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 the gaseous molecule is

For any ideal gas, we use the relation between density ($$\rho$$), pressure ($$P$$), molar mass ($$M$$) and temperature ($$T$$):

$$\rho = \frac{PM}{RT}$$

Here $$R$$ is the universal gas constant. Notice that at the same temperature $$T$$ and with the same units for pressure, the factor $$\dfrac{1}{RT}$$ is common for every gas we compare.

We are told that at 300 K, the density of the unknown gas (let us call it X) at 2 bar is double the density of nitrogen ($$N_2$$) at 4 bar. Writing this mathematically,

$$\rho_X = 2\,\rho_{N_2}$$

Using the density formula for each gas and keeping $$RT$$ common we have

$$\frac{P_X M_X}{RT} = 2 \left(\frac{P_{N_2} M_{N_2}}{RT}\right)$$

The $$RT$$ terms cancel out immediately:

$$P_X M_X = 2\,P_{N_2}\,M_{N_2}$$

Now substitute the given pressures and the known molar mass of dinitrogen ($$M_{N_2}=28\ \text{g mol}^{-1}$$):

$$\bigl(2\ \text{bar}\bigr)\,M_X = 2 \times \bigl(4\ \text{bar}\bigr)\times (28\ \text{g mol}^{-1})$$

Divide both sides by the 2 bar on the left to isolate $$M_X$$:

$$M_X = \frac{2 \times 4 \times 28}{2}\ \text{g mol}^{-1}$$

Simplify step by step:

$$M_X = \frac{8 \times 28}{2}\ \text{g mol}^{-1}$$

$$M_X = 4 \times 28\ \text{g mol}^{-1}$$

$$M_X = 112\ \text{g mol}^{-1}$$

Hence, the correct answer is Option B.