Question

180 ml of hydrocarbon having the molecular weight 16 diffuses in 1.5 min. Under similar conditions time taken by 120 ml of $S{O}_2$ to diffuse is:

• A. $2$ min
• B. $1.5$ min
• C. $1$ min
• D. $1.75$ min

Answer 1

The correct option is A $2$ min

• According to Graham's law, $\frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$
  

where $r_1$ and $r_2$ are the rates of diffusion of gases 1 and 2 respectively

and, $M_1$ and $M_2$ are the molecular weights of gases 1 and 2 respectively

• $\text{rate}=\frac{\text{volume of gas}}{\text{time required for diffusion}}$

$r_1=\frac{180}{1.5}=120$

$\frac{120}{r_2}=\sqrt{\frac{64}{16}}$

$r_2=60$

• $r_2=\frac{120}{t_2}$
  
• $t_2=\frac{120}{60}$

$t_2=2min$