# 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

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/NA g
4. 12 × NA g

12/NA 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 (NA) are related as

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

n = N / NA

Explanation

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

N = n × NA

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

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

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

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. NA mol

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

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 MA > MB. The molar volume (Vm) at given pressure and temperature, are like

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

Vm(A) < Vm(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:

Vm(A) < Vm(B)

Question 10

A sample contains m g of a compound X (Mx = 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

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.

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

Question 2

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

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]

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]

6.022 × 1023

Explanation

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

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]

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.

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

Question 8

Which scientist gave the concept of a molecule ?

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]

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]

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]

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?

Molar mass of N2 = 2 × Atomic mass of N

= 2 × 14 g

= 28 g or 28 g/mol

[Each of Mark-2]

Question 1

What is Avogadro number? What is its value?

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.

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.

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]

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]

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]

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]

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]

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]

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]

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.

[Each of Mark-3]

Question 1

Define molecular mass and gram molecular mass.

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]

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]

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/NA, where N is the number of N2 molecules and NA 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)(NA)

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?

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]

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

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