A container holds an unknown gas, Y. It takes 15\text{ seconds} for a specific volume of g… — JAMB UTME

A container holds an unknown gas, $Y$ . It takes $15\text{ seconds}$ for a specific volume of gas $Y$ to effuse through a fine orifice. Under identical temperature and pressure conditions, the same volume of oxygen gas ( $O_2$ ) requires $30\text{ seconds}$ to effuse through the same orifice. What is the molar mass of gas $Y$ ? [Molar mass of $O_2 = 32\text{ g/mol}$ ]

$8\text{ g/mol}$ Answer
B
$16\text{ g/mol}$
C
$64\text{ g/mol}$
D
$128\text{ g/mol}$

Answer

The molar mass of gas

Y

8\text{ g/mol}

By Graham's Law of Effusion, the time required for equal volumes of two gases to effuse under identical conditions is directly proportional to the square root of their molar masses:

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

. Substituting

t_Y = 15\text{ s}

t_{O_2} = 30\text{ s}

, and

M_{O_2} = 32\text{ g/mol}

gives

\frac{15}{30} = \sqrt{\frac{M_Y}{32}}

. Squaring both sides yields

\frac{1}{4} = \frac{M_Y}{32}

, which simplifies to

M_Y = 8\text{ g/mol}

Step-by-Step Solution

State Graham's Law of Effusion relating effusion time (

t

) to molar mass (

M

\frac{t_Y}{t_{O_2}} = \sqrt{\frac{M_Y}{M_{O_2}}}

Effusion time for a fixed volume of gas is directly proportional to the square root of its molar mass because rate of effusion is inversely proportional to time.

Substitute the given values into the equation.

\frac{15}{30} = \sqrt{\frac{M_Y}{32}} \implies 0.5 = \sqrt{\frac{M_Y}{32}}

Given

t_Y = 15\text{ s}

t_{O_2} = 30\text{ s}

, and

M_{O_2} = 32\text{ g/mol}

Square both sides of the equation to eliminate the square root.

(0.5)^2 = \frac{M_Y}{32} \implies 0.25 = \frac{M_Y}{32}

Squaring isolates the ratio of molar masses.

Solve for

M_Y

M_Y = 0.25 \times 32 = 8\text{ g/mol}

Multiplying both sides by

32\text{ g/mol}

yields the molar mass of gas

Y

Key Concept

Graham's Law of Diffusion and Effusion relating time of effusion to molar mass

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