# Class 11 | Chapter 1 NCERT Chemistry Solutions

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Question 1

Calculate the molar mass of the following:
(i) H2O

Molar Mass of H2O = 2 × Molar Mass of Hydrogen(H) + Molar Mass of Oxygen(O)

Molar Mass of Hydrogen = 1 u
Molar Mass of Oxygen = 16 u

⇒ Molar Mass of H2O = 2 × 1 + 16 = 2 + 16

Molar Mass of H2O = 18 u

(ii) CO2

Molar Mass of CO2 = Molar Mass of Carbon(C) + 2 × Molar Mass of Oxygen(O)

Molar Mass of Carbon = 12 u
Molar Mass of Oxygen = 16 u

⇒ Molar Mass of CO2 = 12 + 2 × 16 = 12 + 32

Molar Mass of CO2 = 44 u

(iii) CH4

Molar Mass of CH4 = Molar Mass of Carbon(C) + 2 × Molar Mass of Hydrogen(H)

Molar Mass of Carbon = 12 u
Molar Mass of Hydrogen = 1 u

Molar Mass of CH4 = 12 + 2 × 1 = 12 + 2 = 14

Molar Mass of CH4 = 14u

Question 2

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

Formula for Calculating Mass per cent of an element in a compound is

So in order to calculate mass per cent of each element in sodium sulphate (Na2SO4) we firstly need to calculate Molar Mass of the compound sodium sulphate (Na2SO4).

Molar Mass of sodium sulphate (Na2SO4)
= 2 × Molar Mass of Sodium (Na)
+ Molar Mass of Sulphur(S)
+ 4 × Molar Mass of Oxygen (O)

Molar Mass of Sodium (Na) = 23 u
Molar Mass of Sulphur(S) = 32 u
Molar Mass of Oxygen (O) = 16 u

⇒ Molar Mass of sodium sulphate (Na2SO4) = 2 × 23 + 32 + 4 × 16 = 46 + 32 + 64 = 142 u

Molar Mass of sodium sulphate (Na2SO4) = 142 u

Question 3

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

Relative Moles of Iron(Fe) in the iron oxide = Percentage of iron / Atomic Mass of iron

Percentage of iron = 69.9%
Atomic Mass of iron = 55.85

⇒ Relative Moles of Iron(Fe) in the iron oxide = 69.9/55.85 = 1.25

Relative Moles of Iron(Fe) in the iron oxide = 1.25

Relative Moles of Oxygen(O) in the iron oxide = Percentage of oxygen / Atomic Mass of oxygen

Percentage of oxygen = 30.1%
Atomic Mass of oxygen = 16

Relative Moles of Oxygen(O) in the iron oxide = 30.1/16 = 1.88

Relative Moles of Oxygen(O) in the iron oxide = 1.88

Simplest Molar Ratio of iron to oxygen
1.25 : 1.88 = 125 : 188 = 2 : 3

Thus given iron oxide have Simplest Molar Ratio of iron to oxygen as 2 : 3

⇒ Empirical Formula of iron oxide is Fe2O3 which is known as Ferric Oxide.

Question 4

Calculate the amount of Carbon Dioxide that could be produced when

Balanced Chemical Equation for reaction of Carbon with Oxygen

(i) 1 mole of carbon is burnt in air
As per above Balanced Chemical Equation it’s clear the burning of 1 mole of carbon produces 1 mole(44g) of Carbon Dioxide.

(ii) 1 mole of carbon is burnt in 16g of dioxygen
As per above Balanced Chemical Equation, burning of 1 mole of carbon require 32g of Oxygen but only 16g is available, thus only 0.5 mole of carbon will be burned which will produce 0.5 mole(22g) of Carbon Dioxide.

(iii) 2 moles of carbon are burnt in 16g of dioxygen
As per Balanced Chemical Equation, 1 mole of carbon is required to react with 1 mole(32g) of oxygen. But only 16g of oxygen is available which can only react with 0.5 mole of carbon producing 0.5 mole(22g) of Carbon Dioxide.

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.

0.375 molar aqueous solution means
1000 mL of solution contains 0.375 moles of Sodium Acetate (CH3COONa)

⇒ 500 mL of solution contains 0.1875 moles of Sodium Acetate (CH3COONa)

As per question, Molar mass of Sodium Acetate is 82.0245 g mol-1

Mass of 0.1875 moles of Sodium Acetate (CH3COONa)
= 0.1875 × 82.0245
= 15.38 g

Thus 15.38g of Sodium Acetate is required to make 500 mL of aqueous solution of Sodium Acetate.

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%.

Mass percent of Nitric Acid in the sample = 69%
⇒ 100g of sample contains 69g of Nitric Acid

Density of sample = 1.41g mL-1
Mass of sample = 100g

Volume of sample = Mass of sample / Density of sample = 100/1.41

Volume of sample = 70.92 mL = 70.92 × 10-3 L

Volume of sample = 70.92 × 10-3 L

⇒ 70.92 × 10-3 L of sample contains 69g of Nitric Acid

Number of moles of Nitric Acid = Mass of Nitric Acid / Molar Mass of Nitric Acid

Molar Mass of Nitric Acid(HNO3)
= Molar Mass of Hydrogen(H) + Molar Mass of Nitrogen(N) + 3 × Molar Mass of Oxygen(O)

= 1 + 14 + 3 × 16 = 1 + 14 + 48 = 63

Molar Mass of Nitric Acid(HNO3) = 63

Number of moles of Nitric Acid = 69/63 = 1.095
Volume of sample = 70.92 × 10-3 L

Concentration of nitric acid in moles per litre in sample
= Number of moles of Nitric Acid / Volume of sample

= 1.095 / 70.92 × 10-3

= 15.44 moles L-1

Concentration of nitric acid in moles per litre in sample = 15.44 moles L-1

Question 7

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

1 mole of copper sulphate (CuSO4) contains 1 mole of copper (Cu)

Molar Mass of copper sulphate (CuSO4)
= Molar Mass of copper(Cu) + Molar Mass of Sulphur(S) + 4 × Molar Mass of Oxygen(O)

= 63.5 + 32 + 4 × 16

= 63.5 + 32 + 64

= 159.5

Molar Mass of copper sulphate (CuSO4) = 159.5 g

⇒ 159.5 g copper sulphate (CuSO4) contains 63.5 g of copper (Cu)

⇒ 100 g copper sulphate (CuSO4) contains 63.5/159.5 × 100 g of copper (Cu)

Thus 100 g copper sulphate (CuSO4) contains 39.81g of copper(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.

Mass percent of iron(Fe) = 69.9%
Mass percent of oxygen(O) = 30.1%

Number of moles of Fe present in oxide = 69.9/Molar Mass of Fe = 69.9/55.85

Number of moles of Fe present in oxide = 1.25

Number of moles of O present in oxide = 30.1/Molar Mass of O = 30.1/16

Number of moles of O present in oxide = 1.88

Ratio of number of moles of Fe to O in oxide
= 1.25 : 1.88

= 125 : 188

= 2 : 3

⇒ Formula of oxide is Fe2O3

Question 9

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

Average Atomic Mass of chlorine
= Fractional abundance of 35Cl × Molar mass of 35Cl
+ Fractional abundance of 37Cl × Molar mass of 37Cl

= 75.77/100 × 34.9689 + 24.23/100 × 36.9659

= 0.7577 × 34.9689 + 0.2423 × 36.9659

= 26.50 + 8.96

= 35.46 u

Thus Average Atomic Mass of chlorine = 35.46 u

Question 10

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

(i) Number of moles of carbon atoms
1 mole of ethane (C2H6) has 2 moles of carbon(C)
⇒ 3 moles of ethane (C2H6) has 6 moles of carbon

(ii) Number of moles of hydrogen atoms
1 mole of ethane (C2H6) has 6 moles of hydrogen(H)
⇒ 3 moles of ethane (C2H6) has 18 moles of hydrogen

(iii) Number of molecules of ethane
1 moles of ethane (C2H6) contains 6.023 × 1023 molecules of ethane

⇒ 3 moles of ethane (C2H6) contains 3 × 6.023 × 1023 molecules of ethane
= 18.023 × 1023 molecules of ethane

Question 11

What is the concentration of sugar (C12H22O11) in mol L-1 if its 20g are dissolved in enough water to make a final volume up to 2L?

Concentration of sugar(solute) in water(solvent) is same as Molarity of solution.

Formula for calculating Molarity is
Molarity of Solution = Number of moles of solute / Volume of solution in Litres

Number of moles of solute = Mass of solute / Molar Mass of solute

Molar mass of solute sugar (C12H22O11)
= 12 × Molar mass of Carbon(C)
+ 22 × Molar mass of Hydrogen(H)
+ 11 × Molar mass of Oxygen(O)

= 12 × 12 + 22 × 1 + 11 × 16
= 144 + 22 + 176
= 342

⇒ Molar mass of solute sugar (C12H22O11) = 342g

⇒ Number of moles of solute = Mass of solute(20g) / Molar Mass of solute(342g)
= 20/342
= 0.06 (approximately)

Number of moles of solute = 0.06 moles
Volume of solution = 2 L

Molarity of Solution = Number of moles of solute / Volume of solution in Litres
= 0.06 / 2
= 0.03 moles per L

Thus concentration of solution is 0.03 moles per L if 20g of sugar (C12H22O11) is dissolved in 2L of water.

Question 12

If 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) = 32 g mol-1

Molarity of given solution = 0.25 M

Formula
Molarity of a solution = Number of moles of solute (methanol) / Volume of solution

Formula
Number of moles of solute (methanol) = Mass of solute (methanol) / Molar mass of solute (methanol)

Using both of these two formulas
Molarity of a solution = Mass of solute (methanol) / Molar mass of solute (methanol) × Volume of solution

Density = Mass of solute (methanol) / Volume of solution

⇒ Molarity of a solution = Density / Molar mass of solute (methanol)

Density = 0.793 kg L-1
Molar mass of solute (methanol) = 32 g mol-1 = 0.032 kg mol-1

⇒ Molarity of a solution = 0.793 / 0.032 = 24.78 mol L-1 or 24.78 M

Using formula
M1V1 (For Given Solution) = M2V2 (For New Solution which we want to prepare)

Replacing
M1 = 24.78 M
M2 = 0.25 M
V2 = 2.5 L

⇒ 24.78 × V1 = 0.25 × 2.5

24.78 × V1 = 0.63

V1 = 0.63 / 24.78 = 0.0254 L

V1 = 24.4 mL

Thus 24.4 mL of methanol having density 0.793 kg L-1 is required to make 2.5 L of 0.25 M solution

Question 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-2

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

Pressure is defined as force per unit of area
Which can be written as following mathematically
Pressure = Force / Area

In this case, Force is the weight of air
Which can be calculated as Force = Mass × Acceleration (which is equals to g = 9.8 m s-2)

⇒ Pressure = Force / Area = Mass × 9.8 / Area

As per question Mass / Area = 1034 g cm-2
Converting this to SI units

Mass / Area = 1034 g cm-2 = 10340 kg m-2

Replacing this value in
⇒ Pressure = Force / Area = Mass × 9.8 / Area = 10340 × 9.8

Pressure = 101332 Pa or 1.01332 × 105 Pa

Thus if mass of air at sea level is 1034 g cm-2 then it’s pressure is 1.01332 × 105 Pa

Question 14

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

The S.I unit of mass is kilogram (kg). A kilogram is equal to the mass of a platinum-iridium cylinder kept at the International Bureau of Weights and Measures in France.

Question 15

Match the following prefixes with their multiples:

Below is the answer table

Question 16

What do you mean by Significant Figures?

All certain digits plus one doubtful digit represent Significant Figures. These depend purely upon precision of scale or instrument used for measurement.

For example – Let’s say oscillation of a pendulum is 3.45 second, in this number digits 3 and 4 are reliable while 5 is uncertain. So in total this number have 3 significant figures.

Question 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 per cent by mass
1ppm = 1 part out of 1 million parts

⇒ Mass percent of 15 ppm chloroform in H2O

= 15/106 × 100

= 1.5 × 10-3 %

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

Formula for calculating Molality is
Molality = Number of moles of solute / Weight of solvent(in Kg)

Number of moles of solute (chloroform) = Mass of chloroform / Molar mass of chloroform

Molar mass of chloroform(CHCl3)
= Molar mass of Carbon(C) + Molar mass of Hydrogen(H)
+ 3 × Molar mass of Chlorine(Cl)

= 12 + 1 + 3 × 35

= 12 + 1 + 105 = 119.5

Molar mass of chloroform(CHCl3) = 119.5

⇒ Number of moles of solute (chloroform) = Mass of chloroform / Weight of solvent(in Kg)
= 15/119.5 = 0.125

Number of moles of solute (chloroform) = 0.125

But in order to calculate Molality we need to find out Weight of solvent(in Kg)

Weight of Solution = Weight of Solvent + Weight of Solute
As per question
Weight of Solution(water + chloroform) = 106
Weight of Solute(chloroform) = 15

⇒ 106 = Weight of Solvent + 15

Weight of Solvent = 106 – 15

Molality = Number of moles of solute / Weight of solvent(in Kg)
= 0.125 / 106 – 15
= 0.125 / 999985

= 1.25 × 10−7

Molality = 1.25 × 10−7

Question 18

Express the following in the scientific notion:

(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 × 100

Question 19

How many significant figures are present in the following?

(i) 0.0025
2 significant figures

(ii) 208
3 significant figures

(iii) 5005
4 significant figures

(iv) 2.0034
5 significant figures

Question 20

Round up the following upto three significant figures:

(i) 34.216
34.22

(ii) 10.4107
10.4

(iii) 0.04597
0.0460

(iv) 2808
2810

Question 21

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

(a) Which law of chemical combination is obeyed by the above experimental data? Give its statement.

If we fix mass of dinitrogen in the reaction to 14 g then as per above data 16 g, 32 g, 16 g, 40 g of dinitrogen will be required.

Which is in proportion 16 : 32 : 16 : 40 = 1 : 2 : 2 : 5

As this ration bears a whole number ratio of 1:2:2:5
Hence this reaction obeys Law of Multiple Proportions.

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

(i) 1 km = ………. mm = ……. pm
1 km = 1000 m
1 m = 100 cm
1 cm = 10 mm
⇒ 1 km = 1000 × 100 × 10 mm = 106 mm
1 km = 106 mm

1 km = 1000 m
1 m = 1012 pm
⇒ 1 km = 1000 × 1012 pm = 1015 pm
1 km = 1015 pm

1 km = 106 mm = 1015 pm

(ii) 1 mg = ………… kg = ………. ng
1 kg = 1000 g
1g = 1000 mg

⇒ 1 mg = 10-3 g
⇒ 1 g = 10-3 kg

⇒ 1 mg = 10-3 × 10-3 = 10-6 kg

1 mg = 10-6 kg

1 ng = 10-9 g
1 mg = 10-3 g

⇒ 1 mg = 10-3 g
⇒ 1 g = 109 ng

⇒ 1 mg = 10-3 × 109 ng = 106 ng

1 mg = 106 ng

1 mg = 10-6 kg = 106 ng

(iii) 1 mL = …………… L = ………….. dm3
1 L = 1000 mL

1 mL = 10-3 L

1 dm3 = 10-3 m3
1 m3 = 1000 L
1 L = 1000 mL

⇒ 1 mL = 10-3 L
⇒ 1 L = 10-3 m3
⇒ 1 m3 = 103 dm3

⇒ 1 mL = 10-3 × 10-3 × 103 dm3 = 103 dm3

⇒ 1 mL = 103 dm3

1 mL = 10-3 L = 103 dm3

Question 22

If the speed of light is 3.0 × 108 m s-1, calculate the distance covered by light in 2.00 ns.

Speed of light = 3.0 × 108 m s-1
Time = 2.00 ns = 2 × 10-9 s

Distance = Speed × Time
= 3.0 × 108 m s-1 × 2 × 10-9 s

= 3.0 × 2 × 108 × 10-9 m

= 6.0 × 10-1 m

= 0.60 m

Thus light travel 0.60 m in 2.00 ns

Question 23

In a reaction
A + B2 → AB2

Identify the limiting agent if any in the following reaction mixtures:

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

As per chemical reaction A + B2 → AB2
1 atom of A ⇒ 1 molecule of B2

⇒ 300 atoms of A react with 300 molecules of B but only 200 molecules are given thus B is limiting agent.

(ii) 2 mol A + 3 mol B

As per chemical reaction A + B2 → AB2
1 mol of A ⇒ 1 mol of B

⇒ 2 mol of A will react with 2 mol of B but 3 moles of B are given. But only 2 are required thus A is limiting agent.

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

As per chemical reaction A + B2 → AB2
1 atom of A ⇒ 1 molecule of B2

⇒ 100 atoms of A can react with 100 molecules of B. And 100 molecules of each of A and B are given thus there’s no limiting agent in this reaction.

(iv) 5 mol A + 2.5 mol B

As per chemical reaction A + B2 → AB2
1 mol of A ⇒ 1 mol of B

⇒ 5 mol of A can react with 5 mol of B but as per question only 2.5 mol of B is given thus B is limiting agent in this reaction.

(v) 2.5 mol A + 5 mol B

As per chemical reaction A + B2 → AB2
1 mol of A ⇒ 1 mol of B

⇒ 2.5 mol of A can react with 2.5 mol of B but as per question 5 mol of B is given(B is in excess as compared to what is needed) thus in this reaction A is limiting agent.

Question 24

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

N2 (g) + H2 (g) → 2NH3 (g)

(i) Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen

Let’s first balance the given chemical equation
N2 (g) + 3H2 (g) → 2NH3 (g)

As per this balanced chemical equation
1 mol of N2 reacts with 3 mol of H2

Converting moles to grams
28 g of N2 reacts with 6 g of H2

⇒ 1 g of N2 reacts with 6/28 g of H2

As per question we’re given 2.00 × 103 g dinitrogen
So let’s calculate amount of H2 which can react with this much of dinitrogen

2.00 × 103 g of N2 reacts with 6/28 × 2.00 × 103 g of H2

⇒ 2.00 × 103 g of N2 reacts with 0.429 × 103 g of H2

⇒ 2000g of N2 reacts with 429 g of H2

But as per question only 1.00 × 103 g = 1000g of dihydrogen is given which is substantially more as compared to what’s needed to react with dinitrogen.

Thus in this reaction dinitrogen(N2) is Limiting Reagent
So the amount of ammonium formed will depend on amount of dinitrogen.

N2 (g) + 3H2 (g) → 2NH3 (g)

As per this balanced chemical equation 1 mol of N2 produces 2 mol of NH3

Writing this in terms of grams

⇒ 28 g of N2 produces 34 g of ammonia

⇒ 1 g of N2 produces 34/28 g of ammonia

As per question, 2.00 × 103 g of dinitrogen(N2) is given

⇒ 2.00 × 103 g of N2 produces 34/28 × 2.00 × 103 g of ammonia

Simplifying this

⇒ 2.00 × 103 g of N2 produces 2.428 × 103 g of ammonia

Thus if 2.00 × 103 g dinitrogen reacts with 1.00 × 103 g of dihydrogen then 2.428 × 103 g of ammonia will be formed.

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

Yes some amount of hydrogen will be left, as explained earlier in above question hydrogen is available in excess as compared to what’s needed.

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

Hydrogen will left unreacted, please see (i) of this question above where I did exact calculations.
So 2.00 × 103 g of N2 need only 429 g of H2 but as per question we’re given 1.00 × 103 g = 1000 g

Thus amount of hydrogen left will be 1000 – 429 = 571 g

Question 25

How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?

Calculating Molar Mass of Na2CO3

Molar Mass of Na2CO3
= 2 × Molar Mass of Na
+ Molar Mass of C
+ Molar Mass of O

= 2 × 23 + 12 + 16

= 46 + 12 + 16 = 70

Molar Mass of Na2CO3 = 70

Thus 0.50 mol of Na2CO3 = 0.50 × 70 = 35 g of Na2CO3

0.50 M Na2CO3 means 0.50 mol of Na2CO3 in 1 L of water(or some other solvent)

⇒ 35 g of Na2CO3 in 1 L of water(or some other solvent)

⇒ 0.50 mol of Na2CO3 means 35 g of Na2CO3 while 0.50 M Na2CO3 means 35 g of Na2CO3 dissolved in some liquid(like water or some other solvent)

Question 26

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

Chemical equation of reaction between dihydrogen and dioxygen
H2 (g) + O2 (g) → H2O (g)

Balancing above chemical equation
2H2 (g) + O2 (g) → 2H2O (g)

Thus

2 volumes of H2 + 1 volume of O2 produces 2 volumes of H2O vapours

Multiplying this by 5

⇒ 10 volume of H2 + 5 volumes of O2 produces 10 volumes of H2O vapours

Hence if 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, then 10 volumes of water vapour will be produced.

Question 27

Convert the following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg

Question 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) 1 g of Cl2 (g)

Question 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)

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

0.040 = Number of moles of C2H5OH / Number of moles of solution

Number of moles of solution = Number of moles of C2H5OH + Number of moles of water

⇒ 0.040 = Number of moles of C2H5OH / Number of moles of C2H5OH + Number of moles of water

Number of moles present in 1L water

Number of moles of water = 1000g / 18 g mol-1 = 55.55 mol

Number of moles of water = 55.55 mol

⇒ 0.040 = Number of moles of C2H5OH / Number of moles of C2H5OH + 55.55

Let’s suppose that Number of moles of C2H5OH = y

0.040 = y / y + 55.55

0.040(y + 55.55) = y

0.040y + 2.222 = y

2.222 = y – 0.040y = 0.96y

0.96y = 2.222

y = 2.222 / 0.96 = 2.314

y = 2.314

⇒ Number of moles of C2H5OH = y = 2.314 mol

⇒ Volume of solution = 1 L

⇒ Molarity = 2.314 mol / 1 L = 2.314 mol L-1

Thus molarity of solution(containing ethanol and water) will be 2.314 mol L-1 if mole fraction of ethanol is 0.040

Question 30

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

1 mole of any substance contains 6.022 × 1023 atoms

⇒ 1 mole of 12C contains 6.022 × 1023 atoms

1 mole of 12C = 12 g of 12C

⇒ 12 g of 12C contains 6.022 × 1023 atoms

⇒ 1 atom of 12C contains 12/6.022 × 1023 = 2 × 10-23 g

Thus mass of one 12C atom in g is 2 × 10-23

Question 31

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

\begin{equation} \textbf{(i) } \frac { 0.02856 × 298.15 × 0.112 } { 0.5785 } \end{equation}

As this calculation involves just multiplication and division thus final answer should contain significant numbers equal to least precise number in the calculation which is 0.112

Thus final answer should have 3 significant figures

(ii) 5 × 5.364

As this calculation involves just multiplication and division thus final answer should contain significant numbers equal to least precise number in the calculation which is 5.364

Thus final answer should have 4 significant figures

(iii) 0.0125 + 0.7864 + 0.0215

As this calculation involves just addition thus final answer must contain decimal numbers equal to least number of decimal numbers amongst all the numbers involved in calculations.

0.0125 have 4 decimal numbers
0.7864 have 4 decimal numbers
0.0215 have 4 decimal numbers

Thus least number of decimal places amongst all the number involved in the calculation is 4.

Final answer of calculation should have 4 decimal places.

Question 32

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

Molar Mass of Argon
= Σ (Isotopic Molar Mass × Proportion of it’s Abundance in nature)

= 35.96755 × 0.337/100 + 37.96272 × 0.063/100 + 39.9624 × 99.600/100

= 0.1212106435 + 0.0239165136 + 39.8025504

= 39.94767756 ≈ 39.947 g mol-1

Thus Molar Mass of Argon is 39.947 g mol-1 based upon proportion of abundance of it’s three isotopes(36Ar, 38Ar, 40Ar) in nature.

Question 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

Formula used in this question
Number of atoms = Number of moles × 6.022 × 1023

⇒ Number of atoms (in 52 moles of Ar)
= 52 × 6.022 × 1023 = 313 × 1023

⇒ Number of atoms (in 52 u of He)
1 atom of He – 4 u of He

⇒ 13 atoms of He – 52 u of He

⇒ Number of atoms (in 52 g of He)
Firstly we need find out how many moles is 52 g of He

Number of moles = Mass/Molar Mass = 52/4 = 13

⇒ Number of atoms (in 13 moles of He)
= 13 × 6.022 × 1023
= 78 × 1023

Question 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 in STP) of this welding gas is found to weigh 11.6 g

Calculate (i) Empirical Formula
(ii) Molar Mass of the gas
(iii) Molecular Formula

Question 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?

As per given chemical equation

1 mol of CaCO3 (s) reacts with 2 mol of HCl (aq)

25 mL of 0.75 M HCl contains = 0.75 × 25/1000
= 0.75 × 0.025 = 0.01875 moles

Thus 25 mL of 0.75 M HCl contains 0.01875 moles of HCl

1 mol of CaCO3 (s) reacts with 2 mol of HCl (aq)

⇒ 1 mol of HCl (aq) reacts with 1/2 mol of CaCO3 (s)

⇒ 0.01875 moles of HCl (aq) reacts with 1/2 × 0.01875 mol of CaCO3 (s)
= 0.009375 mol of CaCO3 (s)

Thus 0.01875 moles of HCl (aq) reacts with 0.009375 mol of CaCO3 (s)

Molar Mass of CaCO3 (s) = 100 g

⇒ Converting 0.009375 mol of CaCO3 (s) to grams
= 0.009375 × 100
= 0.9375 g

Thus for chemical reaction CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)
0.9375 g of CaCO3 is required to react completely with 25 mL of 0.75 M HCl

Question 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) → 2 H2O (l) + MnCl2 (aq) + Cl2 (g)

How many grams of HCl react with 5.0 g of Manganese Dioxide?

As per given balanced chemical equation
4 moles of HCl (aq) reacts with 1 mol of MnO2 (s)

Molar Mass of HCl (aq) = 36.5 g
Molar Mass of MnO2 (s) = 87 g

⇒ 4 × 36.5 g of HCl (aq) reacts with 87 g of MnO2 (s)

146 g of HCl (aq) reacts with 87 g of MnO2 (s)

⇒ 1 g of MnO2 (s) reacts with 146/87 g of HCl (aq)

Thus
5 g of MnO2 (s) reacts with 146/87 × 5 g of HCl (aq)
= 8.39 g of HCl (aq)

Thus 8.39 g of HCl (aq) reacts with MnO2 (s) in chemical reaction 4 HCl (aq) + MnO2 (s) → 2 H2O (l) + MnCl2 (aq) + Cl2 (g)