Chemistry NCERT Exemplar Solutions Class 11th Chapter One

This is a Matching Type Questions as classified in NCERT Exemplar

(i) → (e) (ii) → (d) (iii) → (b) (iv) → (g) (v) → (c), (h) (vi) → (f) (vii) → (a) (viii) → (i)

Explanation:

(i) It is defined as the number of moles of solute dissolved in 1 litre of solution. So, the unit of molarity is mol L-1
(ii) Mole fraction is defined as the number of moles of a constituent divided by the total number of moles. So, it is unitless.

(iii) Mole is the unit to measure the amount of an atom or molecule in SI system. it is symbolized as "mol".

(iv) It is defined as the number of moles of solute dissolved in 1 kg of solvent, so, the unit of morality is mol kg-1.

(v) It is defined as the force per unit area. so, the unit of pressure is Pascal.

(vi) It is the amount of visible light released per unit solid angle in unit time. so, it's unit will be Candela.

(vii) It is defined as the mass of a substance (in kg) divided by the volume of that substance (in mL). The unit of density is g ml-1

(viii) Mass is used to measure the amount of a substance or an object. The unit of mass is kg.

This is a Matching Type Questions as classified in NCERT Exemplar

(i)- (b); (ii) - (c); (iii) - (a); (iv) - (e); (v) - (d)

(i) 88 g of CO₂

The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  - (1)

So, the number of moles of CO₂ is calculated by using equation (1) as follows

Moles of CO₂ = $\frac{88\mathrm{}\mathrm{g}}{44\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}}$ = 2 mol

Thus, option (i) from column I is matched with (b) from column II

(ii)  1 mol of H₂O gives 6.022*10²³  molecules. So, 6.022*10²³  molecules contain 1 mol of H₂O . Thus, option (ii) from column I is matched with (b) from column II

(iii) 5.6 liters of O₂ at STP

1 mol of gas occupies 22.4 liters of O2 . For 5.6 liters of O₂, the number of moles of oxygen gas is calculated as

Moles of O₂ = $\frac{88\mathrm{}\mathrm{g}}{22.4\mathrm{}\mathrm{L}\mathrm{}\mathrm{}\mathrm{}}$  * 5.6 L = 0.25 mol

(iv) 96 g of O₂

The number of moles of O₂ is calculated by using equation (1) as follows,

Moles of O₂ = $\frac{96\mathrm{}\mathrm{g}}{32\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}}$  = 3 mol

Thus, option (iv) from column I is matched with (e) from column II.

(v) 1 mol of any gas 1 mole of any gas contains 6.022*10²³  molecules.

Thus, option (v) from column I is matched with (d) from column II.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (A), (D)

Law of conservation of mass: According to this law, the mass is neither created nor destroyed during a chemical reaction in an isolated system. It can only be transformed from one form to another. Burning of wood is an example of conservation of mass. Law of definite proportion:

 According to this law, a compound contains exactly the same elements in the fixed proportion by mass. For example, pure water consists of 11.196 H and 88.9% o by mass.

Law of multiple proportion: According to the law of multiple proportion, when two elements react to form two or more than two chemical compounds, the ratio between different masses of one of the elements combining with a fixed mass of the other is always in the ratio of tiny numbers. For example, compounds of carbon and oxygen. Avogadro's law: This law states that the number of moles of a gas is directly related to the volume enclosed by the gas at constant pressure and temperature.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (C), (D)

Molality: It is defined as the number of moles of solute divided by the mass of solvent in kg . It is denoted by m . The formula is expressed as

m = $\frac{\mathrm{n}}{\mathrm{W}\mathrm{}\mathrm{}}$

The SI unit of molality is 1 mol/kg

Molarity: It is defined as the number of moles of solute divided by the volume of solution in latice. It is represented by M  The formula of molarity is expressed as

M = $\frac{\mathrm{n}}{\mathrm{V}\mathrm{}\mathrm{}}$

Here, n is the number of moles of solute and V is the volume of solution expressed in liters. The SI unit of molarity is mol/ L .

Mole fraction (x ) : The mole fraction of a substance is defined as the number of moles of that substance divided by the total number of moles of all substances present in the solution.

Mass percent: it is defined as the weight of the solute divided by the total weight of the solution and the obtained result is multiplied by 100 in order to give a percent. Thus, mole fraction and mass percent are unitless.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option(C), (D)

The number of moles is given by the following formula,

Moles = $\frac{\mathrm{M}\mathrm{a}\mathrm{s}\mathrm{s}}{\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  ---------------(1)

The number of molecules can be calculated as, number of moles

= $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{A}\mathrm{v}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{}\mathrm{s}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}}$  ----------(2)

(C) The number of moles of N₂ is calculated by using equation (1) as follows,

Moles of N₂ = $\frac{14\mathrm{}\mathrm{g}}{28\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}}$  = 0.5 mol

The number of molecules can be calculated by using equation (2) as follows,

0.5 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{6.022*\mathrm{}{10}^{23}\mathrm{}\mathrm{}}$

number of molecules = 0.5 $*$  6.022 $*$  10²³

(D) The number of moles of H₂ is calculated by using equation (1) as follows,

Moles of H₂ = $\frac{1\mathrm{g}}{2\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}}$  = 0.5 mol

The number of molecules can be calculated by using equation (2) as follows,

0.5 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{e}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{s}}{6.022*\mathrm{}{10}^{23}\mathrm{}\mathrm{}}$

number of molecules = 0.5 $*$ 6.022 $*$  10²³