Question

3b. The mole fraction He is 0.4 in gaseous mixture He
and CH4. If both the gaseous are effusing through
the constant area of the orifice of the container, then
what will be % composition by volume of CH4, gas
effusing out initially?
A) 50% B) 40% C) 43% D) 75%​

Answer 1

43% composition by volume of CH4 gas effusing out initially.

Explanation:

• $\frac{The rate of diffusion of He}{rate of diffusion of CH_ 4 } = \sqrt{\frac{molar mass of CH_4}{molar mass He} }$ $= \sqrt{\frac{16}{4} } = 4/2 = 2$
• From the above calculation it is inferred that Helium diffuses at twice the rate of $CH_4$.
• As per the given condition, the mole fraction of helium is 0.40. To find the rate of diffusion assume 10 moles as total in which  4 mole He and 6 mole $CH_4$
• Moles of He that diffuse is twice the moles of $CH_4$.
• Thus 4 moles of He will diffuse in the same period of time as 3 moles or 4.0 liters of He will diffuse in the same period of time as 3.0 liters of $CH_4$.

Percent by volume of $CH_4$ = $\frac{ volume of CH_4 }{total volume} \times 100$

% volume $CH_4 = \frac{3.0}{4.0+3.0} \times 100 = \frac{3}{7}\times 100 = 42.8%$

Thus it is inferred that 42.8% of $CH_4$ gas effuses out initially.

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