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Q.
At $300\, K$, the density of a certain gaseous molecule at $2$ bar is double to that of dinitrogen $(N_2)$ at $4$ bar. The molar mass of gaseous molecule is :
For unknown gas :
$ PM =\rho RT $
$ 2 \text { bar } \times M=\rho_{\text {unknown }} R T \ldots \ldots . .(1) $
For nitrogen gas :
$ 4 \text { bar } \times 28 g / mol =\rho_{\text {nitrogen }} RT \ldots \ldots . .(2) $
But,
$\rho_{\text {unknown }}=2 \rho_{\text {nitrogen }}$
Substitute this in equation (1)
$ 2 \text { bar } \times M=2 \rho_{\text {nitrogen }} R T \ldots \ldots .(3) $
Divide equation (3) with equation (2)
$ \frac{2 \operatorname{bar} \times M}{4 \operatorname{bar} \times 28 g / mol }=\frac{2 \rho_{\text {nitrogen }} R T}{\rho_{\text {nitrogen }} R T} $
$\frac{ M }{2 \times 28 g / mol }=2$
$M =4 \times 28 g / mol =112 g / mol$
The molar mass of the gaseous molecule is $112 g / mol$