Class 11 - Chemistry - Concepts of Chemistry

Question 1.1

Calculate the molar mass of the following:

• H2O (ii) CO2 (iii) CH4

• Molecular mass of water, H2O = (2 x At. Mass of Hydrogen) + (1 x At. Mass of oxygen)= [(2 x 1.0084) + (1 x 16)]= 2.016 + 16 = 18.016 u
• Molecular mass of carbon dioxide, CO2 = (2 x At. Mass of C) + (2 x At. Mass of O)= [(1 x 12.011) + (2 x 16)] = 12.011 + 32 = 44.01 u
• Molecular mass of methane, CH4= [(1 x 12.011) + (4 x 1.008)] = 12.011 + 4.032 = 16.043

Question 1.2

Calculate the mass per cent of different elements present in sodium sulphate (Na2SO4).

Molecular mass of sodium sulphate Na2SO4

Molar mass of Na2SO4 = [(2 x 23) + (1 x 32.066) + (4 x 16)]

= 46 + 32.066 + 64

= 142.066 g

Mass % of Sodium = 46.0  x 100

142.066

= 32.379

= 32.4 %

Mass % of Sulphur = 32.066 x 100

142.066

= 22.57

= 22.6 %

Mass % of oxygen = 64 x 100

142.066

= 45.049

= 45.05 %

Question 1.3

Determine the empirical formula of an oxide of iron, which has 69.9% iron and 30.1% dioxygen by mass.

% of iron by mass = 69.9 %

% of oxygen by mass = 30.1 %

Relative moles of iron in iron oxide = (% of iron by mass)/(At. Mass of iron)

= (69.9/55.85)  = 1.25

Relative moles of oxygen in iron oxide

=(% of oxygen by mass)/(  At. Mass of oxygen)

= (30.1/16)      = 1.88

Simplest Molar Ratio of iron to oxygen = 1.25 : 1.88 = 1 : 1.5 = 2 : 3

Empirical formula of iron oxide is Fe2O3.

Question 1.4

Calculate the amount of carbon dioxide that could be produced when

• 1 mole of carbon is burnt in air.
• 1 mole of carbon is burnt in 16 g of dioxygen.
• 2 moles of carbon are burnt in 16 g of dioxygen.

• 1 mole of carbon is burnt in air

C + O2 --> CO2

1 mole of carbon reacts with 1 mole of O2 to form one mole of CO2.

Amount of CO2 produced = 44 g

• 1 mole of carbon is burnt in 16 g of O2

1 mole of carbon burnt in 32 g of O2 it forms 44 g of CO2.

Therefore, 16 g of O2 will form (44 x 16)/32

= 22 g of CO2

• 2 moles of carbon are burnt in 16 g of O2

1 mole of carbon are burnt in 16 g of oxygen it forms 22 g of CO2.

2 moles of carbon are burnt it will form (2 x 22) /1 = 44 g of CO2.

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.

0.375 M aqueous solution of CH3COONa  = 1000 mL of solution containing 0.375 moles of CH3COONa

Number of moles of CH3COONa in 500 mL = 0.375 x 1000 = 0.1875 mole

Molar mass of sodium acetate = 82. 0245 g mol-1

Mass required for CH3COONa = (82.0245 g mol-1) (0.1875 mole)

= 15.38 g

Question 1.6

Calculate the concentration of nitric acid in moles per litre in a sample which has a density 1.41 g mL-1 and the mass per cent of nitric acid in it being 69 %.

Mass % of HNO3 in sample = 69 %

100 g of HNO3 contains 69 g of HNO3 by mass

Molar mass of HNO3 = [1 + 14 = 3(16)] g mol-1

= 1 + 14 + 48 = 63 g mol-1

Number of moles in 69 g of HNO3 = (69 /63) = 1.095 mol

Volume of 100 g HNO3 solution = (Mass of solution) / (Density of solution)

= (100 /1.41) = 70.92 mL

= 70.92 x 10-3 L

Concentration of HNO3 = (1.095 mole)/(70.92 x 10-3 L)

Concentration of HNO3 = 15.44 mol/L

Question 1.7

How much copper can be obtained from 100 g of copper sulphate (CuSO4)?

1 mole of CuSO4 contains 1 mole of Cu

Molar mass of CuSO4 = (63.5 + 32 + (4 x 16)) = 159.5 g

159.5 g of CuSO4 contains 63.5 g of Cu.

So, 100 g of CuSO4 will contain [(63.5 x 100) /159.5] of Cu = 39.81 g

Question 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.

Mass % of Fe = 69.9 %

Mass % of O = 30.1 %

Number of moles of iron present in oxide = (69.90)/(55.85) = 1.25

Number of moles of oxygen present in oxide = (30.1)/(16) = 1.88

Ratio of Iron to oxygen in oxide = 1.25: 1.88

= 1.25 : 1.88

=  1.25    1.25

=1 : 1.5

= 2:3

Empirical formula of oxide is Fe2O3.

Empirical formula mass of Fe2O3 = [(2 x 55.85) + (3 x 16)]

Molar mass of Fe2O3 = 159.69 g

n = (Molar mass) /(Empirical formula mass) = (159.69)/(159.7) = 0.999

= 1 (approx)

Molecular formula of the oxide is Fe2O3.

Question 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

Average atomic mass of Cl = [(Fractional abundance of 35Cl) (Molar mass of 35Cl) + (fractional abundance of 37Cl) (Molar mass of 37Cl)]

= [{(75.77/(100) (34.9689)} + {(24.23/(100) (34.9659)}]

= 26.4959 + 8.9568

= 35.4527 u

Question 1.10

In three moles of ethane (C2H6), calculate the following:

• Number of moles of carbon atoms.
• Number of moles of hydrogen atoms.
• Number of molecules of ethane.

• 1 mole C2H6 contains two moles of C-atoms

Number of moles of C- atom in 3 moles of C2H6

= 2 x 3 = 6

• 1 mole of C2H6 contains six moles of H-atoms

Number of moles of C-atom in 3 moles of C2H6

= 3 x 6 =18

• 1 mole of C2H6 contains six moles of H-atoms

Number of molecules in 3 moles of C2H6

= 3 x 6.023 x 1023

= 18.069 x 1023

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 2L?

Molarity (M) = (Number of moles of solute) /( Volume of solution in Litres)

Molarity (M) = (Mass of sugar/Molar mass of sugar)/2L

Let us first find, Mass of sugar/Molar mass of sugar

= (20) / [(12 x 12) + (1 x 22) + (11 x 16)]

= (20)/ (342)

= 0.0585 mol

Put this value in the equation above,

M = (0.0585)/(2) = 0.02925 mol L-1

Molar concentration = 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?

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 solution = 0.793 kg L-1/0.032 kg mol-1 = 24.78 mol L-1

Now, M1V1 = M2V2

24.78 V1 = 2.5 x 0.25

V1 = 0.0252 L

V1 = 25.22 mL

Question 1.13

Pressure is determined as force per unit area of the surface. The SI unit of pressure, pascal is as shown below:

1 Pa = 1 Nm-2

If mass of air at sea level is 1034 g cm-2, calculate the pressure in pascal.

Pressure is force per unit area of the surface.

P = F/A

P = ( 1034 g x 9.8 ms-2/ cm2) x (1 Kg/1000g) x (100)2 cm2/1m2

= 1.01332 x 105 Kg m-1 s-2

1 N = 1 Kg m s-2

1 Pa = 1 Nm-2 = 1 Kg m-2 s2

1 Pa  = 1 kg m-1 s-2

Pressure = 1.01332 x 105 Pa

Question 1.14

What is the SI unit of mass? How is it defined?

SI unit of mass is kilogram (Kg).

The mass equal to the international prototype of kilogram is known as mass.

Question 1.15

Match the following prefixes with their multiples:

Prefixes Multiples

• Micro 106
• Deca 109
• Mega 10-6
• Giga 10-15
• Femto 10

Prefixes Multiples

• Micro 10-6
• Deca 10
• Mega 106
• Giga 109
• Femto 10-15

Question 1.16

What do you mean by significant figures?

Significant figures are meaningful digits which are known with certainty.

Let us take an example:

The result of an experiment is 14.6 mL. In this case, 14 is certain and 6 is uncertain. The total significant figures are 3.

Significant figures are 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).

• Express this in per cent by mass.
• Determine the molality of chloroform in the water sample.

• 1 ppm is equivalent to 1 part out of 1 million parts.

Mass percent of 15 ppm chloroform in water = (15/ 106) x(100)

= 1.5 x 10-3 % (approx)

• 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 of CHCl3 in water = (1.5 x 10-2 g)/Molar mass of CHCl

Molar mass of CHCl3 = 12 + 1 + 3 x35.3 = 119.5 g mol-1

Molality of chloroform in water = 0.0125 x 10-2 m = 1.25 x 10-4 m

Question 1.18

Express the following in the scientific notation:

• 0.0048
• 234,000
• 8008
• 500.0
• 6.0012

• 0.0048 = 4.8 x 10-3
• 234,000 = 2.34 x105
• 8008 = 8.008 x 103
• 500.0 = 5.000 x102
• 6.0012 = 6.0012

Question 1.19

How many significant figures are present in the following?

• 0.0025
• 208
• 5005
• 126,000
• 500.0
• 2.0034

• 2 significant figures
• 3 significant figures
• 4 significant figures
• 3 significant figures
• 4 significant figures
• 5 significant figures

Question 1.20

Round up the following upto three significant figures:

• 34.216
• 10.4107
• 0.04597
• 2810

• 34.2
• 10.4
• 0.0460
• 2810

Question 1.21

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

Mass of dinitrogen      Mass of dioxygen

• 14 g                                  16 g
• 14 g                                  32 g
• 28 g                                  32 g
• 28 g                                  80 g
• Which law of chemical combination is obeyed by the above experimental data? Give its statement.
• Fill in the blanks in the following conservation:
• 1 Km = ____ mm = ____ pm
• 1 mg = ____ kg = ____ ng
• 1 mL = ____ L = ____dm3

• If we set 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.
• The masses of dioxygen bear a whole number ratio of 1:2:2:5. This is proved that the given experimental data obeys law of multiple proportions.
• The law of multiple proportions states, “If 2 elements combine to form more than 1 compound, then the masses of one element that combines with the fixed mass of another element are in the ratio of small whole numbers.”

• 1 Km = 106 mm = 1015 pm
• 1 mg = 10-6 kg = 106 ng
• 1 mL = 10-3 L = 10-3 dm3

Question 1.22

If the speed of light is 3.0 x 108 ms-1, calculate the distance covered by light in 2.00 ns.

Time taken to cover the distance = 2 ns

= 2 x 10-9 s

Speed of light = 3 x 108 m/s

Distance travelled by light in 2 ns = speed of light x time taken

Distance = 3 x 108 x 2 x 10-9

Distance = 6 x 10-1 m = 0.600 m

Question 1.23

In a reaction

A + B2 --> AB2

Identify the limiting agent, if any, in the following reaction mixture.

• 300 atoms of A + 200 molecules of B
• 2 mol A + 3 mol B
• 100 atoms of A + 100 molecules of B
• 5 mol A + 2.5 mol B
• 5 mol A + 5 mol B

A limiting reagent in a reaction determines the extent of a reaction.

This is the first to get consumed during the reaction.

• If 1 atom of A reacts with 1 molecule of B. 200 molecules of B will react with 200 atoms of A. Thus, 200 molecules of B will get consumed earlier and 100 atoms of A will remain unused. B is the limiting reagent.
• 1 atom of A combines with 1 molecule of B. 2 moles of A will react with only 2 moles of B. 1 mole of B will not be consumed. Thus, A will be limiting reagent.
• 1 atom of A combines with 1 molecule of B. Thus, all the atoms of A will combine with all 100 molecules of B. Hence, there is no limiting reagent.
• 1 mole of atom A combines with 1 mole of molecule B. 2.5 moles of B will combine with 2.5 moles of A. 2.5 moles of A will be left. Hence, B is a limiting reagent.
• 1 mole of atom A combines with 1 mole of molecule B. 2.5 moles of A will combine with 2.5 moles of B. 2.5 moles of B will be left. Hence, A is the limiting reagent.

Question 1.24

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:

N2 (g) + H2 (g) à 2NH3(g)

• Calculate the mass of ammonia produced if 2.00 x 103 g dinitrogen reacts with 1.00 x 103 g of dihydrogen.
• Will any of the two reactants remain unreacted?
• If yes, which one and what would be its mass?

• Let us first balance the given equation:

N2 (g) + 3H2 (g) -->2NH3(g)

1 mole (28 g) of Nitrogen reacts with 3 moles of dihydrogen to give 2 moles (34 g) of ammonia.

2 x 103 g of dinitrogen will react with (6/28) x 2 x 103 g dihydrogen.

2 x 103 g of dinitrogen will react with 428.6 g of dihydrogen.

Amount of dihydrogen = 1 x 103 g

N2 is the limiting reagent.

Hence, 28 g of Nitrogen produces 34 g of ammonia.

Mass of ammonia produced by 2000 g of nitrogen = (3428) x(2000) = 2428.57 g

• Nitrogen is the limiting reagent and hydrogen is in excess. Hence, some of the hydrogen will remain unreacted.
• Mass of dihydrogen left unreacted = (1 x 103)g – (428.6) g = 571.4 g

Question 1.25

How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?

Molar mass of Na2CO3 = (2 x 23) + 12 + (3 x 16) = 106 g mol-1

1 mole of Na2CO3 means 106 g of Na2CO3

0.5 mole of Na2CO3 = (106 / 1) x 0.5 = 53 g of Na2CO3

0.5 M of Na2CO3 = 0.50 mol/L Na2CO3

Hence, 0.5 mol of Na2CO3 is in 1 Litre of water or 53g of Na2CO3 is in 1L of water.

Question 1.26

If 10 volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water vapour would be produced?

Reaction of dihydrogen with dioxygen is

2H2(g) + O2(g) à 2 H2O (g)

Two volumes of dihydrogen react with one volume of dihydrogen to producetwo volumes of water vapour.

Ten volumes of dihydrogen will react with five volumes of dioxygen to produce ten volumes of water vapour.

Question 1.27

Convert the following into basic units:

• 28.7 pm
• 15.15 pm
• 25365 mg

• 28.7 pm

1 pm = 10-12 m

28.7 pm = 28.7 x 10-12 m

2.87 x 10-11 m

(ii)  15.15 pm

1 pm = 10-12 m

15.15 pm = 15.15 x 10-12 m

= 1.515 x 10-12 m

(iii)25365 mg

1 mg = 10-3 g

25365 mg = 2.5365 x 104 x 10-3 g

1 g = 10-3 kg

2.5365 x 101 g = 2.5365 x 10-1 x 10-3 Kg

25365 mg = 2.5365 x 10-2 kg

Question 1.28

Which one of the following will have the largest number of atoms?

• 1 g Au (s)
• 1 g Na (s)
• 1 g Li (s)
• 1 g of Cl2 (g)

• 1 g of Au

= 1/197 mol of Au

= (6.022 x 1023)/(197 atoms of Au)

= 3.06 x 1021 atoms of Au

• 1 g of Na

= (1/23) mol of Na

= 6.022 x 1023/(23) atoms of Na

= 0.262 x 1023 atoms of Na

= 26.2 x 1021 atoms of Na

• 1 g of Li

= (1/7) mole of Li

= 6.022 x 1023/(7) atoms of Li

= 0.86 x 1023 atoms of Li

• 1 g of Cl2

= (1/71) mole of Cl2 [Molar mass of Cl2 molecule = 35.5 x 2 = 71 g mol-1]

=  6.022 x 1023/(71) atoms of Cl2

= 0.0848 x 1023 atoms of Cl2

= 8.48 x 1021 atoms of Cl2

Hence, 1 g of Li has largest number of atoms.

Question 1.29

Calculate the molarity of a solution of ethanol in water, in which the mole fraction of ethanol is 0.40 (assume the density of water to be one).

Mole fraction of C2H5OH = Number of moles of C2H5OH/ Number of moles of solution

0.040 = n C2H5OH/ (n C2H5OH + n H2O)  ------- (i)

Number of moles present in 1 L water.

nH2O = 1000 g / 18 g mol-1

n H2O = 55.55 mole

Substituting the value of nH2O in equation (1)

(n C2H5OH)/ (n C2H5OH + 55.55)  = 0.040

n C2H5OH = 0.040  n C2H5OH + (0.040 x 55.55)

0.96 n C2H5OH = 2.222 mol

n C2H5OH = 2.222/0.96 mole

n C2H5OH = 2.314 mole

Molarity of solution = (2.314 mole)/(1 L) = 2.314 M

Question 1.30

What will be the mass of one 12C atom in g?

1 mole of C-atoms = 6.023 x 1023 atoms of C

= 12 g of C

Mass of one C-12 atom = 12 g / 6.022 x 1023 = 1.993 x 10-23 g = 1.993 x 10-23 g

Question 1.31

How many significant figures should be present in the answer of the following calculations?

• (0.02856 x 298.15 x 0.112)/(0.5785)
• 5 x 5.365
• 0.0125 + 0.7864 + 0.0215

• Least precise number of calculation = 0.112

Number of significant figures = Number of significant figures in the least precise number = 3

• Least precise number of calculation = 5.364

Number of significant figures = Number of significant figures in 5.364 = 4

• Least number of decimal places in each term = 4

Also, number of significant figure = 4

Question 1.32

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

Isotopes Isotopic molar mass Abundance

36Ar 35.96755 g mol-1 0.337%

38Ar 37.96272 g mol-1 0.063 %

40Ar 39.9624 g mol-1 99.600 %

Molar mass of Argon (Ar)

= [{35.96755 x (0.337/100)}] + [{37.96272 x (0.063/100)}] + [{39.9624 x (90.60/100)}]

= [0.121 + 0.024 + 39.802]

= 39.947 g mol-1

Question 1.33

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.

• 1 mole of Ar = 6.022 x 1023 atoms of Ar

52 mole of Ar = 52 x 6.022 x 1023 of Ar = 3.131 x 1025 atoms of Ar

• 1 atom of He = 4 u of He

1 u of He = ¼ atom of He

52 u of He = 52/4 atoms of He = 13 atoms of He

• 4 g of He = 6.022 x 1023 atoms of He

52 g of He = (6.022 x 1023 x 52)/4 atoms of He = 7.8286 x 1024 atoms of He

Question 1.34

A wielding 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 wielding gas is found to weigh 11.6 g.

Calculate (i) empirical formula (ii) molar mass of the gas and (iii) molecular formula

• 1 mole of CO2 (44 g) contains 12 g of C.

3.38 g of CO2 contains carbon = (12/44) x 3.38 g = 0.9217 g

18 g of water contains 2 g of hydrogen

0.690 g of water contains hydrogen = (2/18) x 0.690 = 0.0767 g

Since, C and H are the constituents of the compounds, total mass of compound = 0.9217 g + 0.0767 g = 0.9984 g

Percent of C in the compound = (0.9217/0.9984) x 100 = 92.32 %

Percent of H in the compound = (0.0767/0.9984) x 100 = 7.68 %

Moles of C in the compound = 92.32/12 = 7.68

Ratio of C to H in the compound = 7.69 : 7.68 = 1:1

Empirical formula comes out to be CH.

• Weigh of 10 L of the gas at S.T.P. = 11.6 g

Weight of 22.4 L of gas at S.T.P. = (11.6 / 10) x 22.4 L = 25.984 g = 26 g (approx)

Molar mass of the gas is 26 g.

• Empirical formula mass of CH = 12 + 1 = 13 g

n = Molar mass of gas/Empirical formula mass of gas = 26/13

n = 2

Molecular formula of gas = (CH)n = C2H2

Question 1.35

Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3 (s) + 2 HCl (aq) -->CaCl2 (aq) + CO2 (g) + H2O (l)

What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

0.75 M of HCl = 0.75 mol of HCl are present in 1 L of water

= [(0.75 mole) x (36.5 g mol-1)] HCl is present in 1 L of water

27.375 g of HCl is present in 1 L of water.

1000 mL of solution contains 27.375 g of HCl

Amount of HCl present in 25 mL of solution = (27.375g/1000 mL) x 25 mL = 0.6844 g

Chemical reaction:

CaCO3 (s) + 2 HCl (aq) --> CaCl2 (aq) + CO2 (g) + H2O (l)

2 moles of HCl (71 g) react with 1 mole of CaCO3 (100 g).

Amount of CaCO3 that will react with 0.6844 g = (100/71) x 0.6844 g = 0.9639 g

Question 1.36

Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction

4 HCl (aq) + MnO2 (s) --> 2H2O (l) + MnCl2 (aq) + Cl2 (g)

How many grams of HCl react with 5.0 g of manganese dioxide?

1 mol MnO2 reacts completely with 4 mole.

Molar Mass MnO2 = [55 + (2 x 16)] = 87 g.

Molar mass of HCl = [(4 x 1) + 36.5] =146 g

5 g of MnO2 will react with HCl.

= (146/87) x 5 of HCl

= 8.4 g of HCl