From gas equation, $ {{P}_{1}}{{V}_{1}}={{n}_{1}}\cdot R{{T}_{1}}=\frac{{{w}_{1}}}{{{m}_{1}}}\cdot R{{T}_{1}} $ ?(i) $ {{P}_{2}}{{V}_{2}}=\frac{{{w}_{2}}}{{{m}_{2}}}\cdot R{{T}_{2}} $ ?(ii) As, $ {{P}_{1}}={{P}_{2}}, $ $ {{V}_{1}}={{V}_{2}} $ hence, $ {{T}_{1}}={{T}_{2}} $ Hence, $ \frac{{{w}_{1}}}{{{m}_{1}}}=\frac{{{w}_{2}}}{{{m}_{2}}} $ $ \therefore $ $ \frac{15}{30}=\frac{75}{{{m}_{2}}} $ ( $ \because $ molecular weight of ethane = 30) $ \therefore $ $ {{m}_{2}}=150 $ Hence, vapour density of $ {{X}_{2}} $ gas is $ =\frac{M}{2}=\frac{150}{2}=75 $