Question

A certain gas takes three times as long to effuse out as helium. Its molecular mass will be:

• A. 36 u
• B. 64 u
• C. 9 u
• D. 27 u

Answer 1

The Correct Option is A

To determine the molecular mass of the unknown gas based on the effusion rate, we can use Graham's Law of Effusion which states:

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

Where:

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

In this question, helium is gas 1 and the unknown gas is gas 2. Given that the unknown gas takes three times as long to effuse out as helium, the rate of effusion for the unknown gas r_2 is \frac{1}{3} of the rate of helium r_1.

Substituting the given values into Graham’s Law, we have:

\frac{1}{3} = \sqrt{\frac{M_2}{4}}

Here, the molar mass of helium M_1 is 4 u. Squaring both sides of the equation, we get:

\left(\frac{1}{3}\right)^2 = \frac{M_2}{4}

\frac{1}{9} = \frac{M_2}{4}

Solving for M_2 gives:

M_2 = 4 \times \frac{1}{9}

M_2 = \frac{4}{9} \times 9

M_2 = 36 u

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

The correct answer is therefore 36 u.

Conclusion: Based on Graham's Law of Effusion, we determined that the gas with a molecular mass of 36 u effuses at a rate that makes it take three times as long as helium to escape the container. This matches with option A (36 u), confirming it is the correct choice.