Class 11th

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

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option(C), (D)

(A) The number of moles is given by the following formula,

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

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

Moles of O₂ = $\frac{16\mathrm{}\mathrm{g}}{32\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  = 0.5 mol

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

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

On substituting the values in the equation (2)

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

number of atoms = 0.5 $*\mathrm{}$ 6.022 $*\mathrm{}$ 10²³

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

Moles of H₂ = $\frac{4\mathrm{g}}{2\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

= 2 mol

On substituting the values in the equation (2):

2 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022*\mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 2 $*\mathrm{}$ 6.022 $*\mathrm{}$ 10²³

(B) The number of moles of CO₂ is calculated by using equation (1) as follows

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

= 1 mol

On substituting the values in the equation (2), the number of atoms can be calculated as ,

1 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022*\mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 1 $*\mathrm{}$ 6.022 $*\mathrm{}$ 10²³

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

Moles of N₂ = $\frac{28\mathrm{g}}{28\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  =1 mol

On substituting the values in the equation (2), the number of molecules can be calculated as

1 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022*\mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 1 $*\mathrm{}$ 6.022 $*\mathrm{}$ 10²³

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

Moles of O₂ = $\frac{32\mathrm{}\mathrm{g}}{32\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$  = 1

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

$\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{\mathrm{A}\mathrm{v}\mathrm{o}\mathrm{g}\mathrm{a}\mathrm{d}\mathrm{r}\mathrm{o}\mathrm{}\mathrm{s}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{}}\mathrm{}$ -

On substituting the values in the equation (2)

1 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022*\mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 1 $*\mathrm{}$ 6.022 $*\mathrm{}$ 10²³

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

Moles of C = $\frac{12\mathrm{}\mathrm{g}\mathrm{}}{12\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}}$ =1 mol

On substituting the values in the equation (2), the number of molecules can be calculated as

1 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022*\mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 1 $*\mathrm{}$ 6.022 $*\mathrm{}$ 10²³

The number of moles of Na is calculated by using equation (1) as fo

Moles of Na = $\frac{23\mathrm{}\mathrm{g}\mathrm{}}{23\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}}$ =1 mol

On substituting the values in the equation (2), the number of molecules can be calculated as

1 mol = $\frac{\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}\mathrm{}}{6.022*\mathrm{}{10}^{23}\mathrm{}\mathrm{}\mathrm{}\mathrm{}}$

number of atoms = 1 $*\mathrm{}$ 6.022 $*\mathrm{}$ 10²³

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (A), (D)

One mole of oxygen gas at STP is equal to  6.022 x 10²³  molecules of oxygen The number of moles is given by the following formula,

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

The number of moles of O₂ is calculated by using equation (1) as follow.

Moles of O₂  = $\frac{16\mathrm{}\mathrm{g}}{32\mathrm{}\mathrm{g}\mathrm{}/\mathrm{}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{}\mathrm{}}$

                       = 0.5 mol

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)

On substituting the values in the above equation:

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

The number of moles is given by the following formula,

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

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

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

=1 mol

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)

On substituting the values in the above equation:

1 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 = 1 $*$  6.022 $*$ 10²³

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (B)

Law of multiple proportion: According to the law of multiple proportion, whe form two or more than two chemical compounds, the ratio between different elements combining with a fixed mass of the other is always in the ratio of tin Compounds of carbon and oxygen:

Carbon and oxygen react to form two different compounds CO and CO2. In Carbon react with 16 parts by mass of oxygen.

 In CO₂ ,12  parts by mass of Carbon react with 32 parts by mass of oxygen. If the mass of Carbon is fixed at 12 parts of mass then the ratio in the masses with the fixed mass of carbon is 16: 32, that is, 1: 2. Therefore, the mass of oxygen contains a simple ratio of 2: 1.to each other