If 50\text{ cm}^3 of sulfur dioxide (SO_2) gas diffuses through a porous partition in 20\t… — JAMB UTME

If $50\text{ cm}^3$ of sulfur dioxide ( $SO_2$ ) gas diffuses through a porous partition in $20\text{ seconds}$ , and the same volume of an unknown hydrocarbon gas $Z$ diffuses under identical conditions of temperature and pressure in $10\text{ seconds}$ , what is the molar mass of gas $Z$ in $\text{g/mol}$ ? (Relative atomic masses: $\text{S} = 32$ , $\text{O} = 16$ )

Answer: 16 / 16 g/mol / 16g/mol / 16 g mol^-1

Answer

16 g/mol

According to Graham's Law, the time required for equal volumes of two gases to diffuse under the same conditions is directly proportional to the square root of their molar masses:

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

. Sulfur dioxide (

SO_2

) has a molar mass of

64\text{ g/mol}

and takes

20\text{ seconds}

. Gas

Z

takes

10\text{ seconds}

, which is half the time, meaning its molar mass must be

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

of the molar mass of

SO_2

. Therefore,

M_Z = \frac{64}{4} = 16\text{ g/mol}

Step-by-Step Solution

Determine the relative molar mass of sulfur dioxide (

SO_2

\text{Molar mass of } SO_2 = 32 + 2(16) = 64\text{ g/mol}

The molar mass of the reference gas is needed to apply Graham's Law.

Set up the relationship between diffusion time (

t

) and molar mass (

M

) according to Graham's Law for equal volumes of gases.

\frac{t_Z}{t_{SO_2}} = \sqrt{\frac{M_Z}{M_{SO_2}}}

For equal volumes of gas, the time taken to diffuse is directly proportional to the square root of the molar mass.

Substitute the known diffusion times and molar mass of

SO_2

into the ratio.

\frac{10}{20} = \sqrt{\frac{M_Z}{64}} \implies \frac{1}{2} = \sqrt{\frac{M_Z}{64}}

Insert

t_Z = 10\text{ s}

t_{SO_2} = 20\text{ s}

, and

M_{SO_2} = 64\text{ g/mol}

into the equation.

Square both sides of the equation and solve for

M_Z

\left(\frac{1}{2}\right)^2 = \frac{M_Z}{64} \implies \frac{1}{4} = \frac{M_Z}{64} \implies M_Z = 16\text{ g/mol}

Eliminate the radical to calculate the molar mass of the unknown gas.

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, which implies that the time taken for a given volume to diffuse is directly proportional to the square root of its molar mass.

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