Given that the molar mass of methane (CH_4) is 16\text{ g/mol} and that of sulfur(IV) oxid… — JAMB UTME

Given that the molar mass of methane ( $CH_4$ ) is $16\text{ g/mol}$ and that of sulfur(IV) oxide ( $SO_2$ ) is $64\text{ g/mol}$ , what is the ratio of the rate of diffusion of methane to that of sulfur(IV) oxide under identical conditions of temperature and pressure?

$2 : 1$ Answer
B
$4 : 1$
C
$1 : 2$
D
$1 : 4$

Answer

2 : 1

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

). Comparing methane (

CH_4

M = 16\text{ g/mol}

) to sulfur(IV) oxide (

SO_2

M = 64\text{ g/mol}

), the ratio of their rates is

\sqrt{\frac{64}{16}} = \sqrt{4} = 2

. Therefore, methane diffuses twice as fast as sulfur(IV) oxide, yielding a ratio of

2 : 1

Step-by-Step Solution

State Graham's Law of Diffusion formula

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

The rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

Substitute the given molar masses into the formula

\frac{r_{CH_4}}{r_{SO_2}} = \sqrt{\frac{64}{16}}

Molar mass of

SO_2 = 64\text{ g/mol}

and molar mass of

CH_4 = 16\text{ g/mol}

Simplify the ratio

\frac{r_{CH_4}}{r_{SO_2}} = \sqrt{4} = 2

Taking the square root of 4 gives 2, which corresponds to a ratio of

2 : 1

Key Concept

Graham's Law of Diffusion

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