Question

$5\text{mL}$ 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.

• A. $2.5\text{mL}$
• B. $2.5\text{L}$
• C. $3.5\text{L}$
• D. $4\text{L}$

Answer 1

The correct option is A $2.5\text{mL}$

$\text{Rate of diffusion for a gas}\left(r\right)=\frac{\text{Volume of gas coming out}}{\text{Time taken}}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$r_{H_2}=\frac{5\text{mL}}{1\text{s}}=5\text{mL/s}$
$\frac{r_{O_2}}{r_{H_2}}=\sqrt{\frac{M_{H_2}}{M_{O_2}}}=\sqrt{\frac{2}{32}}=\frac{1}{4}$
$\therefore r_{O_2}=r_{H_2}\times \frac{1}{4}=5\text{mL/s}\times \frac{1}{4}=\frac{5}{4}mL/s$
Hence, the volume of $O_2$ diffused in 2 seconds $=\frac{5}{4}mL/s\times 2\text{s}=2.5\text{mL}$