Find the molecular mass of a gas that takes three times more time to effuse as compared to He with the same volume

The correct option is 36 u.

To find the molecular mass of the unknown gas, we can use Graham's Law of Effusion. According to Graham's Law, the rate of effusion of a gas is inversely proportional to the square root of its molecular mass (or molar mass) at constant temperature and pressure.
This relationship can be mathematically written as:

$$r\propto \frac{1}{\sqrt{M}}$$

Here, $r$ represents the rate of effusion and $M$ represents the molecular mass of the gas.

The rate of effusion is defined as the volume of gas effused ($V$) divided by the time taken ($t$):

$$r=\frac{V}{t}$$

Since the question states that both gases effuse with the same volume ($V_{\text{unknown}}=V_{\text{He}}$), the rate of effusion is inversely proportional to the time taken:

$$r\propto \frac{1}{t}$$

Therefore, we can relate the effusion times of the two gases directly to their molecular masses:

$$\frac{t_1}{t_2}=\sqrt{\frac{M_1}{M_2}}$$

Let gas 1 be the unknown gas and gas 2 be Helium (He).
From the given information:
- The molecular mass of Helium, $M_{\text{He}}=4\text{u}$.
- The time taken by the unknown gas is three times the time taken by Helium, which means $t_{\text{unknown}}=3\times t_{\text{He}}$, or $\frac{t_{\text{unknown}}}{t_{\text{He}}}=3$.

Substituting these values into the formula, we get:

$$3=\sqrt{\frac{M_{\text{unknown}}}{4}}$$

To solve for $M_{\text{unknown}}$, we square both sides of the equation:

$$3^2=\frac{M_{\text{unknown}}}{4}$$

$$9=\frac{M_{\text{unknown}}}{4}$$

Multiplying both sides by 4:

$$M_{\text{unknown}}=9\times 4=36$$

Thus, the molecular mass of the unknown gas is 36 u.