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 :

Question

Statement-1: Sublimation is a purification technique that is used to separate those solid substances which change from solid to vapor state without passing through liquid state.
Statement-2: If external atmospheric pressure is reduced, then boiling point of substance decreases.
Which of the following is correct?

Answer 1

To solve this problem, we need to understand the relationships between the density of gases, pressure, molar mass, and temperature using the ideal gas equation.

The ideal gas law is given as:

PV = nRT

Here, P is the pressure, V is the volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin.

Density (\rho) is mass per unit volume. For an ideal gas:

\rho = \frac{m}{V} = \frac{PM}{RT}

where M is the molar mass.

According to the problem, the density of the gaseous molecule at 2 bar is double that of nitrogen at 4 bar. Let's write the relation:

For dinitrogen (N_2):
\rho_{N_2} = \frac{P_{N_2}M_{N_2}}{RT_{N_2}}
\rho_{N_2} = \frac{4 \times 28}{R \times 300}

Now, for the given gaseous molecule with density double to that of (N_2):

\rho_{gas} = \frac{P_{gas} M}{RT_{gas}}
\rho_{gas} = \frac{2M}{R \times 300} (since pressure is given as 2 bar)

Given that \rho_{gas} = 2 \rho_{N_2}, we equate:

\frac{2M}{R \times 300} = 2 \times \frac{4 \times 28}{R \times 300}

Solving for M:

2M = 8 \times 28
2M = 224
M = \frac{224}{2} = 112 \, g \, mol^{-1}

Thus, the molar mass of the gaseous molecule is 112 \, g \, mol^{-1}, aligning with the provided options.

Answer 2