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Class-XI-CBSE-Chemistry Some Basic Concepts Of Chemistry

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CBSE NCERT Solutions for Class 11 Chemistry Chapter 1

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1. Calculate the molar mass of the following:

(i) H2O

(ii) CO2

(iii) CH4s

Solution:

(i) The molecular mass of water( H2O)

We know : Atomic mass of hydrogen =1.008 u

Atomic mass of oxygen = 16.00 u

molecular mass of H2O = (2 × Atomic mass of hydrogen)+(1 × Atomic mass of

oxygen)

= [2(1.008 u) + 1(16.00 u)]

= 2.016 u + 16.00 u

= 18.016 u

(ii) The molecular mass of carbon dioxide (CO2)

Atomic mass of carbon =12.0 u

Atomic mass of oxygen = 16.00 u

=(1 × Atomic mass of carbon)+(2 × Atomic mass of oxygen)

= [1(12.0u) + 2(16.00u)]

= 12.0u + 32.00u

= 44.0u

(iii) The molecular mass of methane ( CH4)


Atomic mass of hydrogen = 1.008 u

=(1 × Atomic mass of carbon)+(4 × Atomic mass of hydrogen)

= [1(12.0u) + 4(1.008 u)]

= 12.0u + 4.032 u



= 16.032 u

2. Calculate the mass percent of different elements present in sodium sulphate (Na2SO4).

Solution:

We know : Atomic mass of sodium =23.0 u

Atomic mass of Sulphur = 32.0 u


Calculate : The molecular formula of sodium sulphate is Na2SO4

Molar mass of Na2SO4 = [(2 × 23.0) + (32.0) + 4(16.00)]

= 142 u or 142 g

Formula : Mass percent of an element =Mass of that element in the compound

Molar mass of the compound× 100

∴ Mass percent of sodium==Mass of sodium element in the compound

Molar mass of the Na2SO4× 100

= 46.0 g

142.0 g × 100

= 32.39

= 32.4%

Mass percent of Sulphur =Mass of sulphur element in the compound


= 32.0 g

142.0 g × 100

= 22.54 %

Mass percent of oxygen =Mass of oxygen element in the compound


= 64.0 g

142.0 g × 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:

Given : % of iron by mass = 69.9%

% of oxygen by mass = 30.1%

Atomic mass of iron = 55.85

Atomic mass of oxygen = 16.00

Formula used= Relative moles of atom in compound = % by mass of atom

Atomic mass of atom

Relative moles of iron in iron oxide = % by mass of iron

Atomic mass of iron

=69.9

55.85

= 1.25

Relative moles of oxygen in iron oxide:

=% of oxygen by mass

Atomic mass of oxygen

= 30.1

16.00

= 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

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 is burnt in 16 g of dioxygen.



Solution:

The balanced reaction of combustion of carbon can be written as:

(i) C + O2 → CO2

Initial mole 1 1 −

Mole

stoichiometry coefficient

1

1=1

1.0

1= 1 -

Left mole 1 − 1 = 0 1 − 1 = 0 1

As per stoichiometry of the balanced equation, 1 mole of carbon burns in 1 mole of

dioxygen (air) to Produced 1 mole of carbon dioxide.

(ii) 1 mole of carbon is burnt in 16 g of dioxygen.

Given in question : dioxygen= 16 g ; mole of carbon = 1

We know : Moles of oxygen =moles

molecular weight=

16

32= 0.5

C + O2 → CO2

Initial mole 1 0.5 0.5

Mole


1

1=1

0.5

1= 0.5 0

Left mole 1 − 0.5 = 0.5 0 0.5

According to the stoichiometry of the balanced equation

Hence, dioxygen will react with 0.5 moles of carbon to give 0.5 mole of carbon dioxide

(22 g). Hence, dioxygen is a limiting reactant.

(iii) C + O2 → CO2

Given : dioxygen (𝑂2)= 16 g ; mole of carbon = 2

We know : molecular weigth of O2 = 32

moles of dioxygen =𝑤𝑒𝑖𝑔ℎ𝑡

𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡=

16

32= 0.5

C + O2 → CO2

Initial mole 2 0.5 0.5

Mole


2

1=2

0.5

1= 0.5 0



Left mole 2 − 0.5 = 1.5 0 0.5

According to the question, only 0.5 𝑚𝑜𝑙𝑒 of dioxygen or is available. It is a limiting

reactant. Thus, 0.5 mole of dioxygen can combine with only 0.5 mole of carbon dioxide

Weight of carbon dioxide = moles × molecular weight of CO2 = 0.5 × 44 = 22 gram

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 gmol−1.

Solution:

given : mass of sodium acetate (CH3COONa) = ?

Molar mass of sodium acetate = 82.024 gmole‐1

Molarity of aqueous solution of sodium acetate = 0.375M

Volume of solution = 500 ml

we know ∶ ∴ A number of moles of sodium acetate in 500 mL. = Molarity × volume of

solution (litre)

= 0.375

1000× 500

= 0.1875 mole

∴ Required mass of sodium acetate = moles × molecular weight

⇒ (82.0245gmo1−1) × (0.1875 mole)

⇒ 15.38 g

6. Calculate the concentration of nitric acid in moles per liter in a sample which has a

density, 1.41gmL−1 and the mass percent of nitric acid in it being 69%.

Solution:

Given : density = 1.41gmL−1

Mass percent of nitric acid in the sample = 69%



We know, the mass percent of nitric acid in it being 69% = 100g of nitric acid contains

69g of nitric acid by mass.

Molar mass of nitric acid (HNO3) = {1 + 14 + 3(16)} gmo1−1

= (1 + 14 + 48) gmo1−1

= 63 gmo1−1

∴ Number of moles in 69 g of HNO3

= weight (g)

molecular mass (gmo1−1)=

69g

63 gmo1−1

= 1.095mol

Volume of 100g nitric acid solution=Mass of solution

density of solution

=100 g

1.41gmL−1

= 100g

141gmL−1× 100

= 70.92mL ≡ 70.92 × 10−3L (∴ 1 𝑚𝑙 = 10−3 𝑙𝑖𝑡𝑒𝑟) )

Concentration of nitric acid= mole

litre

= 1.095 mo1e

70.92 × 10−3L

= 15.44 mol L⁄

∴ Concentration of nitric acid = 15.44 mol L⁄

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

Solution:

Given : mass of CuSO4 = 100 g

We know : Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00)

= 63.5 + 32.00 + 64.00

= 159.5 g

159.5g of CuSO4 contains 63.5 g of copper.



⇒ 100g of CuSO4 will contain copper =63.5×100g

159.5 = 39.81 g

∴ Amount of copper that can be obtained from 100 g CuSO4 = 39.81 g

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.

Solution:

Given :Mass percent of iron (Fe) = 69.9%

Mass percent of oxygen (O) = 30.1%

We know : The atomic mass of iron = 55.85

The atomic mass of oxygen = 16

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

The ratio of iron to oxygen in the oxide,

= 1.25: 1.88

=1.25

1.25:1.88

1.25

= 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 = 159.69g

∴ n =Molar mass

Empirical formula mass=

159.69g

159.7g

= 0.999

= 1(approx)

Molecular formula of a compound = (empirical formula of a compound) n.

Thus , molecular formula = ( Fe2O3) × 1.

Hence, the molecular formula of the oxide is Fe2O3.



9. Calculate the atomic mass (average) of chlorine using the following data:

% Natural

Abundance

Molar Mass

Cl 35 75.77 34.9689

Cl 37 24.23 36.9659

Solution:

given in question :

for Cl 35 : % Natural Abundance =75.77

Molar Mass=34.9689

Cl 37 : % Natural Abundance = 24.23

Molar Mass = 36.9659

The average atomic mass of chlorine.

= [(Fractional abundance

of Cl 35

) (Molar mass

of Cl 35 ) + (

Fractional abundance of Cl

37 ) (Molar mass

of Cl 37 )]

= [{(75.77

100) (34.9689u)} + {(

24.23

100) (36.9659u)}]

= 26.4959 u + 8.9568 u

= 35.4527u

∴ The average atomic mass of chlorine = 35.4527u

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 mole of C2H6 contains 2 moles of carbon atoms.



∴ Number of moles of carbon atoms in 3 moles of C2H6

= 2 × 3 = 6 moles of carbon

(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms.

∴ Number of moles of carbon atoms in 3 moles of C2H6

= 3 × 6 = 18 moles of hydrogen atoms

(iii) 1 mole of C2H6 contains= NA ×molecules of ethane= 6.023 × 1023 × molecules of

ethane.

∴ Number of molecules in 3 moles of C2H6 = 3 × NA = 3 × 6.023 × 1023

= 18.069 × 1023 molecules of ethane

11. What is the concentration of sugar (C12H22O11) in molL−1 if its 20 g are dissolved in

enough water to make a final volume up to 2L?

Solution:

Given mass of glucose = 20 g

We know : molar mass of sugar (C12H22O11) = [(12 × 12) + (1 × 22) + (11 × 16)] =

342

Formula : Molarity of a solution (M) =Number of moles of solute

Volume of solution in Litres

By putting the values in formula we get

M= Mass of sugar molar mass of sugar⁄

2L

= 20g 342⁄

2L

= 0.0585mol

2L

= 0.02925molL−1

∴ Molar concentration of sugar = 0.02925 molL−1



12. lf the density of methanol is 0.793 kgL−1 what is its volume needed for making 2.5 L of

its 0.25 M solution?

Solution:

Given:

density of methanol is = 0.793 kgL−1

volume of solution (V2) = 2.5 L

molarity of solution (M2) = 0.25 M

we know Molar mass of methanol (CH3OH) = (1 × 12) + (4 × 1) + (1 × 16) =

32gmo1−1 = 0.032 kgmol−1

density of methanol is 0.793 kgL−1 means one litre of solution contain

0.793 kg of methanol =793 gm of methanol =793

32 moles of methanol

i. e. molarity of methanol volume (𝑀1)= 24.78molL−1

formula used : M1V1 = M2V2

(24.78molL−1) V1 = (2.5L)(0.25molL−1)

V1 = 0.0252L (1 liter= 1000 mililiter)

V1 = 25.22mL

13. Pressure is determined as force per unit area of the surface. The Sl unit of pressure, pascal

is as shown below:

1Pa = 1Nm−2

If the mass of air at sea level is 1034 gcm−2 calculate the pressure in Pascal.

Solution:

Given : mass of air at sea level = 1034 gcm−2

The pressure is defined as force acting per unit area of the surface.

P =F

A

=1034g × 9.8ms−2

cm2×

1kg

1000g×

(100)2cm2

1m2



= 1.01332 × 105kgm−1s−2

We know,

1N = 1kgms−2

Then,

1Pa = 1Nm−2 = 1kgm−2s−2

1Pa = 1kgm−1s−2

∴ Pressure = 1.01332 × 105Pa

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

Solution:

The SI unit of mass is a kilogram (kg) .

1 kilogram is defined as the mass equal to the mass of the international

prototype(Platinum-Iridium cylinder in Paris) of the kilogram.

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

(i) Micro 10−6

(ii) Deca 10

(iii) Mega 106

(iv) Giga 109

(v) Femto 10−15



16. What do you mean by significant figures?

Solution:

Significant figures are those meaningful digits that are known with certainty. They

indicate uncertainty in an experiment or calculated value. For example, if 25.8 mL is the

result of an experiment, then 25 is certain, while 8 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.

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:

(i) 1 ppm = 1 part out of 1 million (106) parts.

Let assume Mass of chloroform CHCl3 =15

Mass of solution = 106

Formula :

Mass % of solute =𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 × 100

𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛

∴ Mass percent of l5 ppm chloroform in water.

= 15

106 × 100

≈ 1.5 × 10−3%

(ii) 100 g of the sample contains 1.5 × 10−3g of CHCl3.

Given : weight of CHCl3 = 1.5 × 10−3g

Weight of solvent =100 g



We know : Molar mass of CHCl3 = 12.00 + 1.00 + 3(35.5) = 119.5gmo1−1

Moles of CHCl3 =𝑤𝑒𝑖𝑔ℎ𝑡(𝑔𝑚)

𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 =

1.5×10−2g

Molar mass of CHC13=

1.5×10−2g

119.5gmo1−1

Formula : Molaity =

weight of solute (gram)

Molar mass of solute

𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝑘𝑔)

Put all the above values in formula we get

Molality of chloroform in water.

=

weight (gram)Molar mass of CHCl3

𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝑘𝑔)=

1.5 × 10−2gMolar mass of CHCl3

𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 (𝑘𝑔)=

1.5 × 10−2g119.5 gmo1−1

1

= 0.0125 × 10−2

= 1.25 × 10−4molal

18. Express the following in the scientific notation:

(i) 0.0048

(ii) 234,000

(iii) 8008

(iv) 500.0

(v) 6.0012

Solution:

(i) 0.0048 = 4.8 × 10−3

(ii) 234,000 = 2.34 × 105

(iii) 8008 = 8.008 × 103

(iv) 500.0 = 5.000 × 102

(v) 6.0012 = 6.0012

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:

(i) There are 2 significant figures.

(ii) There are 3 significant figures.

(iii) There are 4 significant figures.

(vi) There are 3 significant figures.

(v) There are 4 significant figures.

(vi) There are 5 significant figures.

20. Round up the following upto three significant figures:

(i) 34.216

(ii) 10.4107

(iii) 0.04597

(iv) 2808

Solution:

(i) 34. 2

(ii) 10.4

(iii) 0.0460

(iv) 2810

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

different compounds:

Mass of dinitrogen Mass of dioxygen

(i) 14 g 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.

(b) Fill in the blanks in the following conversions:

(i) 1 km = ____________ mm = ____________ pm

(ii) 1 mg = ____________ kg = ____________ ng

(iii) 1 mL = ____________ L = ____________ dm3

Solution:

(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, 64g, 32g, and 80 g.

Mass of dinitrogen Mass of dioxygen Ration of oxygen

(i) 14 g ×2=28 16 g ×2=23 32

(ii) 14 g ×2=28 32 g ×2=62 64

(iii) 28 g 32 g 32

(iv) 28 g 80 G 80

The masses of dioxygen bear a whole number ratio of 1: 2: 1: 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 ration of small whole numbers.

(b) (i) 1 km = 1 km ×1000m

1km×

100cm

1m×

10mm

1cm

∴ 1 km = 106mm

1 km = 1 km ×1000m

1km×

1pm

10−12m

∴ 1 km = 1015pm

Hence, 1 km = 106mm = 1015pm

(ii) 1 mg = 1 mg ×1g

1000mg×

1kg

1000g

⇒ 1 mg = 10−6kg

1 mg = 1 mg ×1g

1000mg×

1ng

10−9g



⇒ 1 mg = 106ng

∴ 1 mg = 10−6kg = 106ng

(iii) 1mL = 1mL ×1L

1000mL

⇒ 1mL = 10−3L

1mL = 1cm3 = 1cm31dm × 1dm × 1dm

10cm × 10cm × 10cm

⇒ 1mL = 10−3dm3

∴ 1mL = 10−3L = 10−3dm3

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

Solution:

Given : speed of light = 3.0 × 108 ms−1

distance covered by light =?

Time taken to cover the distance = 2.00 ns = 2.00 × 10−9s

We know: velocity of light = 3.0 × 108ms−1

Formula : Distance traveled by light = Speed of light × Time taken

= (3.0 × 108ms−1) (2.00 × 10−9s)

= 6.00 × 10−1m

= 0.600 m

23. In a reaction A + B2 → AB2

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 reagent determines the extent of a reaction. It is the reactant which is the first to get

consumed during a reaction, thereby causing the reaction to stop and limiting the amount

of products formed.

(i) 300 atoms of A + 200 molecules of B

According to the stoichiometry of reaction, 1 atom of A reacts with 1 molecule of B.

A + B2 → AB2

Initial moles 300 atoms = 300 NA 200 molecule =200 NA

−

Mole


300NA

1=300NA

200NA

1= 200NA -

Left mole (300-200) NA=100NA 0 200NA

Thus, 200 NA moles of B2 will react with 300 NA moles of A, thereby leaving 100NA

moles of A unused. Hence, B2 is the limiting reagent.

(ii) 2 mol A + 3 mol B

According to the reaction, 1 mol of A reacts with 1 mol of 3.

A + B2 → AB2

Initial moles 2 3 −

Mole


2

1=2

3

1= 3 -

Left mole 2-2=0 3-2=1 2

Thus, 2 mol of A will react with only 2 mol of B. As a result, 1 mol of A will not be

consumed. Hence, A is the limiting reagent.

(iii) 100 atoms of A + 100 molecules of B

According to the given reaction, 1 atom of A combines with 1 molecule of B.

A + B2 → AB2

Initial moles 100 atoms = 100 NA 100 molecule =100 NA

−

Mole


100NA

1=100NA

100NA

1= 100NA -



Left mole 0 0 100NA

Thus, all 100NA moles of A will combine with all 100NA moles of B. Hence, the mixture

is stoichiometric where no limiting reagent is present.

(ii) 5 mol A + 2.5 mol B

A + B2 → AB2

Initial moles 5 2.5 −

Mole


5

1=5

2.5

1= 2.5 -

Left mole 5-2.5=2.5 2.5-2.5=0 2.5

1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of B will combine

with only 2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the

limiting reagent.

(iii) 2.5 mol A + 5 mol B2

A + B2 → AB2

Initial moles 2.5 5 −

Mole


2.5

1=2.5

5

1= 5 _

Left mole 2.5-2.5=0 5-2.5=2.5 2.5

According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus,

2.5 mol of A will combine with only 2.5 mol of B and the remaining 2.5 mol of B will be

left as such. Hence, A is the limiting reagent.

24. Dinitrogen and dihydrogen react with each other to produce ammonia according to the

following chemical reaction:

N2(g) + H2(g) → 2NH3(g)

(i) Calculate the mass of ammonia produced if 2.00 × 103g dinitrogen reacts with

1.00 × 103g of dihydrogen.

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass?



Solution:

Given : mass of dinitrogen = 2.00 × 103g

mass of dihydrogen = 1.00 × 103g

we should know : 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 of N2 =28

𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 of H2 =2

𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑤𝑒𝑖𝑔ℎ𝑡 of NH3= 34

Moles of dinitrogen = 𝑤𝑒𝑖𝑔ℎ𝑡(𝑔)


2.00×103g

28= 71.43

moles of dihydrogen = = 𝑤𝑒𝑖𝑔ℎ𝑡(𝑔)


1.00×103g

2= 500

Balancing the given chemical equation,

N2(g) + 3H2(g) → 2NH3(g)

Initial moles 71.43 500 −

Mole


71.43

1=71.43

500

3= 166.66 _

From the equation, 1 mol of nitrogen reacts with 3 mol of dihydrogen to give 2 moles of

ammonia.

Mole

stoichiometry coefficient, the ratio is less for N2 Hence, N2 is the limiting reagent.

Hence, the moles of ammonia produced by 71.43 moles of N2 = 2 × 71.43

mass of ammonia produced by 71.43 moles of N2 = 2 × 71.43 ×

molecular weight of ammonia

= 2 × 71.43 × 17 g

= 2428.62g

(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will remain

unreacted.

(iii) moles of dihydrogen left unreacted = (500 − 71.43 × 3)moles

Mass of dihydrogen left unreacted = (500 − 71.43 × 3)2 𝑔= 285.71 ×2 g= 571.4g

25. How are 0.50 mol Na2CO3 and 0.50M Na2CO3 different?



Solution:

Given: mols of Na2CO3= 0.50

Molarity of Na2CO3 = 0.50 M

We know : Molar mass of Na2CO3 = (2 × 23) + 12.00 + (3 × 16)

= 106gmol−1

Now, 1 mole of Na2CO3 means 106g of Na2CO3

∴ 0.5 mol of Na2CO3 =106g

1mole× 0.5molNa2CO3

= 53 g Na2CO3

⇒ 0.50M of Na2CO3 = 0.50mol

LNa2CO3

Hence, 0.50 mol of Na2CO3 is present in 1 L of water

or (0.5 × 106) = 53 g of Na2CO3 is present in 1 L of water.

26. If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how many

volumes of water vapour would be produced?

Solution:

Reaction of dihydrogen with dioxygen can be written as:

2H2 (g) + O2(g) → 2H2O(g)

Mole ratio 2 1 2 ∴ moles ∝ volume

According to stiochimerty of reaction two volumes of dihydrogen react with one volume

of dioxygen to produce two volumes of water vapour. Hence, ten volumes of dihydrogen

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

27. Convert the following into basic units:

(i) 28.7 pm

(ii) 15.15 pm

(iii) 25365 mg



Solution:

(i) 28.7 pm:

1pm = 10−12 m

∴ 28.7pm = 28.7 × 10−12 m

= 2.87 × 10−11 m

(ii) 15.15 pm:

1 pm = 10−12 m

∴ 15.15 pm = 15.15 × 10−12 m

= 1.515 × 10−12 m

(iii) 25365 mg :

1 mg = 10−3g

25365 mg = 2.5365 × 104 × 10−3g

Since,

1 g = 10−3 kg

2.5365 × 101g = 2.5365 × 10−1 × 10−3kg

∴ 25365 mg = 2.5365 × 10−2 kg

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

(i) 1 g Au (s)

(ii) 1 g Na (s)

(iii) 1 g Li (s)

(iv) 1g Cl2(g)

Solution:

we know ∶ NA = 6.022 × 1023

Atomic mass of Au = 107

Atomic mass of Na = 23

Atomic mass of Li = 7

Molar mass of Cl2 molecule = Atomic mass of chlorine × 2 = 35.5 × 2 = 71gmol−1



(i) 1g Au(s) =1

197 mol of Au(s)

1

197× NA =

6.022×1023

197 atoms of Au (s)

= 3.06 × 1021 atoms of Au(s)

(ii) 1g of Na(s) =1

23 mol of Na(s)

1

23× NA =

6.022×1023

23 atoms of Na(s)

= 0.262 × 1023 atoms of Na(s)

= 26.2 × 1021 atoms of Na(s)

(iii)1 g Li(s) =1

7 mol of Li(s)

=1

7× NA =

6.022×1033

7 atoms of Li(s)

= 0.86 × 1023 atoms of Li (s)

= 86.0 × 1021 atoms of Li (s)

(iv)1g Cl2(g) =1

71 mol of Cl2(g)

=1

71× NA =

6.022×1033

71 atoms of Cl2(g)

= 0.0848 × 1023 atoms of Cl2(g)

= 8.48 × 1021 atoms of Cl2(g)

Hence, 1 g of Li(s) will have 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:

Given : mole fraction of ethanol = 0.040

We know : density of water = 1 gm/ml



the molecular mass of water (H2O) =18

Mole fraction of C2H5OH =Number of moles of C2H5OH

Number of moles of solution

Number of moles of solution

= Number of moles of solute + Number of moles of solvent

Number of moles of solute = nC2H5OH

Number of moles of solvent = nH2O =1000 g

18 gmo𝑙−1 = 55.55mol

0.040 =nC2H5OH

nC2H5OH + nH2O … (1)

Substituting the value of nH2O in equation (1),

nC2H5OH

nC2H5OH + 55.55= 0.040

nC2H5OH = 0.040 nC2H5OH + (0.040) × (55.55)

0.96nC2H5OH = 2.222mol

nC2H5OH =2.222

0.96 mol

nC2H5OH ≡ 2.314mol

𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒

𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 (𝑙𝑖𝑡𝑒𝑟)

Molarity of solution =2.314mo1

1L

= 2.314M

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

Solution:

1 mole of carbon atoms = 6.023 × 1023 atoms of carbon

= 12g of carbon

∴ Mass of one C 12 = 12 × 1.67 × 10−24 g = 1.993 × 10−23g



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

calculations?

(i) 0.02856×298.15×0.112

0.5785

(ii) 5 × 5.364

(iii) 0.0125 + 0.7864 + 0.0215

Solution:

(i) 0.02856×298.15×0.112

0.5785

Least precise number of calculation = 0.112

∴ Number of significant figures in the answer.

= Number of significant figures in the least precise number.

= 3

(ii) 5 × 5.364

Least precise number of calculation = 5.364

∴ Number of significant figures in the answer = Number of significant figures in 5.364

= 4

(iii) 0.0125 + 0.7864 + 0.0215

Since the least number of decimal places in each term is four, the number of significant

figures in the answer is also 4.

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

occuring argon isotopes:

Isotope Isotopic molar mass Abundance

Ar 36 35.96755 g mol−1 0.337%

Ar 38 37.96272 g mol−1 0.063%

Ar 40 39.9624 g mol−1 99.600%



Solution:

Given :

for Isotope Ar 36 ∶Isotopic molar mass

35.96755 g mol−1 , percentage Abundance 0.337%

for Isotope Ar 38 ∶

Isotopic molar mass 37.96272 g mol−1 percentage Abundance0.063%

for Isotope

Ar ∶ Isotopic molar mass 40 39.9624 g mol−1 percentage Abundance99.600%

Average molar mass

=

molar mass of isotope 1×% abundance of isotope 1+ molar mass of isotope 2×% abundance of isotope 2+molar mass of isotope 3×% abundance of isotope3

100

Average molar mass of argon

= [ (35.96755 × 0.337

100) + (37.96272 ×

0.063

100) + (39.9624 ×

90.60

100)] gmo1−1

= [0.121 + 0.024 + 39.802] gmo1−1

= 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:

We know : atomic mass of He=4 ; NA = 6.022 × 1023

(i) 1 mole of Ar = NA atoms of Ar = 6.022 × 1023 atoms of Ar

∴ 52 mol of Ar = 52 × 6.022 × 1023 atoms of Ar

= 3.131 × 1025 atoms of Ar

(ii) 1 atom of He = 4u of He

Or,

4u of He = 1 atom of He

1u of He =1

4 atom of He



52u of He =52

4𝑢 atom of He

= 13u atoms of He

(ii) I mole of He atom =4 g He

I mole of He atom = 𝑁𝐴 atoms of He = 6.022 × 1023 atoms of He

So 4 g of He = 6.022 × 1023 atoms of He

∴ 52 g of He =6.022×1023×52

4 atoms of He

= 7.8286 × 1024 atoms of He

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 of 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:

Given: a mass of carbon dioxide in welding fuel gas =3.38 g

Mass of Water welding fuel gas =0.690 g

The volume of welding fuel gas = 10 litre

Weight od welding fuel gas=11.6 g

(i) We know 1 mole (44g) of CO2 contains 12g of carbon.

∴ 3.38g of CO2 will contain carbon =12g

44g× 3.38g

= 0.9217g

we know 18 g of water(H2O) contains 2 g of hydrogen.

∴ 0.690 g of water will contain hydrogen =2g

18g× 0.690

= 0.0767g



Since carbon and hydrogen are the only constituents of the compound, the total mass of

the compound is:

= 0.9217g + 0.0767g

= 0.9984g

Formula : Percent of atom in the compound =weight of atom

weight of sample × 100

∴ Percent of C in the compound =0.9217g

0.9984g× 100

= 92.32%

Percent of H in the compound =0.0767g

0.9984g× 100

= 7.68%

Formula : 𝑚𝑜𝑙𝑒𝑠 =𝑤𝑒𝑖𝑔ℎ𝑡

𝑎𝑡𝑜𝑚𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡

Moles of carbon in the compound =92.32

12

= 7.69

Moles of hydrogen in the compound =7.68

1

= 7.68

∴ Ratio of carbon to hydrogen in the compound = 7.69: 7.68

= 1: 1

Hence, the empirical formula of the gas is CH.

(ii) Given,

Weight of 10.0L of the gas (at S.T.P) = 11.6g

∴ Weight of 22.4 L of gas at STP =11.6 g

10 L× 22.4L

= 25.984g

≈ 26g

Hence the molar mass of the gas is 26 g.

(iii) Empirical formula mass of CH = 12 + 1 = 13 g

n =Molar mass of gas

Empirical formula mass of gas



=26g

13g

n = 2

formula : Molecular formula of gas= ( emperical formula of gas)× 𝑛 n

∴ Molecular formula of gas = (CH)n

= C2H2

35. 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 : volume of HCl solution = 25 mL

molarity of HCl solution = 0.75 M

we know molecular formula of CaCO3 = 100

formula : moles = molarity × volume (litre)

moles of HCl = 0.75 M ×25 mL

1000 = 0.01875

From the given chemical equation,

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

According to stoichiometry of reaction :

∴ 2 mol of HCl react with = 1 mol of CaCO3

0.01875 mol HCl of react with =1

2× 0.01875 mol of CaCO3

(∴ weight = moles × molecular weight )

Amount of CaCO3 that will react =1

2× 0.01875 × 100

= 0.9375 g



36. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with

aqueous hydrochloric acid according to the reaction

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

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

Solution:

given : weight of manganese dioxide= 5 g

we know molecular mass of MnO2 = [55 + 2 × 16 = 87g]

molecular mass of HCl = [35.5 + 1 = 36.5]

moles of MnO2 =weight

molecular weight=

5

87= 0.057 mole

according to stoichiometry

1 mole MnO2 reacts completely with 4 mol of HCl.

∴ 0.057 mole of MnO2 will react with HCl =4

1× 0.057 =

0.2298 mole of HCl formed

weight of HCl = moles molecular × moles of HCl = 36.5 × 0.2298 =8.4g

Hence, 8.4g of HCl will react completely with 5.0g of manganese dioxide.

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