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Graham's Law of Effusion

5 mL of H 2 g...

Question

$5 mL$ of $H_{2}$ gas diffuses out in $1$ second from a hole. Find the volume of $O_{2}$ that will diffuse out from the same hole under identical conditions in $2$ seconds.

2.5 mL

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2.5 L

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3.5 L

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4 L

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Solution

The correct option is A $2.5 mL$
$Rate of diffusion for a gas (r) = \frac{Volume of gas coming out}{Time taken} \propto \frac{1}{\sqrt{Molar mass of the gas}}$

$r_{H_{2}} = \frac{5 mL}{1 s} = 5 mL/s$
$\frac{r_{O_{2}}}{r_{H_{2}}} = \sqrt{\frac{M_{H_{2}}}{M_{O_{2}}}} = \sqrt{\frac{2}{32}} = \frac{1}{4}$
$∴ r_{O_{2}} = r_{H_{2}} \times \frac{1}{4} = 5 mL/s \times \frac{1}{4} = \frac{5}{4} m L / s$
Hence, the volume of $O_{2}$ diffused in 2 seconds $= \frac{5}{4} m L / s \times 2 s = 2.5 mL$

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Graham's Law of Effusion

Standard XII Chemistry

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5 mL of H2 gas diffuses out in 1 second from a hole. Find the volume of O2 that will diffuse out from the same hole under identical conditions in 2 seconds.

The correct option is A 2.5 mLRate of diffusion for a gas(r)=Volume of gas coming outTime taken∝1√Molar mass of the gas rH2=5 mL1 s=5 mL/s rO2rH2=√MH2MO2=√232=14 ∴rO2=rH2×14=5 mL/s×14=54 mL/s Hence, the volume of O2 diffused in 2 seconds =54 mL/s×2 s=2.5 mL

$5 mL$ of $H_{2}$ gas diffuses out in $1$ second from a hole. Find the volume of $O_{2}$ that will diffuse out from the same hole under identical conditions in $2$ seconds.