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Question-1

Calculate the molecular mass of the following :
(i) H2O
(ii) CO2
(iii) CH4

Solution:
i) The molecular mass of H2O =2 × (Atomic mass of H2) + 1 ×(Atomic mass of O2)

Atomic mass of H2 = 1

Atomic mass of O2 = 16

=(2×1) + 1×16

= 18 g mol–1

ii) The molecular mass of CO2 = 1× (Atomic mass of carbon) + 2 ×(Atomic mass of O2)

Atomic mass of carbon = 12

Atomic mass of oxygen =16

= (1×12) + (2×16)

= 12 + 32

= 44g mol–1

iii) Molecular mass of CH4 = 1×(Atomic mass of C) + 4×(Atomic mass of H2)

= (1
×12) + (4×1)

=12 + 4

= 16g mol-1.

Question-2

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

Solution:
Mass of Na2SO4 = 2 × 23 + 32 + 4 × 16 = 46 + 32 + 64 = 142

Mass % of Na = × 100 = 32.39.

Mass % of S= 32/142 * 100 = 22.53

Mass % of oxygen = 64/142 * 100 = 45.07

Question-3

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

Solution:

 

Elements

At mass

% age

M. Ratio

R. Ratio

Simple Ratio

1

Fe

56

69.9

1.0 × 2 = 2

2 O 16 30.1 1.50 × 2= 3

Hence molecular formula is Fe2O3.

Question-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.

Solution:
(i) 1 mole of carbon is burnt in air.

C + O2
CO2

1 mole of carbon reacts with1 mole of dioxygen to form one mole of carbon dioxide

Amount of CO2 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 CO2


16 grams of dioxygen will form

= 22g of CO2

iii) If 1 mole of carbon are burnt in 16 grams of dioxygen it forms

22 grams of CO2


If 2 moles of carbon are burnt it will form =

= 44g of CO2

Question-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.

Solution:
Molar mass of CH3COONa = 82.0245 g mol-1 1000 ml of 1M solution of CH3COONa = 82.0245 of the substance 

500 ml of 0.375 M solution will contain = =15.38 grams.

Question-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%.

Solution:
When %age by mass is given then molarity of solution is given as

Molarity =

Hence, conc. of HNO3(M) = = 15.44M.

Question-7

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

Solution:
Molecular mass of CuSO4 = 1×(Atomic mass of copper) + 1× (Atomic mass of sulphur) + 4×(Atomic mass of oxygen)

=1×63.6 + 1×16 + 4×16

= 95.6 + 64

= 159.6 g mol-1

159.6 grams of CuSO4 gives 63.66 grams of Cu


100 grams of CuSO4 gives = 39.85 % of Cu

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.

Solution:

Elements

At mass

% age

M. Ratio

R. Ratio

Simple Ratio

Fe

56

69.9

1.0 × 2 = 2

O 16 30.1 1.50 × 2= 3

Hence molecular formula is Fe2O3.

Question-9

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.

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 ×6.023 ×1023 = 18.023 ×1023 molecules.

Question-10

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?

Solution:
The molecular formula of sugar is C12H22 O11

Molecular mass of sugar = 12 × carbon + 22 × Hydrogen + 11 × Oxygen

= (12 × 12) + 22(1) + (11 × 16)

= 144 + 22+ 176

= 342

Concentration of sugar = = 0.0292 mol L-1

Question-11

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?

Solution:
Amount of methanol required m(g) =

= = 20g

since, density =

Volume of CH3OH required = = = 25.22 mL.

Question-12

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.

Question-13

 


Solution:
Micro is 10-6; deca is 10; mega is 106; giga is 109 and femto is 10-15.

Question-14

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.

Question-15

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 percent by mass.

(ii) Determine the molality of chloroform in the water sample.

Solution:
(i) 15ppm means 15 parts by mass of CHCL3 in 106 parts by mass of water

Hence, percent by mass = × 100 = 1.5 × 10-3 %

(ii) moles of chloroform = = 0.1255

Hence, molality (m) of CHCL3 = × 103 = 1.25 × 10-4 m.

Question-16

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 × 10-3

(ii) 2.34 × 105

(iii) 8.008 × 103

(iv) 5 × 102

(v) 6.0012

Question-17

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

Question-18

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

Question-19

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

Solution:
Given that the speed of the light is 3 × 108 ms-1 is 2.00ns

2.00 ns = 2 × 10-9 sec.

Distance = speed × time

= 2 × 10-9 ×3
× 108

= 60 m.

Question-20

How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?

Solution:
0.50 mol of Na2Co3 where as 0.50 × 106 = 53 grams of Na2Co3 where as 0.50 MNa2Co3 means 1 litre solution of sodium carbonate contains 0.5 moles of sodium carbonate.

Question-21

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

Solution:
2 H2 + O2 H2O

Given that ten volumes of dihydrogen gas reacts with five volumes of dioxygengas.

Consider that the volume is expressed in litres.

2 H2 + O2 H2O

2L + 1L 1L


10L + 5L 5L

If 10 litres of hydrogen is mixed with 5 litres of oxygen 5 litres of water vapour is produced.

Question-22

Convert the following into basic units:

Solution:
(i) 28.7 pm = 28.7 × 10-12 m

(ii) 15.15 pm = 15.15 × 10-12 m

(iii) 25365 mg = 25.365 × 103 mg

Question-23

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 Cl2(g)

Solution:
Lithium being small atom in the given set will have largest number of atoms.

Question-24

Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040.

Solution:
Xethanol = 0.040; XH2o = 0.960

Molality(m) = = = 2.315m.

Question-25

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

Solution:
12 grams of carbon has 6.023 × 1023 atoms.

1 atom has

= 1.99 × 10-23 gram of carbon.

Question-26

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

Isotope Isotopic molar mass Abundance


36Ar 35.96755g mol–1 0.337%

38Ar 37.96272 g mol–1 0.063%

40Ar 39.9624 g mol–1 99.600%

Solution:

Molar mass of Argon =

=


= =


= 39.95

Question-27

(iii) 52 g of He.

 


Solution:
(i) Number of atoms of Ar = 52 × 6.02 × 1023 = 3.132 × 1025 atoms

(ii) 1 Atom of He has a mass of 4u


Number of atoms of He = = 13 atoms

(iii) Number of moles of He = = 13 moles


Number of atoms of He = 13 × 6.02 ×1023 = 7.826 × 1024 atoms.

Question-28

(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 = C2H2.

Question-29

Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction, CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?

Solution:
Given CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l)
                               100g 2 moles

Moles present in 25mL of 0.75 M HCI =
× 25 = 0.01875 moles

Since 2 moles react completely with CaCO3 = 60g


0.01875 moles react completely with CaCO3 = × 0.01875 = 0.5625g.

Question-30

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

Solution:
4 HCl (aq) + MnO2(s) 2H2O (l) + MnCl2(aq) + Cl2(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

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.





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