Question1
Calculate the molecular mass of the following :
(i) H2O
(ii) CO2
(iii) CH_{4}
Solution:
i) The molecular mass of H_{2}O =2 Ã— (Atomic mass of H_{2}) + 1 Ã—(Atomic mass of O_{2})
Atomic mass of H_{2} = 1
Atomic mass of O_{2 }=_{ }16
=(2Ã—1) + 1Ã—16
= 18 g mol^{â€“1}
ii) The molecular mass of CO_{2 }= 1Ã— (Atomic mass of carbon) + 2 Ã—(Atomic mass of O_{2})
Atomic mass of carbon = 12
Atomic mass of oxygen =16
= (1Ã—12) + (2Ã—16)
= 12 + 32
= 44g mol^{â€“1}
iii) Molecular mass of CH_{4 }=_{ }1Ã—(Atomic mass of C) + 4Ã—(Atomic mass of H_{2})
= (1 Ã—12) + (4Ã—1)
=12 + 4
= 16g mol^{1}.
Question2
Calculate the mass per cent of different elements present in sodium sulphate (Na_{2}SO_{4}).
Solution:
Mass of Na_{2}SO_{4} = 2 Ã— 23 + 32 + 4 Ã— 16 = 46 + 32 + 64 = 142
Mass % of Na = Ã— 100 = 32.39.
Solution:
Mass of Na_{2}SO_{4} = 2 Ã— 23 + 32 + 4 Ã— 16 = 46 + 32 + 64 = 142
Mass % of Na = Ã— 100 = 32.39.
Mass % of S= 32/142 * 100 = 22.53
Mass % of oxygen = 64/142 * 100 = 45.07
Question3
Determine the empricial formula of an oxide of iron which has 69.9% iron and
30.1% dioxygen by mass.
Solution:
30.1% dioxygen by mass.
Solution:

Elements 
At mass 
% age 
M. Ratio 
R. Ratio 
Simple Ratio 
1 
Fe 
56 
69.9 
1.0 Ã— 2 = 2 

2  O  16  30.1  1.50 Ã— 2= 3 
Hence molecular formula is Fe_{2}O_{3}.
Question4
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.
Solution:
(i) 1 mole of carbon is burnt in air.
C + O_{2} â†’_{ }CO_{2}
1 mole of carbon reacts with1 mole of dioxygen to form one mole of carbon dioxide
Amount of CO_{2} produced = 44 g
ii) 1 mole of carbon is burnt in 16 grams of dioxygen.
1 mole of carbon burnt in 32 grams of dioxygen it forms 44 grams of CO_{2}
âˆ´16 grams of dioxygen will form
= 22g of CO_{2}
iii) If 1 mole of carbon are burnt in 16 grams of dioxygen it forms
22 grams of CO_{2}
âˆ´If 2 moles of carbon are burnt it will form =
= 44g of CO_{2}
(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.
Solution:
(i) 1 mole of carbon is burnt in air.
C + O_{2} â†’_{ }CO_{2}
1 mole of carbon reacts with1 mole of dioxygen to form one mole of carbon dioxide
Amount of CO_{2} produced = 44 g
ii) 1 mole of carbon is burnt in 16 grams of dioxygen.
1 mole of carbon burnt in 32 grams of dioxygen it forms 44 grams of CO_{2}
âˆ´16 grams of dioxygen will form
= 22g of CO_{2}
iii) If 1 mole of carbon are burnt in 16 grams of dioxygen it forms
22 grams of CO_{2}
âˆ´If 2 moles of carbon are burnt it will form =
= 44g of CO_{2}
Question5
Calculate the mass of sodium acetate (CH_{3}COONa) required to make 500 mL of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^{â€“1}.
Solution:
Molar mass of CH_{3}COONa = 82.0245 g mol^{1} 1000 ml of 1M solution of CH_{3}COONa = 82.0245 of the substance
âˆ´500 ml of 0.375 M solution will contain = =15.38 grams.
Solution:
Molar mass of CH_{3}COONa = 82.0245 g mol^{1} 1000 ml of 1M solution of CH_{3}COONa = 82.0245 of the substance
âˆ´500 ml of 0.375 M solution will contain = =15.38 grams.
Question6
Calculate the concentration of nitric acid in moles per litre in a sample which
has a density, 1.41 g mL^{â€“1} and the mass per cent of nitric acid in it being 69%.
Solution:
When %age by mass is given then molarity of solution is given as
Molarity =
Hence, conc. of HNO_{3}(M) = = 15.44M.
has a density, 1.41 g mL^{â€“1} and the mass per cent of nitric acid in it being 69%.
Solution:
When %age by mass is given then molarity of solution is given as
Molarity =
Hence, conc. of HNO_{3}(M) = = 15.44M.
Question7
How much copper can be obtained from 100 g of copper sulphate (CuSO_{4})?
Solution:
Molecular mass of CuSO_{4 }=_{ }1Ã—(Atomic mass of copper) + 1Ã— (Atomic mass of sulphur) + 4Ã—(Atomic mass of oxygen)
=1Ã—63.6 + 1Ã—16 + 4Ã—16
= 95.6 + 64
= 159.6 g mol^{1}
159.6 grams of CuSO_{4 }gives 63.66 grams of Cu
âˆ´ 100 grams of CuSO_{4 }gives = 39.85 % of Cu
Solution:
Molecular mass of CuSO_{4 }=_{ }1Ã—(Atomic mass of copper) + 1Ã— (Atomic mass of sulphur) + 4Ã—(Atomic mass of oxygen)
=1Ã—63.6 + 1Ã—16 + 4Ã—16
= 95.6 + 64
= 159.6 g mol^{1}
159.6 grams of CuSO_{4 }gives 63.66 grams of Cu
âˆ´ 100 grams of CuSO_{4 }gives = 39.85 % of Cu
Question8
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.
Solution:
Solution:
Elements 
At mass 
% age 
M. Ratio 
R. Ratio 
Simple Ratio 
Fe 
56 
69.9 
1.0 Ã— 2 = 2 

O  16  30.1  1.50 Ã— 2= 3 
Hence molecular formula is Fe_{2}O_{3}.
Question9
In three moles of ethane (C_{2}H_{6}), calculate the following:
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Solution:
(i) Number of moles of carbon atoms.
No: of moles of carbon atoms = 6
(ii) Number of moles of hydrogen atoms.
No: of moles of hydrogen atoms = 18
(iii) Number of molecules of ethane.
No: of molecules of ethane = 3 Ã—6.023 Ã—10^{23 }=^{ }18.023 Ã—10^{23} molecules.
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Solution:
(i) Number of moles of carbon atoms.
No: of moles of carbon atoms = 6
(ii) Number of moles of hydrogen atoms.
No: of moles of hydrogen atoms = 18
(iii) Number of molecules of ethane.
No: of molecules of ethane = 3 Ã—6.023 Ã—10^{23 }=^{ }18.023 Ã—10^{23} molecules.
Question10
What is the concentration of sugar (C_{12}H_{22}O_{11}) in mol L^{â€“1} if its 20 g are dissolved in enough water to make a final volume up to 2L?
Solution:
The molecular formula of sugar is C_{12}H_{22} O_{11}
Molecular mass of sugar = 12 Ã— carbon + 22 Ã— Hydrogen + 11 Ã— Oxygen
= (12 Ã— 12) + 22(1) + (11 Ã— 16)
= 144 + 22+ 176
= 342
Concentration of sugar = = 0.0292 mol L^{1}
Solution:
The molecular formula of sugar is C_{12}H_{22} O_{11}
Molecular mass of sugar = 12 Ã— carbon + 22 Ã— Hydrogen + 11 Ã— Oxygen
= (12 Ã— 12) + 22(1) + (11 Ã— 16)
= 144 + 22+ 176
= 342
Concentration of sugar = = 0.0292 mol L^{1}
Question11
If the density of methanol is 0.793 kg L^{â€“1}, what is its volume needed for making 2.5 L of its 0.25 M solution?
Solution:
Amount of methanol required m(g) =
= = 20g
since, density =
âˆ´ Volume of CH_{3}OH required = = = 25.22 mL.
Solution:
Amount of methanol required m(g) =
= = 20g
since, density =
âˆ´ Volume of CH_{3}OH required = = = 25.22 mL.
Question12
What is the SI unit of mass? How is it defined?
Solution:
The S.I unit of mass is kilogram.
Definition
One kilogram is equal to the mass of a platinum iridium cylinder kept at the international bureau of weights and measures near paris and nearly equal to 100 Cubic cms of water at the temperature of its maximum density.
Solution:
The S.I unit of mass is kilogram.
Definition
One kilogram is equal to the mass of a platinum iridium cylinder kept at the international bureau of weights and measures near paris and nearly equal to 100 Cubic cms of water at the temperature of its maximum density.
Question13
Solution:
Micro is 10^{6}; deca is 10; mega is 10^{6}; giga is 10^{9} and femto is 10^{15}.
Question14
What do you mean by significant figures?
Solution:
Significant figures also called significant digits is a method of expressing Errors in measurements. They begins with the first digit to the left that is not zero and end with the last digit to the right that is corrected. For example 0.002540 has 4 significant figures.
Solution:
Significant figures also called significant digits is a method of expressing Errors in measurements. They begins with the first digit to the left that is not zero and end with the last digit to the right that is corrected. For example 0.002540 has 4 significant figures.
Question15
A sample of drinking water was found to be severely contaminated with chloroform, CHCl_{3}, supposed to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.
Solution:
(i) 15ppm means 15 parts by mass of CHCL_{3 }in 10^{6} parts by mass of water
Hence, percent by mass = Ã— 100 = 1.5 Ã— 10^{3} %
(ii) moles of chloroform = = 0.1255
Hence, molality (m) of CHCL_{3 = }Ã—_{ }10^{3} = 1.25 Ã—_{ }10^{4} m.
(i) Express this in percent by mass.
(ii) Determine the molality of chloroform in the water sample.
Solution:
(i) 15ppm means 15 parts by mass of CHCL_{3 }in 10^{6} parts by mass of water
Hence, percent by mass = Ã— 100 = 1.5 Ã— 10^{3} %
(ii) moles of chloroform = = 0.1255
Hence, molality (m) of CHCL_{3 = }Ã—_{ }10^{3} = 1.25 Ã—_{ }10^{4} m.
Question16
Express the following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Solution:
(i) 4.8 Ã— 10^{3}
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
Solution:
(i) 4.8 Ã— 10^{3}
(ii) 2.34 Ã— 10^{5}
(iii) 8.008 Ã— 10^{3}
(iv) 5 Ã— 10^{2}
(v) 6.0012
Question17
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
Solution:
(i) 0.0025  2
(ii) 208  3
(iii) 5005  4
(iv) 126,000  6
(v) 500.0  4
(vi) 2.0034  5
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034
Solution:
(i) 0.0025  2
(ii) 208  3
(iii) 5005  4
(iv) 126,000  6
(v) 500.0  4
(vi) 2.0034  5
Question18
Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Solution:
(i) 34.216 = 34.2
(ii) 10.4107 = 10.4
(iii) 0.04597 = 0.0460
(iv) 2808 = 281
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Solution:
(i) 34.216 = 34.2
(ii) 10.4107 = 10.4
(iii) 0.04597 = 0.0460
(iv) 2808 = 281
Question19
If the speed of light is 3.0 Ã— 10^{8} ms^{â€“1}, calculate the distance covered by light in 2.00 ns.
Solution:
Given that the speed of the light is 3 Ã— 10^{8} ms^{1} is 2.00ns
2.00 ns = 2 Ã— 10^{9} sec.
Distance = speed Ã— time
= 2 Ã— 10^{9} Ã—3 Ã— 10^{8}
= 60 m.
Solution:
Given that the speed of the light is 3 Ã— 10^{8} ms^{1} is 2.00ns
2.00 ns = 2 Ã— 10^{9} sec.
Distance = speed Ã— time
= 2 Ã— 10^{9} Ã—3 Ã— 10^{8}
= 60 m.
Question20
How are 0.50 mol Na_{2}CO_{3} and 0.50 M Na_{2}CO_{3} different?
Solution:
0.50 mol of Na_{2}Co_{3} where as 0.50 Ã— 10^{6} = 53 grams of Na_{2}Co_{3 }where as 0.50 MNa_{2}Co_{3 }means 1 litre solution of sodium carbonate contains 0.5 moles of sodium carbonate.
Solution:
0.50 mol of Na_{2}Co_{3} where as 0.50 Ã— 10^{6} = 53 grams of Na_{2}Co_{3 }where as 0.50 MNa_{2}Co_{3 }means 1 litre solution of sodium carbonate contains 0.5 moles of sodium carbonate.
Question21
If ten volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of water vapour would be produced?
Solution:
2 H_{2} + O_{2} H_{2}O
Given that ten volumes of dihydrogen gas reacts with five volumes of dioxygengas.
Consider that the volume is expressed in litres.
2 H_{2} + O_{2} H_{2}O
2L + 1L 1L
âˆ´10L + 5L 5L
If 10 litres of hydrogen is mixed with 5 litres of oxygen 5 litres of water vapour is produced.
Solution:
2 H_{2} + O_{2} H_{2}O
Given that ten volumes of dihydrogen gas reacts with five volumes of dioxygengas.
Consider that the volume is expressed in litres.
2 H_{2} + O_{2} H_{2}O
2L + 1L 1L
âˆ´10L + 5L 5L
If 10 litres of hydrogen is mixed with 5 litres of oxygen 5 litres of water vapour is produced.
Question22
Convert the following into basic units:
Solution:
(i) 28.7 pm = 28.7 Ã— 10^{12} m
(ii) 15.15 pm = 15.15 Ã— 10^{12} m
(iii) 25365 mg = 25.365 Ã— 10^{3} mg
Solution:
(i) 28.7 pm = 28.7 Ã— 10^{12} m
(ii) 15.15 pm = 15.15 Ã— 10^{12} m
(iii) 25365 mg = 25.365 Ã— 10^{3} mg
Question23
Which one of the following will have largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl_{2}(g)
Solution:
Lithium being small atom in the given set will have largest number of atoms.
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl_{2}(g)
Solution:
Lithium being small atom in the given set will have largest number of atoms.
Question24
Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040.
Solution:
X_{ethanol} = 0.040; X_{H2o} = 0.960
Molality(m) = = = 2.315m.
Solution:
X_{ethanol} = 0.040; X_{H2o} = 0.960
Molality(m) = = = 2.315m.
Question25
What will be the mass of one 12C atom in g?
Solution:
12 grams of carbon has 6.023 Ã— 10^{23} atoms.
âˆ´ 1 atom has
= 1.99 Ã— 10^{23} gram of carbon.
Solution:
12 grams of carbon has 6.023 Ã— 10^{23} atoms.
âˆ´ 1 atom has
= 1.99 Ã— 10^{23} gram of carbon.
Question26
Use the data given in the following table to calculate the molar mass of naturally occuring argon isotopes:
Isotope Isotopic molar mass Abundance
^{36}Ar 35.96755g mol^{â€“1} 0.337%
^{38}Ar 37.96272 g mol^{â€“1} 0.063%
^{40}Ar 39.9624 g mol^{â€“1} 99.600%
Solution:
Molar mass of Argon =
=
= =
= 39.95
Isotope Isotopic molar mass Abundance
^{36}Ar 35.96755g mol^{â€“1} 0.337%
^{38}Ar 37.96272 g mol^{â€“1} 0.063%
^{40}Ar 39.9624 g mol^{â€“1} 99.600%
Solution:
Molar mass of Argon =
=
= =
= 39.95
Question27
(iii) 52 g of He.
Solution:
(i) Number of atoms of Ar = 52 Ã— 6.02 Ã— 10^{23} = 3.132 Ã— 10^{25} atoms
(ii) 1 Atom of He has a mass of 4u
âˆ´ Number of atoms of He = = 13 atoms
(iii) Number of moles of He = = 13 moles
âˆ´ Number of atoms of He = 13 Ã— 6.02 Ã—10^{23} = 7.826 Ã— 10^{24} atoms.
Solution:
(i) Number of atoms of Ar = 52 Ã— 6.02 Ã— 10^{23} = 3.132 Ã— 10^{25} atoms
(ii) 1 Atom of He has a mass of 4u
âˆ´ Number of atoms of He = = 13 atoms
(iii) Number of moles of He = = 13 moles
âˆ´ Number of atoms of He = 13 Ã— 6.02 Ã—10^{23} = 7.826 Ã— 10^{24} atoms.
Question28
(ii) molar mass of the gas, and
(iii) molecular formula.
Solution:
Mass of carbon in 3.38g carbon dioxide = = 0.922 g
Mass of hydrogen in 0.690g of water = = 0.077g
The ratio by mass of C and H in the sample = 0.922:0.077
The mole ratio of C and H in the sample =
Thus the empirical formula of the gas is CH.
Calculation of Molar Mass:
10.0 L of gas at S.T.P. weighs =11.6g
22.4 L of gas at S.T.P. weighs = =25.98g
calculation of molecular formula
n = = = 1.998 =2
Thus molecular formula is (CH)_{2} = C_{2}H_{2}.
(iii) molecular formula.
Solution:
Mass of carbon in 3.38g carbon dioxide = = 0.922 g
Mass of hydrogen in 0.690g of water = = 0.077g
The ratio by mass of C and H in the sample = 0.922:0.077
The mole ratio of C and H in the sample =
Thus the empirical formula of the gas is CH.
Calculation of Molar Mass:
10.0 L of gas at S.T.P. weighs =11.6g
22.4 L of gas at S.T.P. weighs = =25.98g
calculation of molecular formula
n = = = 1.998 =2
Thus molecular formula is (CH)_{2} = C_{2}H_{2}.
Question29
Calcium carbonate reacts with aqueous HCl to give CaCl_{2} and CO_{2} according to the reaction, CaCO_{3}(s) + 2HCl(aq) â†’ CaCl_{2}(aq) + CO_{2}(g) + H_{2}O(l) What mass of CaCO_{3} is required to react completely with 25 mL of 0.75 M HCl?
Solution:
Given CaCO_{3}(s) + 2HCl(aq) â†’ CaCl_{2}(aq) + CO_{2}(g) + H_{2}O(l)
100g 2 moles
Moles present in 25mL of 0.75 M HCI = Ã— 25 = 0.01875 moles
Since 2 moles react completely with CaCO_{3 }= 60g
âˆ´ 0.01875 moles react completely with CaCO_{3 }= Ã— 0.01875 = 0.5625g.
Solution:
Given CaCO_{3}(s) + 2HCl(aq) â†’ CaCl_{2}(aq) + CO_{2}(g) + H_{2}O(l)
100g 2 moles
Moles present in 25mL of 0.75 M HCI = Ã— 25 = 0.01875 moles
Since 2 moles react completely with CaCO_{3 }= 60g
âˆ´ 0.01875 moles react completely with CaCO_{3 }= Ã— 0.01875 = 0.5625g.
Question30
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO_{2}) with aqueous hydrochloric acid according to the reaction
Solution:
4 HCl (aq) + MnO_{2}(s) â†’ 2H_{2}O (l) + MnCl_{2}(aq) + Cl_{2}(g) . How many grams of HCl react with 5.0 g of manganese dioxide? (At. mass of Mn = 55)
Solution:
4 HCl (aq) + MnO_{2}(s) â†’ 2H_{2}O (l) + MnCl_{2}(aq) + Cl_{2}(g) . How many grams of HCl react with 5.0 g of manganese dioxide? (At. mass of Mn = 55)
No. of moles of MnO2 = = 0.057
One mole of MnO2 reacts with 4 moles of HCl
âˆ´ 0.057 moles of MnO2 will react with 0.057 X 4 = 0.228 moles
Weight of HCl in grams = 0.228 X 36.5 = 8.322 g.