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**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.

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