Question

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

• A. 7.09
• B. 14.1
• C. 10.1
• D. 28.2

Answer 1

The correct option is B 14.1

According to Graham's Law Diffusion at constant T and P:

$\frac{r_1}{r_2}=\sqrt{\frac{M_1}{M_2}}$

Since rate of diffusion$=\frac{Volumeofgasdiffused\left(V\right)}{Timetakenfordiffusion\left(t\right)}$

$\therefore \frac{r_1}{r_2}=\frac{V_1/t_1}{V_2/t_2}$

$\frac{r_1}{r_2}=\frac{V_1/t_1}{V_2/t_2}=\sqrt{\frac{M_1}{M_2}}$

$=\frac{20/60}{V_2/30}=\sqrt{\frac{32}{64}}=\sqrt{\frac{1}{2}}$

$\frac{10}{V_2}=\sqrt{\frac{1}{2}};V_2=10\sqrt{2}$

$V_2=14.1d{m}^3$