Some Basic Concepts Of Chemistry - shikshahouse...2019/01/01  · Some Basic Concepts Of Chemistry 1) Calculate the molecular mass of the following: i) H 2O ii) CO 2 iii) CH 4 Solution-

Some Basic Concepts Of Chemistry

1) Calculate the molecular mass of the following:

i) H2O ii) CO2 iii) CH4

Solution- The molecular mass of a compound is the sum of the atomic masses of the atoms

present in the compound.

(Atomic mass of H, C and O are 1, 12 and 16 amu respectively)

Molecular mass of H2O = 2 x atomic mass of H + 1 x atomic mass of O

= 2 x 1 + 1 x 16 = 18 amu

Molecular mass of CO2 = 1 x atomic mass of C+ 2 x atomic mass of O

=1 x 12 + 2 x 16 = 44 amu

Molecular mass of CH4 = 1x atomic mass of C + 4 x atomic mass of H

=1 x 12 + 4 x 1= 16amu

2) Calculate the mass percent of different elements present in sodium sulphate ( Na2

SO4 ).

Solution- Mass percent of element = ( Mass of element/ Mass of compound) x 100

Now,

2 x atomic mass of Na + 1 x atomic mass of S + 4 x atomic mass of O

Molecular mass of Na2 SO4 = 2x 23 +32 +4 +16=142

Therefore, Mass % of sodium = (Mass of Na/ Molar mass of Na2SO4 ) x 100

= 46/142 x 100= 32.39 %

Mass % of sulphur = (Mass of S/ Molar mass of Na2 SO4) x100

=32/142 x 100=22.54 %

Mass % of Oxygen = (Mass of O/ Molar mass of Na2 SO4) x 100

= 64/142 x 100

= 45.07 %

3) Determine the empirical formula of an oxide of iron which has 69.9% iron and

30.1% dioxygen by mass.

Solution- Step 1

Calculation of simplest whole number ratio of elements

Element Percentag

e

Atomic

mass

Atomic

Ratio

Simplest

Ratio

Simplest

whole

number

ratio

Fe 69.9 56 69.9/56=

1.25

1.25/1.25

=1

2

O 30.1 16 30.1/16=

1.88

1.88/1.25

=1.5

3

The simplest whole number ratio of elements are

Fe : O = 2:3

Step 2

Writing the empirical formula of the compound.

The empirical formula of the compound= Fe2 O3.

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:

The chemical equation for the combustion of carbon in oxygen is

C + O2 → CO2

1 mole 1 mole 1 mole

12g 32g 44g

i)When 1 mole of carbon is burnt in air

1 mole of carbon will form 1 mole of CO2 = 44 g of CO2

∴ when 1 mole of carbon is burnt in air 44 g of CO2 is formed.

ii) When 1 mole of carbon is burnt in 16g of oxygen

For 1 mole of carbon, oxygen required= 32g

But mass of oxygen available = 16g= ½ mole

∴ Mass of CO2 formed =1/2 mole

=22g

∴ When 1 mole of carbon is burnt in 16 g of oxygen 22g of CO2 is formed.

iii) When 2 moles of carbon are burnt in 16 g oxygen

For 2 moles of carbon, oxygen required = 64g = 2mole

But the mass of oxygen available = 16g =1/2 mole

Now,

1 mole of dioxygen gives 44g (1 mole) of CO2

∴ ½ mole of dioxygen given 22g (1/2 mole) of CO2

So, Mass of CO2 formed= ½ mole= 22g

∴ When 2 moles of carbon are burnt in 16 g of oxygen 22 g of CO2 is formed.

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:

Mole mass of sodium acetate (CH3COONa) = 82 g mol-1

Molarity of solution = 0.375 M

= 0.375 mol L-1

Volume of solution = 500 mL= 500/ 1000 = 0.5 L

Now

Molarity = Mass of sodium acetate/ Molar mass of sodium acetate x 1/volume of

solution (in L)

Let mass of sodium acetate =W

0.375= W/82 x 1/0.5

∴ W = 0.375 x 82 x 0.5 (mol L-1

x gmol-1

x L)

W= 0.375 x 82 x 0.5 (mol-1

x gmol-1

x L)

= 15.375 g

Hence, the mass of sodium acetate (CH3COONa) required to make 500 mL of its 0.375

molar aqueous solution is 15.375 g.

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:

Mass percent 69 means 69 g of nitric acid (HNO3) in 100g of solution

Mass of solution= 100 g

Density of solution = 1.41 g mL-1

Mass of HNO 3 = 69 g

Volume of solution = mass of solution/ Density of solution

= 100/ 1.41 gmL-1

= 70.92 mL

Concentration of HNO3 in moles per litre means Molarity

Molarity (M)= (mass of HNO3 / molar mass of HNO3 ) x 1000/ Volume of solution (mL)

= 69/63 x 1000/70.92

= 15.4 mol L-1

Molarity = 15.4 M

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

Solution:

Molecular mass of CuSO4 =

1 x atomic mass of Cu + 1 x atomic mass of S+4 x atomic mass of O

= 63.5 + 32 + 4 x 16

= 159.5 amu

Molar mass of CuSO4 = 159.5 g

Now,

159.5 g CuSO4 contain 63.5 g Cu

∴ 100 g CuSO4 contain 63.5 x 100/ 159.5 Cu

= 39.81 g

Hence, mass of Cu in 100 g of CuSO4 is 39.81 g.

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:

The simplest whole number ratio of elements are

Fe:O= 2:3

So, the empirical formula of compound = Fe2 o3

Emperical formula mass of Fe2 o3= 2x 56 + 3x 16=160

Let the molecular formula of the compound be (Fe2 o3) n.

Here ‘n’ is a whole number factor

And n= Emperical formula mass/ Molecular formula mass

=160/159.8

≅ 1

Emperical formula of compound = (Fe2 o3) n

=(Fe2 o3)1

=(Fe2 o3)

Molecular formula = Emperical formula = (Fe2 o3)

Element Percen

tage

Atomic

mass

Atomic

Ratio

Simplest

Ratio

Simplest

whole

number ratio

Fe 69.9 56 69.9/56

=1.25

1.25/1.25

=1

2

O 30.1 16 30.1/16

=1.88

1.88/1.25

=1.5

3

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

Solution:

Average atomic mass of an element = Sum of (isotopic mass x % abundance)/100

∴ Average atomic mass of chlorine = 34.9689 x 75.77 + 36.9659 x 24.23/100

=2649.59 + 895.68/100

=35.453 amu

∴ Average atomic mass of chlorine = 35.453 amu

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.

Solution:

i) 1 Molecule of ethane has 2 atoms of carbon.

1 mole of C2H6 has 2 moles of carbon

∴ 3 moles of C2H6 have 2 x 3 moles of carbon

=6 moles

∴ 3 moles of C2H6 has 6 moles of carbon.

ii) 1 molecule of C2H6 has 6 atoms of hydrogen

1 mole of C2H6 has 6 atoms of hydrogen

∴ 3 moles of C2H6 has 6 x 3 moles of hydrogen atoms = 18 noles

∴ Number of moles of hydrogen atom in 3 moles of C2H6 is 18.

iii) 1 mole of C2H6 = 6.022 x 1023

molecule

∴ 3 moles of C2H6 have molecules = 3 x 6.022 x 1023

= 1.8.66 x 1024

molecules

∴ Number of molecule in 3 moles of ethane = 1.8066 x 1024

.

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?

Solution:

Concentration in mol L-1

means Molarity(M)

Given

Mass of sugar = 20 g

Molar mass of sugar (C12H22O11)= 12 x 12 + 22 x 1 + 11 x 16

= 342 mol L-1

Volume of solution = 2L

Molarity = Mass of sugar/ Molar mass of sugar x 1/ volume of solution (L)

= 20/342 x ½

=0.029 M

Molarity of solution – 0.029 M.

Molarity of solution = 0.029 M

Molarity of solution = Concentration of solution

∴ The concentration of the solution is 0.029 M.

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?

Solution:

Step (i)

Calculation of mass of methanol (CH3OH)

Molar mass of methanol (CH3OH)= 12+ 4 x 1+16

= 32 g mol-1

Molarity of solution = 0.25 M

= 0.25 mol L-1

Volume of solution = 2.5 L

Now,

Molarity = Mass of ethanol/Molar mass x 1/volume of solution

Let us consider the mass of methanol as W g

0.25= W/32 x ½2.5 L

∴ W = (0.25 mol L

-1) (32 g mol

-1 ) (2.5 L )

= 20 g

∴ W = Molar mass of methanol = 20 g.

Step (ii)

Calcualtion of volume of amthanol

Mass of methanol=20 g

= 0.02 kg

Density of methanol = 0.793 kg L-1

Volume of methanol = Mass/ Density

= 0.02 kg/ 0.793 kg L-1

= 0.02522 L

=2.522 X 10-2

L

∴ Volume of methanol = 2.522 x 10-2

L

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 N m-1

If the mass of air at sea level is 1034 g cm-2

, calculate the pressure in pascal.

Solution:

Pressure = force/area= Mass x Acceleration due to gravity/ Area

Given

Mass= 1034 g = 1.034 kg

Acceleration due to gravity (g) = 9.8 ms-2

Area = 1 cm2 = 10

-4 m

2

So

Pressure = 1.034 kg x 9.8 ms-2

/ 1 x 10-4

m2

= 1.01 x 105 kg m

-1s

-2

= 1.01 x 10

5N m

-2

= 1.01 x 105 Pa (1 N m

-2= 1 Pa)

Hence, the pressure exerted by air at sea level in pascal= 1.01 x 105 P

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

Solution:

SI Unit of mass is kilogram (kg).

It is defined as the mass equal to the international prototype of the kilogram.

It is the mass of the platinum block stored as the International Bureau of Weights and

Measurements in France.

15) Match the following prefixes with their multiples:

Prefixes Multiples

i) micro 106

ii) deca 109

iii)mega 10-6

iv)giga 10-15

v)femto 10

Solution:

Prefixes Multiples

micro 10-6

deca 10

mega 106

giga 109

femto 10-15

16) What do you mean by significant figures?

Solution:

Significant figures may be defined as follows:

i) The significant figure in a number is the sum of certain digits and one uncertain

(doubtful) digit.

ii) The total number of digits given in a properly recorded measurement are known as

significant figures.

iii) Significant figures are the meaningful digits in a properly measured or calculated

quantity.

Example: In the data 15.0, 15.00, 15.000, the significant figures are 3, 4, and 5,

respectively. The number of certain figures is 2,3 and 4, respectively.

Each data has one uncertain (doubtful ) figure.

Significant figures = certain figure + one doubtful figure.

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 percent by mass.

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

Solution:

Calculation of mass percent 15 ppm level of contamination means

15 g of chloroform (CHCl3)in 106 g of the sample.

Mass percent= Mass of chloroform/ Mass of sample

= 15 g/ 106 g

= 15 x 10-4 %

Calculation of molarity of the solution

Mass of chloroform = 15 g

Molar mass of chloroform = 12+1+3 x 35.5

= 119.5 gmol-1

Mass of sample (solution) = 106 g

Mass of solvent = (106 -15)g

=106 g (15g<<10

6 g)

Now,

Molarity = Mass of chloroform/ Molar mass of chloroform x 1000/ mass of solvent(g)

=15/119.5 x 1000/106 mol kg

-1

= 15/119.5 x10-3

mol kg-1

=1.255 x 10-4

mol kg-1

∴ Molarity of chloroform in water solution = 1.255 x 10-4

mol kg-1

18) Express the following in the scientific notation:

i) 0.0048

ii) 234,000

iii) 8008

iv) 500.0

v) 6.0012

Solution:

In scientific notation, all numbers are expressed as a number between 1,000 and 9,999

multiplies by 10n or 10

-n.

Hence, the scientific notation for

0.0048 = 4.8 x 10-3

234000= 2.34 x 105

8008 = 8.008 x 103

500.0 = 5.000 x 102

6.0012= 6.0012 x 100

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

Solution:

Number of significant figures in

i) 0.0025 =2

It is because all on-zero digits are only significant.

ii) 208 =3

It is because the entire zero between two non-zero digit are significant.

iii) 5005 =4

It is because all the zero between two non- zero digit are significant.

iv) 126, 000=6

It is because all the zeros at the end or the right of a number re significant.

v) 500.0=4

It is because all the zeros at the right of a number and the zero after decimal are significant.

vi) 2.0034=5

It is because all the zeros between two non-zero are significant.

20) Round up the following upto three significant figures:

i) 34.215

ii) 10.4107

iii) 0.04597

iv) 2808

Solution:

i) 34.216 is rounded off as 34.2

It is because the 4th digit is less than 5.

ii) 10.417 is rounded off as 10.4

It is because the 4th digit is less than 5.

iii) 0.04597 is rounded off as 0.046

It is because the 4th digit is greater than 5 , so the 3

rd digit is converted to next higher one.

iv) 280.8 is rounded off as 281.

It is because the 4th digit is greater than 5, so it is rounded figured to 281.

21) The following data are obtained when dinitrogen and dioxygen react together to

form different compounds:

Mass of dinitrogen Mass of dioxygen

i) 14g 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.

Solution:

It is the mass of dinitrogen is fixed at 28 g, the mass of dioxygen combined will be 32 g,

64g, 32 g and 80 g, respectively. These values are in the ratio of 2:4:2:5. It is a simple

whole number ratio. Thus, the given data obeys the law of multiple proportions.

Statement of the law of multiple proportions

When two element combine to form two or more compound, then the different masses of

one element, which combine with the same mass of the other element, are in simple ration

of whole numbers.

Example: CO and CO2

In CO , C:O = 12:16

In CO2, C:O = 12:32

Therefore, the ratio of the masses of oxygen, which combines with 12 g, of carbon is 16:

32 or 1:2. It is the simple whole number ratio.

b) Fill in the blanks in the following conversions:

i) 1km = _________mm = ________pm

ii) 1mg = _______ kg = _________ng

iii) 1mL= _______L = _________dm3

Solution:

1) 1 km = 103 m

1 m = 103 mm

∴ 1 m =103 x 10

3 mm

1 km = 106 mm

1 km = 103 m

1 m = 1012

pm

∴ 1 m = 106 mm = 10

15 pm.

2) 1 mg = 10-3

g

1 g = 109 ng

∴ 1 mg = 10-3 x

x 109 ng

1 mg = 106 ng

1 ng= 10- 9 g

1 g =10-3

kg

1mg =10-3

x 10-3

kg

∴ 1 mg = 106 ng = 10

-6 kg

3) 1 mL = 10-3

L

1 mL = 1 cm3

1 cm3 = 10

-3 dm

3

∴ 1 mL = 10-3

L = 10-3

dm3

22) If the speed of light is 3.0 x 10 8 ms

-1, calculate the distance covered by light in

2.00 ns.

Solution:

Speed = 3.0 x 10 8 ms

-1

Time = 2.00 ns = 2.00 x 10-9

s

Distance = speed x time

=3.0 x 10 8 ms

-1 x 2.00 x 10

-9 s

= 6.00 x 10-1

m

= 0.600 m

Hence, the distance covered by light in 2.00 ns is 0.6 m.

23) In a reaction

A + B2 → A B2

Identify the limiting reagent, if any, in the following reaction mixtures.

i) 300 atoms of A + 200 molecules of B

ii) 2 mol A + 3 mol B

iii) 100 atoms of A + 100 molecules of B

iv) 5 mol A + 2.5 mol B

v) 2.5 mol A + 5 mol B

Solution:

A + B2 → A B2

1 Atom 1 molecule 1 Molecule

1 Mol 1 Mol (2 mole atoms) 1 mol

In the light of the above information, let us find the limiting reactants, if any ;-

I)

1 atom of A will react with 1 molecule of B2

∴ 300 atoms of A will react with 300 molecules of B2

But molecules of B2 actually given = 200

∴ B2 is a limiting reactant.

II)

2 mol A + 3 mol B

1 mol of A reacts with 1 mol of B

2 moles of A react with 2 mol of B

But actually mol of B = 3

∴ A is a limiting reactant in this case.

III) 1 atom of A will react with 1 molecule of B2

100 atoms of A will react with 100 molecules of B2

But molecules of B2 given = 100

IV) 1 mol of A will react with 1 mol of B

5 moles of A will react with 5 moles of B

But moles of B actually available = 2.5

∴ B is limiting reactant in this case.

V)

1 mole of A will react with 1 mol of B

2.5 moles of A will react with 2.5 moles of B.

But moles of B actually available = 5

Hence, A is the limiting reactant in this case.

24) Dinitrigen and dihydrogen react with each other to produce ammonia according to

the following chemical equation:

N2 (g) + 3H2 (g) → 2NH3

28g 3x 23 = 6g 2x 17 =34g

i) 28 g N2 produces 34 g of ammonia

∴ 2000 g N2 produces 34 x 2000/28 g of ammonia

= 2428.57 g

Again,

6 g H2 produces 34g ammonia

∴ 1000 g N2 produces 34X1000/6 g of ammonia= 5666.66 g

Since ammonia obtained from N2 is less in quantity, therefore. ammonia produced =

2428.57 g

ii) Ammonia obtained from N2 is less, hence N2 is the limiting reagent.

Some H2 will remain unreacted

iii) 28 g N2 reacts with 6 g of H2

∴ 2000 g N2 reacts with 6x2000/28 g of H2 = 428.6 g

But H2 actually available = 1000g

Hence, mass of H2 that remains unreacted.

= 1000 -428.6

= 571.4 g

25) How are 0.50 mol Na2CO3 different?

Solution:

0.50 mol of Na2CO3 represent its mass in terms of the number of moles. However, 0.50 M

Na2CO3 represents concentration of it in terms of molarity or mol L-1

, i.e. 0.50 moles of

Na2CO3 present in one litre of its aqueous solution.

26) If ten volumes of dihyrdrogen gas reacts with five volumes of dioxygen gas, how

many volumes of water vapour would be produced?

Solution:

The equation for the reaction of dihydrogen and dioxygen to form water vapour is:

2H2 (g) + O2 → 2 H2 O (g)

2 vol 1 vol 2 vol

10 vol 5 vol 10 vol (according to question)

i.e. 10 vol of H2 will produce 10 vol of H2 O and 5 vol of O2 will produce 10 vol of H2 O.

No reactant will remain behind.

H2 O vapour produced = 10 vol

27) Convert the following into basic units:

i) 28.7 pm

ii) 15.5 pm

iii) 25365 mg

Solution:

i) Basic unit of length is metre(m)

1 pm = 10-12

m

28.7 pm = 28.7 pm x 10-12

m/1 pm

=28.7 x 10-12

m

= 2.87 x 10-11

m

Hence, 28.7 pm is 2.87 x 10-11

m

ii) Basic unit of length is metre(m)

1 pm = 10-12

m

15.15 pm = 15.15 pm x 10-12

m/1pm

= 15.15 x 10-12

m

= 1.515 x 10-11

m

Hence, 15.15 pm is 1.515 x 10-11

m

iii) Basic unit of mass is kg

1mg = 10-6

kg

25365 mg = 25365 x 10-6

kg

= 2.5365 x 10-2

kg

Hence, 25365 mg is 2.5365 x 10-2

kg.

28) 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:

i) 1 g of Au

Atomic mass of gold = 197 g = 1 mole

1 mole of Au is 197 g

Then 1 g of Au = 1/197 moles

1 mole contains 6.022 x 1023

atoms

=3.06 x 1021

atoms.

Hence, 1 g of Au contains 3.06 x 1021

atoms.

ii) 1 g of Na

Atomic mass of Na= 23 g = 1 mole

23g (1 mole) of Na contains 6.022 x 1023

atoms

Then 1 g of Na contains 1/23 x 6.022x 1023

atoms

=2.62 x 1022

atoms

Hence, 1 g of Na contains 2.62 x 1022

atoms.

iii) 1 g of Li

Atomic mass of Li = 7 g = 1 mole

7 g (1mole) of Li contains 6.022 x 1023

atoms

Then 1 g of Li contains 1/7 x 6.022 x 1023

atoms

= 8.6 x 1022

atoms

Hence, 1 g of Na contains 8.6 x 1022

atoms.

iv) 1 g of Cl2

Atomic mass of Cl= 35.5

In 1 molecule of chlorine, there are 2 atoms.

∴ 1 mole of Cl2 = 71 g

71 g of Cl2 contains 2 x 6.022 x 1023

atoms

Then 1 g of Cl2 contains 1/71 mol = 1/71 x 6.022 x 1023

molecules.

2x 1/71 x 6.022 x 1023

atoms

= 1.696 x 1022

atoms.

Hence, 1 g of Cl2 contains 1.696 x 1022

atoms.

∴ Li has the largest number of atoms.

29) Calculate the molarity of a solution of ethanol in water in which the mole fraction

of ethanol is 0.040 (assume the density of water to be one).

Solution:

Our aim is to calculate the number of moles of ethanol in one litre of its solution.

Since the solution is dilute, we can take one litre of water as one kg of it.

i.e. 1 litre of water = 1000g = 1000/ 18 = 55.55 moles

Suppose number of moles of ethanol = a

Hence, the number of moles in the solution = a+ 55.55

Mole fraction of ethanol = 0.04

0.04 = a/ a+55.55

=> 0.04 a+ 2.222 = a

=> 0.96 a = 2.222

∴ a = 2.22/0.96

= 2.31 moles

∴ Number of moles of ethanol on one litre solution= 2.31

Hence, the molarity of the solution = 2.31 mol L-1

30) What will be the mass of one 12

C atom in g?

Solution:

Atomic mass of 12

C= 12 g

12g of C contains 6.022 x 1023

atoms

∴ Mass of 1 atom of 12

C = 12/6.022x 1023

= 1.99 x 10-23

g

∴ Mass of 1 atom of 12

C is 1.99 x10-23

g.

31) How many significant figures should be present in the answer of the following

calculations?

i) 0.02856 x 298.15 x 0.112/0.5785 ii)5 x 5.364 iii)0.0125 + 0.7864 +0.0215

Solution:

1) The least precise term 0.112 has significant figures. Hence , the answer is 3.

2) 5 is an exact number. The other term 5.364 has 4 significant figures.

Hence, the answer is 4.

3) Each term has 4 significant figures. Hence, the answer is 4.

32) Use the data given in the following table to calculate the molar mass of naturally

occurring argon isotopes:

Isotope Isotopic molar mass Abundance

36 Ar 35.96755 g mol

-1 0.337

%

38 Ar 37.96272 g mol

-1 0.063 %

40Ar 39.9624 g mol

-1 99.600%

Solution:

Average molar mass of an element is given as

= Σ Isotopic masses x % abundance/ 100

So, the Average molar mass of argon

= (35.96755 x 0.337) + (37.96272 x 0.063) + (39.9624 x 99.600)/100

= 12.121 + 2.3916 + 3980.26/100

= 39.947 g mol-1

Hence, the molar mass of naturally occurring argon is 39.947 g mol-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

Solution:

i) 1 mole of Ar contains 6.022 x 1023

atoms

∴ 52 moles of Ar contains 52 x 6.022 x 1023

atoms

= 3.13 x 1025

atoms.

ii) 4 u of He = 1 atom of He

∴ 52 u of He = 52x1/4 = 13 atoms

iii) One mole of He = 4 g = 6.022 x 1023

atoms

∴ 52g of He = 6.022 x 5 x 1023

/4

= 7.83 x 1024

atoms.

34) A welding fuel gas contains carbon and hydrogen only. Burning a small sample

of it in oxygen gives 3.38 g carbon dioxide, 0.690 g water and no other products. A

volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g.

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

Solution:

Step 1

Calculating mass percent of carbon and hydrogen.

Molar mass of CO2 = 44 g

Mass of carbon in CO2 =12 g

Mass of carbon in 3.38 g CO2 = 12 x 3.38/44=0.9218 g

Molar mass of H20 = 18 g

Mass of H in H20 = 2 g

Mass of H in 0.690 g of H20= 2 x 0.690/18= 0.9985 g

So, Percentage of C= Mass of C/Mass of fuel gas x 100

=0.9218 x 100/0.9985=92.32

Percentage of H= Mass of H/ Mass of fuel gas x 100

= 0.0767 x 100/0.9985 = 7.68

Step 2

Empirical formula of the fuel gas= CH

Step 3.

Calculating molar mass of fuel gas.

10 litre of gas at NTP weighs= 11.6 g

22.4 litre of gas at NTP weighs = 11.6 x 22.4/10 g

= 25.986 g

=26 g

Molar mass of fuel gas = 26 g

Element Percentage Atomi

c mass

Atomic

Ratio

Simplest

Ratio

Simplest

whole

number ratio

C 92.32 12 92.32/12

=7.69

7.69/7.69

=1

1

H 7.68 1 7.68/1

=7.68

7.68/7.68

=1

1

(Mass of a substance that occupies 22.4 litre by volume is its molar mass).

Step 4.

Calculating molecular formula of gas.

Empirical formula mass of gas CH=12+1=13

Molecular mass of gas=26

n= Molecular mass/ Empirical formula mass x 100

=26/13=2

Molecular formula of the fuel gas = (CH)2

∴ Molecular formula of the fuel gas is C2 H2.

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?

Solution:

Step 1

Calculation of mass of HCl present.

Molarity of HCl = 0.75 M= 0.75 mol L-1

Volume of HCl = 25 mL = 25/1000

Volume of HCl = 0.025 L

Molarity of solution = Mass of HCl / Molar mass of HCl x 1/ Volume of solution

0.75 = Mass of HCl / 36.5 x 1/0/025

Mass of HCl =0.75 mol L-1

x 36.5 g mol-1

x 0.025 L

Mass of HCl = 0.684 g

Step 2

Calculation of mass of CO2 reacted.

CaCO3 + 2HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O(l)

1 mol 2 mol

100 g 2x 36.5 = 73g

73g of HCl requires 100 g of CaCO3 for the reaction

∴ 0.684 g of HCl requires 100x0.684/73 g of CaCO3 for the reaction =0.94 g

∴ Mass of CaCO3 = 0.94 g

∴ Mass of CaCO3 required to react completely with 25 mL of 0.75 M HCl is 0.94 g.

36) 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) + H2 O(i)

4 x 36.5 g 55+2x16g

146 g 87 g

87 g of MnO2 react with 146 g HCl

∴ 5 g of MnO2 react with 146 x 5/ 87 g of HCl

= 8.39 g

∴ Mass in grams of HCl that react with 5 g of manganese dioxide is 8.39 g.