Question

$50\, mL$ of each gas A and of gas B takes $150$ and $200\, s$ respectively for effusing through a pin hole under the similar conditions. If molecular mass of gas B is $36,$ the molecular mass of gas A will be

• A. 96
• B. 128
• C. 32
• D. 20.2

Answer 1

The Correct Option is D

To solve the given problem, we apply Graham's law of effusion. This law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, it can be expressed as:

\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}

where:

• r_1 and r_2 are the rates of effusion of gases A and B, respectively.
• M_1 and M_2 are the molar masses of gases A and B, respectively.

The rate of effusion is given by the volume of gas effusing per unit time. So for gases A and B:

r_1 = \frac{50 \, \text{mL}}{150 \, \text{s}} and r_2 = \frac{50 \, \text{mL}}{200 \, \text{s}}

We can write the expression for the rate of effusion ratio:

\frac{r_1}{r_2} = \frac{\frac{50}{150}}{\frac{50}{200}} = \frac{200}{150} = \frac{4}{3}

Substituting these into Graham's law equation:

\frac{4}{3} = \sqrt{\frac{36}{M_1}}

Squaring both sides to eliminate the square root gives:

\left(\frac{4}{3}\right)^2 = \frac{36}{M_1}

Simplifying the left-hand side:

\frac{16}{9} = \frac{36}{M_1}

Cross-multiplying to solve for M_1:

M_1 = \frac{36 \times 9}{16}

Calculating the right-hand side:

M_1 = \frac{324}{16} = 20.25 \approx 20.2

Therefore, the molecular mass of gas A is 20.2.

Thus, the correct answer is 20.2.