Publisher |
: Santra publication pvt. ltd. |

Book Name |
: Madhyamik Physical Science And Environment |

Class |
: 9 (Madhyamik) |

Subject |
: Physical Science |

Chapter Name |
: Mole Concept |

Table of Contents

Toggle## Multiple Choice Questions (MCQ)

**[Each of Mark – 1]**

**Question 1**

**1 mole is the number of atoms in exactly **

**1 g of**^{12}C**12 g of**^{12}C**1/12 g of**^{12}C**none of these**

**Answer**

12 g of 12C

**Explanation**

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

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 × 10^{23} atoms, which is known as Avogadro’s number.

**Question 2**

**The mass of a C-12 atom is **

**12 g****1/12 g****12/N**_{A}g**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**

**n = N_A\over N****N\over N_A = N****n = N\over N_A****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**

**one oxygen atom****one oxygen molecule****1 mole oxygen molecule****one mole oxygen atom**

**Answer**

one mole of oxygen molecules

**Explanation**

The molar mass of oxygen (O_{2}) is 32 g/mol, which means that one mole of O_{2} has a mass of 32 grams.

**Question 5**

**6.4 g of oxygen is equivalent to**

**0.4 mol of oxygen atoms****0.4 mol of oxygen molecules****0.2 mol of oxygen atoms****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**

**1.2 g of carbon****1\over 12 g of carbon****12.0 g of carbon****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 × 10 ^{23} hydrogen atoms. How many hydrogen atoms does it contain?**

**1 mol****1\over 2 mol****2 mol****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 H _{2}O is**

**33****17****10****18**

**Answer**

18

**Explanation**

The formula mass of water (H_{2}O) 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 H_{2}O = 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**

**V**_{m}(A) = V_{m}(B)**V**_{m}(A) > V_{m}(B)**V**_{m}(A) < V_{m}(B)**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**

**M = m.n****M = m/n****M = n/m****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 × 10^{23}

**Question 2**

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

**Answer**

The molar mass of oxygen (O_{2}) is 32. Therefore, the mass of 1 g mol of oxygen (O_{2}) 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 (O_{2}) contains 32 grams of oxygen.

**Question 3**

**Give the number of molecules present in 1.8 g water. [Ans. 6.022**×** 10 ^{22}]**

**Answer**

Molar mass of water (H_{2}O) 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 × 10^{23}

= 6.022 × 10^{22} molecules

Therefore, there are approximately 6.022 × 10^{22} 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 × 10 ^{23}]**

**Answer**

6.022 × 10^{23}

**Explanation**

One mole of oxygen contains Avogadro’s number (N_{A}) of oxygen molecules.

The value of Avogadro’s number is approximately 6.022 × 10^{23} 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 × 10^{23} 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 ( ^{12}C) 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 × 10^{23} 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 × 10 ^{23}]**

**Answer**

**Step 1: Calculation of Molar mass**

The molar mass of hydrogen (H_{2}) = 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 × 10^{23 }

= 6.022 × 10^{23} molecules

**Question 10**

**64 g O _{2} is equivalent to how many moles of oxygen? [Ans. 2 mol]**

**Answer**

**Step 1: Calculation of molar mass**

The molar mass of oxygen (O_{2}) = 16 × 2 = 32

**Step 2: Calculation of moles**

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

Therefore, 64 g of O_{2} is equivalent to 2 moles of oxygen.

**Question 11**

**How many oxygen atoms are there in 1 mole of CO _{2}? [Ans. 12.044 × 10^{23}]**

**Answer**

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

= 1 mole × 2

= 6.022 × 10^{23} × 2

= 12.044 × 10^{23}

**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 CO_{2}, there are two oxygen atoms. Therefore, the number of moles of oxygen atoms in 1 mole of CO_{2} is:

Number of moles of O atoms = 2 moles CO_{2}

**Step 3: Calculation number of atom**

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

= 2 × 6.022 × 10^{23}

= 12.044 × 10^{23} atoms

Therefore, there are approximately 12.044 × 10^{23} oxygen atoms in 1 mole of CO_{2}.

**Question 12**

**What is gram molecular mass of nitrogen?**

**Answer**

Molar mass of N_{2} = 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 × 10^{23}, 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 H_{2}O = 3.011 × 10^{23} (= n)

and the number of molecules in 44 g of carbon dioxide = 6.022 × 10^{23}

= 2 × 3.011 × 10^{23}

= 2n [∵ 3.011 × 10^{23} = n]

**Question 5**

**SO _{3} is prepared by oxidizing SO_{2} with oxygen. How many grams of SO_{2} will be required to produce 40 g of SO_{3} ? [O = 16, S = 32] [Ans. 32 g] **

**Answer**

The balanced chemical equation for the reaction of SO_{2} with O_{2} to form SO_{3} is:

2 SO_{2} + O_{2} → 2 SO_{3}

Molar mass of SO_{3} = 1 × 32 + 3 × 16 = 80

Number of moles of SO_{3} = 40\over 80 = 0.5 mol

Since the molar ratio of SO_{2} to SO_{3} is 1:1, we need the same number of moles of SO_{2} as SO_{3} to react completely. Therefore, the number of moles of SO_{2} required is also 0.5 mol.

Finally, we can use the molar mass of SO_{2} to calculate the mass of SO_{2} required:

Mass of SO_{2} = 0.5 × 64 = 32 g

Therefore, 32 g of SO_{2} will be required to produce 40 g of SO_{3}.

**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 N_{2} = 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 N_{2} = 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 cm ^{3}. What is the molecular mass of the gas? [Ans. 44 g] **

**Answer**

Mass of gas = 0.44 g

Volume = 224 cm^{3} 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 O _{2} and 44 g of CO_{2}? [Ans. 6.022 × 10^{23}] **

**Answer**

Given Mass of O_{2} = 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 × 10^{23}

∴ Number of molecules in 32 g of O_{2} = 6.022 × 10^{23}

Given Mass of CO_{2} = 44 g

Gram Molecular Mass of CO_{2} = 12 + 16 × 2 = 44 g

Number of Moles = 44\over 44 = 1

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

= 1 × 6.022 × 10^{23}

∴ Number of molecules in 44 g of CO_{2} = 6.022 × 10^{23}

**Question 9**

**What is the volume of 11 g of CO _{2} 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 CO_{2} = Given\ mass \over Molar\ mass

= 11 \over 44 = 0.25 mol

**Step 2: Calculation of number of moles**

volume of CO_{2} = 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 O _{2} at STP? [Ans. 3.011 × 10^{23}]**

**Answer**

**Step 1: Calculation of number of moles**

moles of O_{2} = 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 O_{2} = moles of O_{2} × Avogadro’s number

= 0.5 × 6.022 × 10^{23}

= 3.011 × 10^{23} molecules

Therefore, the number of molecules in 11.2 L of O_{2} at STP is 3.011 × 10^{23} 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 × 10^{23}.

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 N _{2} molecules occupies a volume of V litre. What volume n/2 number of CO_{2} 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 N_{2} molecules will occupy a volume of V litres at STP.

n(N_{2}) = N/N_{A}, where N is the number of N_{2} molecules and N_{A} is Avogadro’s number

The number of moles of N_{2} will be equal to the number of moles of CO_{2}, assuming both gases are at STP. We can calculate the number of CO2 molecules using the following equation:

n(CO_{2}) = (n/2)(N_{A})

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

V(CO_{2}) = n(CO_{2}) × 22.4 L/mol

Substituting the values we get:

n(N_{2}) = 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 × 10 ^{24}]**

**Answer**

Given mass of ammonia = 51 g

Molar mass of ammonia (NH_{3}) = 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 × 10^{23}

= 1.8066 × 10^{24}