Chapter 4.2 – Mole Concept | Chapter Solution Class 9

Publisher	: Santra publication pvt. ltd.
Book Name	: Madhyamik Physical Science And Environment
Class	: 9 (Madhyamik)
Subject	: Physical Science
Chapter Name	: Mole Concept

Multiple Choice Questions (MCQ)

[Each of Mark – 1]

Question 1

1 mole is the number of atoms in exactly 

1. 1 g of ¹²C
2. 12 g of ¹²C
3. 1/12 g of ¹²C
4. none of these

Answer

12 g of 12C

Explanation

1 mole is the number of atoms in exactly 12 g of ¹²C.

This is because a mole is defined as the amount of substance that contains the same number of entities (such as atoms or molecules) as there are atoms in 12 grams of the isotope carbon-12 (12C). Since the atomic mass of carbon-12 is 12, one mole of carbon-12 contains 6.022 × 10²³ atoms, which is known as Avogadro’s number.

Question 2

The mass of a C-12 atom is 

1. 12 g
2. 1/12 g
3. 12/N_A g
4. 12 × N_A g

Answer

12/N_A g

Explanation

The mass of a C-12 atom is approximately 12 atomic mass units (amu), which is equivalent to 1.9926 × 10^-23 grams.

Question 3

The number of mole (n), the number of particles (N) and the Avogadro number (N) and the Avogadro Number (N_A) are related as

1. n = N_A\over N
2. N\over N_A = N
3. n = N\over N_A  
4. n = N.N_A

Answer

n = N / N_A

Explanation

The correct equation relating the number of moles (n), the number of particles (N), and Avogadro’s number (NA) is:

N = n × N_A

Question 4

16 g of oxygen is the mass of oxygen of

1. one oxygen atom
2. one oxygen molecule
3. 1 mole oxygen molecule
4. one mole oxygen atom

Answer

one mole of oxygen molecules

Explanation

The molar mass of oxygen (O₂) is 32 g/mol, which means that one mole of O₂ has a mass of 32 grams.

Question 5

6.4 g of oxygen is equivalent to

1. 0.4 mol of oxygen atoms
2. 0.4 mol of oxygen molecules
3. 0.2 mol of oxygen atoms
4. none of these

Answer

0.4 mol of Oxygen atoms

Explanation

If the 6.4 g corresponds to O atoms, then the number of moles of O atoms is:

n = m/M = 6.4 g / 16 g/mol = 0.4 mol

Therefore, 6.4 g of O is equivalent to 0.4 mol of O atoms.

Question 6

0.10 mol of carbon is equivalent to

1. 1.2 g of carbon
2. 1\over 12 g of carbon
3. 12.0 g of carbon
4. 0.1 g of carbon

Answer

1.2 g of carbon

Explanation

To determine the mass of 0.10 mol of carbon, we need to know the molar mass of carbon. The molar mass of carbon is 12.0 g/mol, which is the average atomic mass of carbon taking into account its isotopes.

Using the molar mass of carbon, we can calculate the mass of 0.10 mol of carbon:

m = n × M

= 0.10 × 12.0 = 1.20 g

Question 7

A sample is known to contain 2 × 6.022 × 10²³ hydrogen atoms. How many hydrogen atoms does it contain?

1. 1 mol
2. 1\over 2 mol
3. 2 mol
4. N_A mol

Answer

2 mol

Explanation

To determine the number of hydrogen atoms in the sample in terms of moles, we can use Avogadro’s number as a conversion factor:

n = N\over N_A

where n is the number of moles, N is the number of particles, and NA is Avogadro’s number.

Substituting the values given in the problem, we get:

n = 2 × 6.022 × 10^{23} \over 6.022 × 10^{23}  

Simplifying, we get:

n = 2

Therefore, the sample contains 2 moles of hydrogen atoms.

Question 8

The formula mass of water H₂O is

1. 33
2. 17
3. 10
4. 18

Answer

18

Explanation

The formula mass of water (H₂O) is 18.

The formula mass is the sum of the atomic masses of all the atoms in a chemical formula. For water, the formula mass is:

Formula mass of H₂O = 2 × (atomic mass of H) + atomic mass of O

= 2 × 1  + 16 = 18

Question 9

Consider two gases A and B or ideal gas with pressure M_A > M_B. The molar volume (V_m) at given pressure and temperature, are like

1. V_m(A) = V_m(B)
2. V_m(A) > V_m(B)
3. V_m(A) < V_m(B)
4. no relation

Answer

V_m(A) < V_m(B)

Explanation

At a given pressure and temperature, the molar volume (Vm) of two ideal gases A and B with pressures MA and MB (where MA > MB) are related as follows:

V_m(A) < V_m(B)

Question 10

A sample contains m g of a compound X (M_x = molar mass). The sample contains n mol of x. Then n, m and M are related as

1. M = m.n
2. M = m/n
3. M = n/m
4. n = m.M

Answer

M = m/n

Explanation

The correct equation relating n, m, and M is:

m = n × M

This equation shows that the mass (m) of a compound X is equal to the product of the number of moles (n) and the molar mass (M) of the compound. Alternatively, we can rearrange the equation to get:

M = m/n

Answer in one word or in one sentence

[Each of Mark-1]

Question 1

Mention the value of the Avogadro number.

Answer

The value of Avogadro’s number, denoted as N_A, is approximately 6.022 × 10²³

Question 2

What is the mass of 1 g mol of oxygen? [Ans. 32 g]

Answer

The molar mass of oxygen (O₂) is 32. Therefore, the mass of 1 g mol of oxygen (O₂) is:

mass = molar mass × amount of substance mass

= 32 × 1

= 32 g

Therefore, the mass of 1 g mol of oxygen is 32 grams. This means that 1 mole of oxygen (O₂) contains 32 grams of oxygen.

Question 3

Give the number of molecules present in 1.8 g water. [Ans. 6.022× 10²²]

Answer

Molar mass of water (H₂O) is = 1 × 2 + 16 = 18

So, the number of moles of water in 1.8 g of water is:

Number of moles = mass/molar mass

= 1.8 / 18

= 0.1 mol

Next, we can use Avogadro’s number to convert from moles to the number of molecules:

Number of molecules = Number of moles × Avogadro’s number

= 0.1 × 6.022 × 10²³

= 6.022 × 10²² molecules

Therefore, there are approximately 6.022 × 10²² water molecules present in 1.8 g of water.

Question 4

Give the number of oxygen molecules in 1 mol of oxygen. [Ans. 6.022 × 10²³]

Answer

6.022 × 10²³

Explanation

One mole of oxygen contains Avogadro’s number (N_A) of oxygen molecules.

The value of Avogadro’s number is approximately 6.022 × 10²³ molecules per mole. Therefore, the number of oxygen molecules in 1 mole of oxygen is:

1 mole of oxygen = Avogadro’s number of oxygen molecules

= 6.022 × 10²³ oxygen molecules

Question 5

Mention the molar volume of a gas (ideal) at STP.

Answer

The molar volume of an ideal gas at standard temperature and pressure (STP) is 22.4 liters.

Question 6

What is the volume of 32 g of a gas (having molar mass of 64), at STP ? [Ans. 11.2 L]

Answer

Step 1: Calculation of Number of moles

Number of moles = mass \over molar\ mass

= 32 \over 64  = 0.5 mol

Step 2: Calculation of volume

At STP, the molar volume of an ideal gas is 22.4 L/mol.

Therefore, the volume of 0.5 mol of the gas at STP is:

volume = moles × molar volume

= 0.5 × 22.4 = 11.2 L

Question 7

Give the number of atoms (¹²C) in 12 g of carbon.

Answer

One mole of carbon atoms contains Avogadro’s number (N_A) of carbon atoms, which is approximately 6.022 × 10²³ atoms.

Question 8

Which scientist gave the concept of a molecule ?

Answer

The concept of a molecule was proposed by the French scientist Amedeo Avogadro in the early 19th century.

Question 9

Give the number of molecules is 2 g hydrogen. [Ans. 6.022 × 10²³]

Answer

Step 1: Calculation of Molar mass

The molar mass of hydrogen (H₂) = 2 × 1 = 2

Step 2: Calculation of number of moles

Number of moles of hydrogen = Mass\ of\ hydrogen \over Molar\ mass\ of\ hydrogen

= 2\over2 = 1

Step 3: Calculation of number of molecules

Number of molecules of hydrogen = Number of moles of hydrogen × Avogadro’s number

= 1 × 6.022 × 10²³

= 6.022 × 10²³ molecules

Question 10

64 g O₂ is equivalent to how many moles of oxygen? [Ans. 2 mol]

Answer

Step 1: Calculation of molar mass

The molar mass of oxygen (O₂) = 16 × 2 = 32

Step 2: Calculation of moles

Number of moles of oxygen = 64 \over 32 = 2 mol

Therefore, 64 g of O₂ is equivalent to 2 moles of oxygen.

Question 11

How many oxygen atoms are there in 1 mole of CO₂? [Ans. 12.044 × 10²³]

Answer

Number of Oxygen atom = Number of mole × Number of O atoms in one molecule

= 1 mole × 2

= 6.022 × 10²³ × 2

= 12.044 × 10²³

Alternative method

Step 1: Calculation of molar mass

Molar mass = 12 + 16 × 2 = 44

Step 2: Calculation of moles

Calculate the number of moles of oxygen atoms: In one mole of CO₂, there are two oxygen atoms. Therefore, the number of moles of oxygen atoms in 1 mole of CO₂ is:

Number of moles of O atoms = 2 moles CO₂

Step 3: Calculation number of atom

Number of oxygen atoms = Number of moles of O atoms × Avogadro’s number

= 2 × 6.022 × 10²³

= 12.044 × 10²³ atoms

Therefore, there are approximately 12.044 × 10²³ oxygen atoms in 1 mole of CO₂.

Question 12

What is gram molecular mass of nitrogen?

Answer

Molar mass of N₂ = 2 × Atomic mass of N

= 2 × 14 g

= 28 g or 28 g/mol

Short answer type questions

[Each of Mark-2]

Question 1

What is Avogadro number? What is its value?

Answer

Avogadro’s number is 6.022 × 10²³, representing the number of particles in one mole of a substance.

Question 2

Define molecular mass on C-12 scale.

Answer

Molecular mass on the C-12 scale is the sum of the atomic masses of the constituent atoms, relative to the mass of a C-12 atom.

Question 3

What is molar volume? Explain.

Answer

Molar volume is the volume occupied by one mole of a substance, typically measured at standard temperature and pressure.

Question 4

If the number of molecules present in 9 g of water is n, what will be the number of molecules in 44 g of carbon dioxide? [H= 1, C = 12, O= 16]. [Ans. 2n]

Answer

Number of molecules in 9 g H₂O = 3.011 × 10²³ (= n)

and the number of molecules in 44 g of carbon dioxide = 6.022 × 10²³

= 2 × 3.011 × 10²³

= 2n [∵ 3.011 × 10²³ = n]

Question 5

SO₃ is prepared by oxidizing SO₂ with oxygen. How many grams of SO₂ will be required to produce 40 g of SO₃ ? [O = 16, S = 32] [Ans. 32 g]

Answer

The balanced chemical equation for the reaction of SO₂ with O₂ to form SO₃ is:

2 SO₂ + O₂ → 2 SO₃

Molar mass of SO₃ = 1 × 32  + 3 × 16 = 80

Number of moles of SO₃ = 40\over 80 = 0.5 mol

Since the molar ratio of SO₂ to SO₃ is 1:1, we need the same number of moles of SO₂ as SO₃ to react completely. Therefore, the number of moles of SO₂ required is also 0.5 mol.

Finally, we can use the molar mass of SO₂ to calculate the mass of SO₂ required:

Mass of SO₂ = 0.5 × 64 = 32 g

Therefore, 32 g of SO₂ will be required to produce 40 g of SO₃.

Question 6

At STP, what is the volume of 7 g of nitrogen gas? [Ans. 5.6 L]

Answer

Step 1: Calculation of number of moles

moles of N₂ = mass\ of\ N_2 \over molar\ mass\ of\ N_2

= 7 \over 28 = 0.25 mol

Step 2: Calculation of Volume

At STP, one mole of any ideal gas occupies 22.4 L. Therefore, the volume of 0.25 mol of nitrogen gas is:

volume of N₂ = 0.25 × 22.4 = 5.6 L

Therefore, the volume of 7 g of nitrogen gas at STP is 5.6 L.

Question 7

At STP, 0.44 g of a gas has a volume of 224 cm³. What is the molecular mass of the gas? [Ans. 44 g]

Answer

Mass of gas = 0.44 g

Volume = 224 cm³ or 0.224 L

Number of Moles (n) = Given\ Volume \over 22.4

= 0.224\over 22.4

= 0.01

Molecular mass = Number\ of\ moles (n) \over Given\ Mass

= 0.44 \over 0.01 = 44 g

Question 8

What is the number of molecules in each of 32 g of O₂ and 44 g of CO₂? [Ans. 6.022 × 10²³]

Answer

Given Mass of O₂ = 32 g

Gram Molecular (Molar) mass = 16 × 2 = 32 g

Number of Moles = 32\over 32 = 1

Number of molecules = Number of moles × Avogadro’s Number

= 1 × 6.022 × 10²³

∴ Number of molecules in 32 g of O₂ = 6.022 × 10²³

Given Mass of CO₂ = 44 g

Gram Molecular Mass of CO₂ = 12 + 16 × 2 = 44 g

Number of Moles = 44\over 44 = 1

Number of molecules = Number of moles × Avogadro’s Number

= 1 × 6.022 × 10²³

∴ Number of molecules in 44 g of CO₂ = 6.022 × 10²³

Question 9

What is the volume of 11 g of CO₂ at STP? [Ans. 5.6 L]

Answer

Given mass = 11 g

Molar mass = 12 + 16 × 2 = 44

Step 1: Calculation of number of moles

Moles of CO₂ = Given\ mass \over Molar\ mass

= 11 \over 44  = 0.25 mol

Step 2: Calculation of number of moles

volume of CO₂ = number of moles of CO2 × molar volume at STP

= 0.25 × 22.4 = 5.6 L

Therefore, the volume of 11 g of CO2 at STP is 5.6 L.

Question 10

What is the number of molecules in 11.2 L of O₂ at STP? [Ans. 3.011 × 10²³]

Answer

Step 1: Calculation of number of moles

moles of O₂ = Given Volume\ of\ O_2 \over 22.4

= 11.2 \over 22.4

= 0.5 mol

Step 2: Calculation of number of molecules

number of molecules of O₂ = moles of O₂ × Avogadro’s number

= 0.5 × 6.022 × 10²³

= 3.011 × 10²³ molecules

Therefore, the number of molecules in 11.2 L of O₂ at STP is 3.011 × 10²³ molecules.

Long answer type questions

[Each of Mark-3]

Question 1

Define molecular mass and gram molecular mass.

Answer

Molecular mass is the sum of the atomic masses of all atoms in a molecule, expressed in atomic mass units or grams per mole.

Gram molecular mass, or molar mass, is the mass of one mole of a substance, expressed in grams per mole. It is numerically equal to the molecular mass in atomic mass units.

Question 2

What is Avogadro’s number? How to determine the mass of one oxygen atom using the Avogadro number? [Ans. 2.66 × 10^-23 g]

Answer

The number of molecules present in one gram molecule of a substance is called Avogadro’s number. It’s value is 6.023 × 10²³.

mass of one oxygen atom = mass\ of\ one\ mole\ of\ O_2 \over Avogadro's\ number × 2

= 32 \over 6.022 × 10^{23} × 2

= 2.66 × 10^-23 g

Question 3

At STP, n number of N₂ molecules occupies a volume of V litre. What volume n/2 number of CO₂ molecules will occupy at STP? [Ans. V/2 L]

Answer

At STP (Standard Temperature and Pressure), the temperature is 273 K and the pressure is 1 atm. One mole of any gas at STP occupies 22.4 liters of volume.

Therefore, n number of N₂ molecules will occupy a volume of V litres at STP.

n(N₂) = N/N_A, where N is the number of N₂ molecules and N_A is Avogadro’s number

The number of moles of N₂ will be equal to the number of moles of CO₂, assuming both gases are at STP. We can calculate the number of CO2 molecules using the following equation:

n(CO₂) = (n/2)(N_A)

The volume occupied by n/2 number of CO₂ molecules can be calculated using the volume of one mole of CO2 at STP:

V(CO₂) = n(CO₂) × 22.4 L/mol

Substituting the values we get:

n(N₂) = N/NA n(CO2) = (n/2)(NA)

V(CO2) = n(CO2) × 22.4 L/mol

We can simplify n(CO2) as follows:

n(CO2) = (n/2)(NA)

n(CO2) = n(N2)/2

Substituting this in the equation for V(CO2), we get:

V(CO2) = (n(N2)/2) x 22.4 L/mol V(CO2) = (n(N2) x 22.4 L/mol)/2 V(CO2) = V/2

Question 4

What is the molar volume of a gas? What is the value of STP?

Answer

The volume occupied by one gram-molecule of any gaseous substance at a fixed temperature and pressure is known as the molar volume.

STP means standard temperature and pressure. Where standard pressure is 1 atm. and the standard temperature is 273 Kelvin (K).

Question 5

51 grams of ammonia is equivalent to how many gram moles of ammonia? How many molecules of ammonia are there? [Ans. 3 mol, 1.806 × 10²⁴]

Answer

Given mass of ammonia = 51 g

Molar mass of ammonia (NH₃) = 14 + 1 × 3 = 17 g

Step 1: Calculation of mole

number of moles = mass \over molar\ mass

= 51 \over 17

= 3 moles of ammonia

Step 2: Calculation of molecules

number of molecules = number of moles × Avogadro’s number

= 3 × 6.022 × 10²³

= 1.8066 × 10²⁴