Under specific laboratory conditions of temperature and pressure, a 50\text{ cm}^3 sample … — JAMB UTME

Under specific laboratory conditions of temperature and pressure, a $50\text{ cm}^3$ sample of sulfur(IV) oxide ( $SO_2$ ) gas diffuses through a tiny orifice in $24\text{ seconds}$ . What is the time, in seconds, required for an equal volume of helium ( $He$ ) gas to diffuse through the same orifice under identical conditions? [Relative atomic masses: $\text{He} = 4$ , $\text{O} = 16$ , $\text{S} = 32$ ]

Answer: 6 s

Answer

The time required for an equal volume of helium gas to diffuse under identical conditions is

6\text{ seconds}

According to Graham's Law of Diffusion, the time required for equal volumes of two gases to diffuse at constant temperature and pressure is directly proportional to the square root of their molar masses (

t_1 / t_2 = \sqrt{M_1 / M_2}

). Because helium (

4\text{ g/mol}

) has one-sixteenth the molar mass of sulfur(IV) oxide (

64\text{ g/mol}

), its diffusion time is

\sqrt{1/16} = 1/4

of the time taken by

SO_2

. Therefore,

t_{He} = 24 \text{ s} \times 0.25 = 6\text{ seconds}

Step-by-Step Solution

Calculate the molar masses of both gases from their chemical formulas and relative atomic masses.

Molar mass of

SO_2 = 64\text{ g/mol}

; Molar mass of

He = 4\text{ g/mol}

Graham's Law requires the molar mass of each gas to determine relative diffusion rates and times.

Set up the Graham's Law expression relating diffusion time to molar mass for equal volumes.

\frac{t_{He}}{t_{SO_2}} = \sqrt{\frac{M_{He}}{M_{SO_2}}}

Diffusion time for a given volume is directly proportional to the square root of the molar mass.

Substitute given values and simplify the square root expression.

\frac{t_{He}}{24} = \sqrt{\frac{4}{64}} = \sqrt{\frac{1}{16}} = \frac{1}{4}

Simplifying the molar mass ratio gives a simple fraction.

Multiply to solve for the diffusion time of helium gas.

t_{He} = 24 \times \frac{1}{4} = 6\text{ seconds}

Helium is lighter and diffuses 4 times faster than

SO_2

, requiring 1/4th of the time.

Key Concept

Graham's Law of Diffusion and Effusion

Estimated Time:1m 30s

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