Question

A and B are two identical vessels. A contains \[15g\] ethane at $1$atm and $298{\text{K}}$. The vessel B contains $75g$ of a gas ${X_2}$ at the same temperature and pressure.
The vapour density of ${X_2}$ is-
A. $75$
B. $150$
C. $37.5$
D. $45$

Answer 1

Hint: Use gas equation PV=nRT where P is the pressure, V is the volume, n is the number of moles of the gas, R is the gas constant and T is the temperature to find the number of moles of both gases. Then we know the number of moles is given by formula-
$ \Rightarrow {\text{n = }}\dfrac{{\text{W}}}{{\text{M}}}$ Where W is given mass and M is molecular mass. Solve for both gases and acquire the molecular mass of gas ${X_2}$. Then use vapour density formula which is given as-
Vapour Density of a gas= $\dfrac{{{\text{Molecular mass}}}}{2}$. Put the obtained value to find the answer.

Step-by-Step Solution-
Given, there are two identical vessels named A and B.
A contains \[15g\] ethane gas at temperature $298{\text{K}}$ and $1$atm. We know that the molecular mass of ethane $\left( {{C_2}{H_6}} \right)$ is =$\left( {12 \times 2} \right) + \left( {6 \times 1} \right) = 30$
Vessel B contains $75g$ of ${X_2}$ gas at the same temperature and pressure. We have to find the vapour density of ${X_2}$ gas.
Now we know the gas equation –
PV=nRT where P is the pressure, V is the volume, n is the number of moles of the gas, R is the gas constant and T is the temperature.
Here the pressure, temperature and volume are constant in this equation for both gases. Then their number of moles ‘n’ will also be constant.
Now we know that n is given as the ratio of given mass and molecular mass of gas-
$ \Rightarrow {\text{n = }}\dfrac{{\text{W}}}{{\text{M}}}$ -- (i) where W is given mass and M is molecular mass.
Then for both gases we can write,
$ \Rightarrow {{\text{n}}_{\text{1}}}{\text{ = }}{{\text{n}}_{\text{2}}}$
Where ${{\text{n}}_{\text{1}}}$=the number of moles of ethane and ${{\text{n}}_2}$= The number of moles of ${X_2}$ gas.
From eq. (i) we can write,
$ \Rightarrow \dfrac{{{{\text{W}}_1}}}{{{{\text{M}}_1}}} = \dfrac{{{{\text{W}}_2}}}{{{{\text{M}}_2}}}$
Where ${{\text{W}}_1}$ and ${{\text{M}}_1}$ are given mass and molecular mass of ethane respectively and ${{\text{W}}_2}$ and ${{\text{M}}_2}$ are given mass and molecular mass of${X_2}$ gas.
On putting the given values in the equation we get,
$ \Rightarrow \dfrac{{15}}{{30}} = \dfrac{{75}}{{{{\text{M}}_2}}}$
We can write it as-
$ \Rightarrow {{\text{M}}_2} = \dfrac{{75 \times 30}}{{15}}$
On solving we get,
$ \Rightarrow {{\text{M}}_2} = 75 \times 2 = 150$
Now we will use formula of vapor density which is given as –
Vapour Density of a gas= $\dfrac{{{\text{Molecular mass}}}}{2}$
On putting the obtained value we get,
Vapour density of ${X_2}$ gas= $\dfrac{{150}}{2} = 75$

Hence the correct answer is A.

Note: Here you can also solve this question by considering the volume for vessel A to be V then putting the given values in the formula of gas equation for vessel A we’ll get,
$ \Rightarrow 1 \times {\text{V}} = \dfrac{{15}}{{30}} \times {\text{RT}}$
On solving we get,
$ \Rightarrow {\text{V}} = 0.5{\text{RT}}$
Now on substituting this value for in the gas equation for vessel B we’ll get-
$ \Rightarrow 1 \times 0.5{\text{RT = }}\dfrac{{75}}{{{{\text{M}}_2}}} \times {\text{RT}}$
On solving we get,
$ \Rightarrow {{\text{M}}_2} = 150$
Now use the vapor density formula and solve, you’ll get the answer.