Question

Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. $20d{m}^3$ of $S{O}_2$ diffuses through the porous partition in $60$ seconds. The volume of $O_2$ in $d{m}^3$ which diffuses under the similar condition in 30 seconds will be:(Atomic mass of sulphur $32u$)

• A. $7.09$
• B. $14.1$
• C. $10.0$
• D. $28.2$

Answer 1

The correct option is B $14.1$

$\frac{r_1}{r_2}=\frac{\frac{V_1}{t_1}}{\frac{V_2}{t_2}}=\sqrt{\frac{M_2}{M_1}}$
$\frac{\frac{20}{60}}{\frac{V_1}{30}}=\sqrt{\frac{32}{64}}$
$V_1=14.1d{m}^3$.