At STP, 0.48g of O₂ diffused through a porous partition in 1200 seconds. The volume of CO₂ diffused under same condition is ?

According to Graham's law of diffusion,

$\frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{\frac{{\mathrm{V}}_1}{\mathrm{t}}}{\frac{{\mathrm{V}}_2}{\mathrm{t}}}=\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\sqrt{\frac{{\mathrm{M}}_2}{{\mathrm{M}}_1}}$

volume of 0.48g of ${\mathrm{O}}_2$ at STP = $\frac{22400}{32}\times 0.48\mathrm{mL}$

(32g of ${\mathrm{O}}_2$ occupies 22400mL at STP)

$\therefore \frac{{\mathrm{V}}_{{\mathrm{O}}_2}}{{\mathrm{V}}_{{\mathrm{CO}}_2}}=\sqrt{\frac{{\mathrm{M}}_{{\mathrm{CO}}_2}}{{\mathrm{M}}_{{\mathrm{O}}_2}}}=\sqrt{\frac{44}{32}}=1.173$

${\mathrm{V}}_{{\mathrm{CO}}_2}=\frac{{\mathrm{V}}_{{\mathrm{O}}_2}}{1.173}=\frac{22400}{32}\times \frac{0.48}{1.173}=286.5\mathrm{mL}$