Detailed explanations in West Bengal Board Class 9 Physical Science Book Solutions Chapter 4.2 Mole Concept offer valuable context and analysis.

## WBBSE Class 9 Physical Science Chapter 4.2 Question Answer – Mole Concept

Very short answer type questions

Question 1.

What is mole of a substance ?

Answer:

Molecular weight of a substance expressed in gram is called mole of the substance.

Question 2.

What is molar volume of a substance ?

Answer:

Molar volume of a substance is the volume of gram molecular weight of any substance in gaseous state.

Question 3.

What is vapour density of a gas ?

Answer:

Vapour density is the ratio of weight of some volume of a gas or vapour to the weight of same volume of hydrogen at same temperature and pressure.

Question 4.

What is Avogadro’s number ?

Answer:

Avogadro’s number is the number of molecules present in 1 gram molecule of a substance.

Question 5.

What is the value of Avogadro’s number ?

Answer:

The value of Avogadro’s number is 6.023 × 10^{23}

Question 6.

How many molecules are present in 44g CO_{2}?

Answer:

6.023 x 10^{23}

Question 7.

What is the gram molecular weight of nitrogen ?

Answer:

28g

Question 8.

What is the atomicity of helium ?

Answer:

Atomicity of helium is 1.

Question 9.

What will be the volume of 22g of CO_{2} at NTP ?

Answer:

\(\frac{22 \cdot 4}{2}\) lit = 11.2 lit

Question 10.

Who postulated the atomic theory ?

Answer:

John Dalton postulated the atomic theory.

Question 11.

What is the molar volume of a gas at NTP ?

Answer:

22.4 lit

Question 12.

Give the name of a gas which has same atom and molecule ?

Answer:

Argon (Ar)

Question 13.

Who gave the first concept of molecule ?

Answer:

idea about the molecules was put forward first by an Italian physicist, Amedeo Avogadro (1811).

Question 14.

Which element has atomicity 4 ?

Answer:

Phosphorus

Question 15.

What is the volume of 1 mole CO_{2} at NTP ?

Answer:

22.4 lit

Question 16.

What is the number of 1 mole electron ?

Answer:

6.023 x 10^{23}

Question 17.

What is the mass of 1 gram-mole of water.

Answer:

18g

Question 18.

If a gas has vapour density 14, what will be its molecular weight ?

Answer:

Molecular weight of the gas = 2 × 14 = 28

Question 19.

How many number of molecules are present in 18g water ?

Answer:

18g water contains 6.023 × 10^{23} molecules of water.

Question 20.

What is the atomicity of Argon gas ?

Answer:

The atomicity of Argon is 1.

Question 21.

What will be the volume of gas of 64g SO_{2} at NTP ?

Answer:

22.4 lit

Question 22.

Calculate the number of molecules in 4.5 g water.

Answer:

The number of molecules present in 4.5 g water

\(=\frac{6.023 \times 10^{23} \times 4.5}{18}=1.505 \times 10^{23}\)

Question 23.

What will be the weight of 44.8 lit CO_{2} at NTP ?

Answer:

The weight of 44.8 lit CO_{2} at NTP

Question 24.

What are the number of electrons in 2 moles electron ?

Answer:

The number of electrons present in 2 moles electron

= 2 x 6.023 x 10^{23} = 12.046 x 10^{23}

Question 25.

Which one is heavier : 2 gram-molecule ammonia and 2 gram molecule carbon dioxide ?

Answer:

2 gram-molecule ammonia = (2 x 17) g = 34g NH_{3}

2 gram-molecule carbon dioxide = (2 x 44)g = 88g CO_{2}

So, 2 gram-molecule carbon dioxide is heavier than 2 gram-molecule ammonia.

Question 26.

If the vapour density of any gas is 35.5, what will be its molecular weight ?

Answer:

The molecular weight of the gas will be = 2 x 35.5 = 71

Question 27.

One atomic mass unit (amu) is how much grams ?

Answer:

1.6603 x 10^{-24}g

Question 28.

What is the volume of 7g nitrogen at STP ?

Answer:

5.6L

Short Answer Type Question

Question 1.

State Avogadro’s hypothesis.

Answer:

Avogadro’s hypothesis : Under the same conditions of temperature and pressure, equal volume of all gases (both elementary and compound) contain equal number of molecules.

Question 2.

What is Gay Lussac’s law ?

Answer:

Gay Lussac’s law : If different gases react chemically under the same condition of temperature and pressure and if the product is gaseous then the reactants and the products maintain a simple ratio in their volumes.

Question 3.

What is Berzelius hypothesis ?

Answer:

Berzelius hypothesis : Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of atoms.

Question 4.

What is vapour density ?

Answer:

Vapour density : Vapour density of a gas is the ratio of the weight of a certain volume of the gas to the weight of the same volume of hydrogen under similar conditions of temperature and pressure.

Question 5.

What is molecule ?

Answer:

Molecule : The smallest particle of an element or compound which can exist in the free state is known as molecule.

Question 6.

What are the types of molecules ?

Answer:

There are two types of molecules :

- Elementary molecule
- Compound molecule

Elementary molecule : The molecule which is formed by one or more atoms of an element is known as elementary molecule, e.g. Na, K, B, C etc.

Compound molecule : The molecule which is formed by atoms of more than one element is known as compound molecule, e.g. H_{2}O, NH_{3}, HCl etc

Question 7.

What is molecular weight ?

Answer:

Molecular weight : The molecular weight of a substance is a number which represents how many times a molecule of the substance is heavier than \(\frac{1}{12}\)part of the weight of a C-12 isotope.

Question 8.

What is atomic weight ?

Answer:

Atomic weight : Atomic weight of an element

\(=\frac{\text { Weight of } 1 \text { atom of the element }}{\frac{1}{12} \text { of the weight of } 1 \text { atom of } \mathrm{C}^{12}}\)

Question 9.

What is gram atomic mass ?

Answer:

Gram atomic mass : When the atomic mass of an element is expressed in gram then the amount of element in gram is known as gram atomic mass or gram-atom of the element.

Question 10.

What is gram molecular mass ?

Answer:

Gram molecular mass : When the molecular mass of an element or compound in expressed in gram then that amount of element or the compound in gram is known as gram molecular mass or gram molecule or gram-mole of the element or the compound.

Question 11.

What is Molar volume ?

Answer:

Molar volume or gram-molar volume : The volume of a gaseous substance (element or compound) under a fixed temperature and pressure is known as molar volume or gram molar volume.

Question 12.

What is Avogadro’s number ?

Answer:

Avogadro’s number : One gram-molecule of any element or compound contains equal number of molecules. This number is known as Avogadro’s number.

It is denoted by N and its value is 6.023 × 10^{23}

Question 13.

What is atomicity ?

Answer:

Atomicity : Atomicity is. the number of atoms by which an elementary molecule is composed of.

e.g. atomicity of He = 1; atomicity of oxygen = 2; atomicity of ozone = 3; atomicity of phosphorous = 4

Question 14.

What do you mean by mole ?

Answer:

Mole : The ‘mole’ is regarded as the amount of the substance in grams which contains Avogadro’s number of elementary entities (such as molecules, atoms, ions) constituting the substance under investigation.

Question 15.

What is atomic mass unit (amu) ?

Answer:

Atomic mass unit (amu) : It is the quantity of mass equal to \(\frac{1}{12}th\) of the mass of carbon atom (c^{12}).

1 amu = 1.6606 x 10^{-24} g

Question 16.

What is the difference between molecular weight and actual weight of a molecule ?

Answer:

Distinction between molecular weight and actual weight of a molecule : Gram molecular weight is the weight of 6.023 × 10^{23} molecules, while the actual weight is the weight of one molecule.

Weight of molecule = \(\frac{\text { Gram-molecular weight }}{6.023 \times 10^{23}}\)

Question 17.

Give statement of two important deductions of Avogadro’s law

Answer:

Two important deductions of Avogadro’s law :

- Molecules of common elementary gases like hydrogen, oxygen, nitrogen etc. are diatomic.
- The gram-molecular (or molar) volume of all gases under the same condition of temperature and pressure is the same and at STP, it is 22.4 litBroad answer type questions

Broad answer type questions

Question 1.

Explain Berzelius hypothesis on the basis of Dalton’s atomic theory.

Answer:

Explanation of Berzelius hypothesis on the basis of Dalton’s atomic theory : Under the same conditions of temperature and pressure one volume of hydrogen and one volume of chlorine combine to form two volumes of hydrogen chloride gas. According to Berzelius hypothesis, suppose that under the same conditions of temperature and pressure each volume of a gas contains

n-number of atoms.

We can say,

n-number of hydrogen atoms + n-number of chlorine atoms

= 2n number of hydrogen chloride atoms

∴ 1 atom of hydrogen + 1 atom of chlorine

= 2 atoms of hydrogen chloride

∴ \(\frac{1}{2}\) atom of hydrogen + \(\frac{1}{2}\) atom of chlorine

= 1 atom of hydrogen chloride

Now, existence of \(\frac{1}{2}\) atom of hydrogen and chlorine is impossible according to Dalton’s atomic theory, so the hypothesis clashed with the basic concept of atomic theory. Hence, it was rejected by Dalton.

Question 2.

Explain Gay-Lussac’s law with the help of Avogadro’s hypothesis.

Answer:

Explanation of Gay-Lussac’s law with the help of Avogadro’s hypothesis :

From experimental result it has been found that under the same conditions of temperature and pressure one volume of hydrogen combines with one volume of chlorine to produce two volumes of hydrogen chloride gas.

Suppose, under the same conditions of temperature and pressure each volume of gas contains

n-number of molecules.

So,

n-number of hydrogen molecules + n-number of chlorine molecules

= 2n-number of hydrogen chloride molecules.

∴ 1 molecule of hydrogen + 1 molecule of chlorine

= 2 molecules of hydrogen chloride gas.

∴ \(\frac{1}{2}\) molecule of hydrogen + \(\frac{1}{2}\) molecule of chlorine

= 1 molecule of hydrogen chloride gas

This is to say,

1 molecule of hydrogen chloride must contain \(\frac{1}{2}\) molecule of hydrogen and \(\frac{1}{2}\) molecule of chlorine.

This fact does not go against Dalton’s atomic theory. Later on, it is known from Avogadro’s hypothesis that each molecule of hydrogen and chlorine contains two atoms.

So, 1 atom of hydrogen + 1 atom of chlorine

= 1 molecule of hydrogen chloride gas. in this way Dalton’s atomic theory and Gay-Lussac’s law were harmonized with the help of Avogadro’s hypothesis.

Question 3.

What are the application of Avogadro’s hypothesis ?

Answer:

Application of Avogadro’s hypothesis :

- Excepting the inert gases, all other active elementary gases are diatomic i.e. molecule containing two atoms.
- Molecular weight of any gas is twice its vapour density.
- Volume of a gram-mole of all elementary or compound gases is 22.4 lit at NTP.
- Number of atoms or molecules of all elementary or compound gases at their gram-atomic or gram-molecular weight is 6.023 x 10
^{23}at NTP. - Molecular formula of a gaseous molecule can be deducted from its volumetric composition.
- Atomic weight of an element can be found out by the application of this hypothesis.

Question 4.

Prove that hydrogen molecule is diatomic in nature.

Answer:

Profit of diatomic nature of hydrogen molecule : From experimental result we know that under the same conditions of temperature and pressure one volume of hydrogen and one volume of chlorine combine chemically to form two volumes of hydrogen chloride gas.

Let us suppose that under the conditions of pressure and temperature at the time of experiment, one volume of hydrogen gas contains re-molecules. So, according to Avogadro’s hypothesis, under the same conditions of temperature and pressure one volume of chlorine contains remolecules and two volumes of hydrogen chloride gas contain 2n molecules.

We can write,

re-molecules of hydrogen + re-molecules of chlorine

= 2re molecules of hydrogen chloride gas.

∴ 1 molecule of hydrogen + 1-molecule of chlorine = 2 molecules of hydrogen chloride gas.

So, 1 molecule of hydrogen chloride

= \(\frac{1}{2}\) molecule of hydrogen + \(\frac{1}{2}\) molecule of chlorine

Now, according to Dalton’s atomic theory, an atom is indivisible.

So, one molecule of hydrogen chloride contains at least one atom of hydrogen and one atom of chlorine. This one atom of hydrogen comes from \(\frac{1}{2}\) molecule of hydrogen.

So one molecule of hydrogen must contain two atoms.

Question 5.

Deduce the relation of density and vapour density of a gas on the basis of hydrogen.

Answer:

Relation of density and vapour density of a gas on the basis of hydrogen

We know,

∴ Density of gas (d) = D x 0.089

(Density of hydrogen gas = 0 089)

∴ d = D x 0.089

Question 6.

Deduce the relation between vapour density and the molecular weight of a gas.

Answer:

Relation between vapour density and the molecular weight of a gas : From the definition of vapour density

Suppose, V volume of the gas contains ‘n’ molecules. Then by Avogadro’s hypotheis, V volume of hydrogen will also contain ‘n’ molecules of hydrogen.

i.e. Molecular weight of a gas 2 x its vapour density.

Question 7.

Gram-molecular volume of all gases at NTP is 224 litres – prove it.

Answer:

If the molecular mass of a gas is M, then we know that the limiting density of the gas 0.0898 × \(\frac{M}{2}\) g/litre

From the definition of limiting density it can be written that mass of 1 litre of a gas at NTP = 00898 x \(\frac{M}{2}\)g

So. at NTP, volume of 0.0898 × g gram of the gas = 1 litre

∴ At NTP volume of 1 gram of the gas = \(\frac{2}{M \times 0.0898}\) litre

∴ At NTP. volume of M g or 1 gram molecule of the gas

\(=\frac{2 \times M}{M \times 0.00898}=\frac{2}{0.0898}\) = 2227 litre

Atomic mass of hydrogen in oxygen scale is 2 016. in that case calculating in a similar way, we can obtain,

Volume of I gram-molecule gas at NTP\(=\frac{2.016}{0.0898}=22.4 \text { litre }\)

So, at NTP, volume of 1 gram-molecule of any gas is 224 litre.

Question 8.

What is the actual weight of a molecule ? What is the actual weight of an atom?

Answer:

Actual weight of a molecule:

Suppose, the molecular mass of any substance M

So, gram-molecular weight of the substance = Mg

We know that there are present 6.023 × 10^{2} molecules in gram-

molecular weight.

∴ Weight of 6.023 x 10^{23} molecules Mg

∴ Weight of 1 molecule = \(=\frac{M}{6.023 \times 10^{23}} \mathrm{~g}\)

Actual weight of an atom:

Suppose, the atomic mass of an element = A

So, gram-atomic weight of the element = Ag

We know that there are 6.023 x 10^{23} atoms present in gram-atomic weight.

∴ Weight of 6.023 × 10^{23} atoms = Ag

∴ Weight of 1 atom \(=\frac{A}{6.023 \times 10^{23}} \mathrm{~g}\)

Question 9.

What are the difference between molecular weight and acutal weight (mass) of a molecule?

Answer:

Difference between molecular weight and actual weight (mass) of a molecule :

Numerical Problem:

Problem 1.

If 200 ml of a gas at NTP weighs 0.27g and the normal density of hydrogen be 0.09 glut, calculate vapour density of the gas.

Answer:

At NTP, vapour density × 0.09 = Normal density

Now, 200 ml of the gas weighs 0.27 g

Problem 2.

Find the actual weight of 1 molecule water. (H = 1, O = 16)

Answer:

Molecular weight of water = (1 × 2 + 16) = 18

gram molecular weight of water = 18 g

So, 18g water contains = 6.023 × 10^{23} molecules

Hence, actual weight of 1 water molecule

= \(\frac{18}{6.023 \times 10^{23}}\) = 2.988 × 10^{-23}g (approx)

Problem 3.

How many gram-moles of chlorine are present in 213 g chlorine and what is the volume of 6.023 x 10^{30} hydrogen molecules at STP ? Given, atomic weight of chlorine is 35.5.

Answer:

Molecular weight of chlorine 355 × 2 = 71

∴ 71g chlorine = 1 gram-mole chlorine

∴ 213 g chlorine = \(\frac{1 \times 213}{71}\) gram-mole = 3 gram mole

At STP, 6.023 × 10^{23} molecules occupy 224 lit

∴ At STP, 6.023 × 10^{30} molecules occupy

\(=\frac{22.4 \times 6.023 \times 10^{30}}{6.023 \times 10^{23}} \mathrm{lit}\)

= 22.4 ×10^{7 }lit

Problem 4.

Find the number of hydrogen and oxygen atoms in 425.6 lit of water vapour at NTP.

Answer:

Number of molecules in 22.4 lit water vapour at NTP

6.023 x 10^{23}

∴ Number of molecules in 425.6 lit water vapour at NTP

\(=\frac{6.023 \times 10^{23} \times 425 \cdot 6}{22 \cdot 4}=114437 \times 10^{20}\)

Now, 1 molecule of water contains number of hydrogen atom = 2

∴ In 114437 x 10^{20 }molecules of water contains number of hydrogen atom

= 2288874 × 10^{20}

Again, 1 water molecule contains 1 oxygen atom

∴ 114437 x 10^{20} water molecules contain 114437× 10^{20} oxygen atom.

Problem 5.

What is the weight of one gram oxygen and how many number of oxygen atoms are there ?

Answer:

One gram-atom of oxygen = 16 g of oxygen

From Avogadro’s hypothesis,

32 g of oxygen contains = 6.023 × 10^{23} molecules

∴ 16 g of oxygen contains \(=\frac{6023 \times 10^{23}}{32} \times 16\) molecules

= 3.0115 × 10^{23} molecules

Again oxygen is a diatomic element,

Hence, it contains 2 × 3.0115 x 10^{23} atoms = 6.023 × 10^{23} atoms.

Problem 6.

Calculate the number of mole in 9 grams of water and calculate the number of hydrogen atoms present in it.

Answer:

18 g water contain = 6.023 × 10^{23} molecules

9 g water contain \(\frac{6.023 \times 10^{23}}{2}\) = 3.0115 x 10^{23} molecules of water

One molecule of water contains 2 hydrogen atoms. So the number of hydrogen atoms =

(3.0115 × 10^{23}) x 2 = 6.023 × 10^{23}

Problem 7.

250 cm^{3} of a gas at NTP weigh 0.7924. What is the molecular weight of the gas ?

Answer:

250 cm^{3} of the gas at NTP weighs = 0.7924 g

∴ 22400 of the gas at NTP weighs

= \(\frac{0.7924 \times 22400}{250} \mathrm{~g}\)

= 70 999 g

∴ Gram-molecule of the gas = 70.999 g

∴ Molecular weight of the gas = 70.999

Problem 8.

The molecular weight of a gaseous substance is 200. What is the volume of 5 g of the substance at NTP?

Answer:

The molecular weight of the substance = 200

∴ 1 gram-molecule of the substance = 200 g

Volume of 1 gram-molecule or 200 g of the substance at NTP = 224 lit.

Volume of 5 g of the substance at NTP = \(\frac{22.4 \times 5}{200}=0.56 \mathrm{lit}\)

Problem 9.

Calculate the number of molecules in a drop of water weighing 0.05 g (H = 1, 0 = 16)

Answer:

Molecular mass of water = (2 × 1 + 1 × 16) = 18

∴ Gram-molecular mass of water = 18 g

Now, 18 g of water contains 6.023 × 10^{23} molecule

∴ 0.05 g of water contains

\(\frac{6.023 \times 10^{23}}{18} \times 0.05\) molecule

= 1.672 × 10^{21}

Problem 10.

Calculate the total number of electrons present in 1.6g of methane (CH_{4}). [H = 1, C = 12]

Answer:

Molecular mass of methane = 16 g

16 g of methane 1 mole

∴ 16 g of methane = 0.1 mole

Now, 1 mole of methane = 6.023 × 10^{23} molecules

∴ 0.1 mole of methane = 6.023 × 10^{22} molecules

We know that one atom of carbon contains six electrons and one atom of hydrogen contains one electron.

∴ One molecule of CH_{4} contains = 6 + 1 × 4 = 10 electrons

∴ 6.023 × 10^{22} molecules contain = 10 x 6.023 x 10^{22}

= 6.023 × 10^{23} electrons

Problem 11.

A tetra-atomic gas occupies 1.4 lit at 0°C and 76cm pressure. Find the number of atoms in the gas.

Answer:

At 0°C and 760 cm pressure, number of molecules present

= 224 lit of the gas.

= 6.023 × 10^{23}

= 6.023 × 10^{23 }x 4 atoms

Number of atoms present in 1.4 lit

= \(\frac{1 \cdot 4}{22 \cdot 4}\) x 6.023× 10^{23} × 4 = 1.505 x10^{23}

Problem 12.

Calclulate the molecular mass of CO_{2}

Answer:

CO_{2} molecule contains one C-atom and two oxygen atoms.

Molecular mass of CO_{2}= (1 x 12 + 2 x 6) = 44U

Problem 13.

Calculate the formula of unit mass of MgCl_{2}

Answer:

MgCl_{2} molecule contains one Mg-atom and two Cl-atoms.

∴ Formula of unit mass of MgCl_{2}

= 24+2 x 35.5 = 24+71=95U

Problem 14.

Calculate the following :

(i) 1 milimole of NH_{3}

(ii) 3.011 x 10^{23} number of \({ }_8^{16} \mathrm{O}\)

Answer:

(i) 1 milimole = 10 mole.

Mass = molar mass x no. of moles

= 17 × 10^{-3}= 0.017g

(ii) Mass = number of moles × atomic mass

\(=\frac{3.011 \times 10^{23}}{6.022 \times 110^{23}} \times 16=8 \mathrm{~g}\)

Example 15 :

Calculate the number of particles in each of the following :

(i) 35.5g of Na atoms.

(ii) 11g of CO_{2}

(iii) 0.2 mole of 2C atoms.

Answer:

(i) the number of atoms

\(=\frac{\text { given mass }}{\text { molar mass }}\)×Avogadro’s number

\(\frac{35 \cdot 5}{23}\)

(ii) Number of Melecules of CO_{2}

(iii) No of ^{12}C atoms

= No of moles of particle x Avogadro’s number.

= 0.2 × 6.022 × 10^{23} = 1.2044 × 10^{23}

Problem 16.

Calculate the mass of 250 molecules of sodium chloride.

Answer:

Molecular mass of NaCl

= 1 × 23 + 1 × 35.5 = 58.5

1 mole of NaCl = 58.5g NaCl

6.022 × 10^{23} molecules of NaCl have mass 58.5g

250 molecules of NaCl have mass = \(\frac{58 \cdot 5 \times 250}{6.022 \times 10^{23}}\) = 2.428 ×10^{-23}g.

Problem 17.

What is the actual weight of one molecule of H_{2} and one atom of hydrogen ? The gram molecule of H_{2} is 2.016g.

Answer:

6.022 x 10^{-23} molecules of hydrogen weigh 2.016g.

1 molecule of hydrogen weigh \(\frac{2 \cdot 016}{6 \cdot 022 \times 10^{23}} \mathrm{~g}\) = 3.34 × 10^{-24}g

Hydrogen is diatomic i.e. one molecule of hydrogen consists of two atoms. Therefore, one atom of hydrogen weighs

\(=\frac{3.34 \times 10^{-24}}{2}\) = 1.67 × 10^{-24}g

Problem 18.

How many moles or milimoles are present in 0.49 g of H_{2}SO_{4}? [H = 1, S=32, O=16]

Answer:

Molecular mass of H_{2}SO_{4}

=2 x 1+1 x 32+4 x 16=98

1g mole of H_{2}SO_{4} is equal to 1 mole of H_{2}SO_{4}

∴ 0.49g of H_{2}SO_{4} is equal to \(\frac{0.49}{98}\) = 0.005 mole

So. no of millimoles

= 0.005 × 103 = 5 millimole

[Since 1 mole 10^{3} millimole]