
= Speed of light × Time taken
= (3.0 × 108 ms –1) (2.00 × 10–9 s)
= 6.00 × 10–1 m
= 0.600 m.

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