Question

A gas has a vapour density 11.2 the volume occupied by 1 gram of gas at ntp is

Answer 1

molecular mass of gas = 2×V.D.

                                 =2×11.2 =22.4

no. of moles of gas = weight÷ molecular mass

                            = 1÷12.4

vol. oF gas = no of moles × V

                = 1÷ 22.4 × 22.4

                = 1 L

Answer 2

Answer: The volume occupied by the gas is 1 L.

Explanation:

To calculate the molecular mass, we use the equation:

$\text{Molecular mass}=2\times \text{Vapor density}$

We are given:

Vapor density = 11.2

Putting values in above equation, we get:

$\text{Molecular mass}=2\times 11.2=22.4g/mol$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of gas = 1 g

Molar mass of gas = 22.4 g/mol

Putting values in above equation, we get:

$\text{Number of moles of gas}=\frac{1}{22.4}mol$

At NTP conditions:

1 mole of a gas occupies 22.4 L of volume

So, $\frac{1}{22.4}mol$ of a gas will occupy = $\frac{22.4}{1}\times \frac{1}{22.4}=1L$ of volume.

Hence, the volume occupied by the gas is 1 L.