Question

The time taken for effusion of 64 mL of oxygen will be same as the time taken for the effusion of which of the following gases under identical conditions?

• A. 64 mL of $H_2$
• B. 100 mL of $N_2$
• C. 64 mL of $C{O}_2$
• D. 45.24mL of $S{O}_2$

Answer 1

The correct option is D 45.24mL of $S{O}_2$

$\text{Rate of diffusion}=\frac{\text{Volume of gas diffused}}{\text{time taken}}$
$r_{O_2}=\frac{64}{t}$
$r_g=\frac{V}{t}$
$\frac{r_{O_2}}{r_g}=\frac{64}{V}$

Option (a)
$\frac{r_{O_2}}{r_{H_2}}=\sqrt{\frac{M_{H_2}}{M_{O_2}}}$
$\frac{64}{V}=\sqrt{\frac{2}{32}}\text{or}V=256\text{mL}$

Option (b)
$\frac{r_{O_2}}{r_{N_2}}=\sqrt{\frac{M_{N_2}}{M_{O_2}}}$
$\frac{64}{V}=\sqrt{\frac{28}{32}}\text{or}V=68.42\text{mL}$

Option (C)
$\frac{r_{O_2}}{r_{C{O}_2}}=\sqrt{\frac{M_{C{O}_2}}{M_{O_2}}}$
$\frac{64}{V}=\sqrt{\frac{44}{32}}\text{or}V=54.58\text{mL}$

Option (d)
$\frac{r_{O_2}}{r_{S{O}_2}}=\sqrt{\frac{M_{S{O}_2}}{M_{O_2}}}$
$\frac{64}{V}=\sqrt{\frac{64}{32}}\text{or}V=\frac{64}{\sqrt{2}}=45.25\text{mL}$
Hence, the correct answer is option (d).