In a gas counter-diffusion experiment using a horizontal glass tube of length 100\text{ cm… — JAMB UTME

In a gas counter-diffusion experiment using a horizontal glass tube of length $100\text{ cm}$ , gas $A$ with a molar mass of $36\text{ g/mol}$ is released from end $P$ , while an unknown gas $B$ is released simultaneously from end $Q$ under identical conditions of temperature and pressure. The two gases meet and form a visible reaction ring at a distance of $40\text{ cm}$ from end $P$ . What is the molar mass of gas $B$ in $\text{g/mol}$ ?

Answer: 16 g/mol

Answer

16 g/mol

According to Graham's Law of Diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (

r_A / r_B = \sqrt{M_B / M_A}

). Because both gases diffuse over the same duration, the distance ratio equals the rate ratio (

d_A / d_B = 40 / 60 = 2/3

). Substituting

M_A = 36\text{ g/mol}

into

2/3 = \sqrt{M_B / 36}

and squaring both sides gives

4/9 = M_B / 36

, which yields

M_B = 16\text{ g/mol}

Step-by-Step Solution

Calculate the distance traveled by gas B

d_B = 100 cm - 40 cm = 60 cm

The total length of the diffusion tube is 100 cm, and gas A traveled 40 cm from end P before meeting gas B.

Determine the ratio of the rates of diffusion of gas A and gas B

r_A / r_B = 40 / 60 = 2/3

The rate of diffusion is directly proportional to the distance traveled in a given time period.

Apply Graham's Law of Diffusion relating diffusion rates to molar masses

2/3 = sqrt(M_B / 36)

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

Solve for the unknown molar mass M_B

4/9 = M_B / 36 => M_B = 16 g/mol

Squaring both sides of the equation eliminates the square root, allowing straightforward algebraic solution.

Key Concept

Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass (

r_1 / r_2 = \sqrt{M_2 / M_1}

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