Question

The pressure in a vessel that contained pure oxygen dropped from 2000 mm to 1500 mm in 47 minutes as the oxygen leaked through a small hole into vacuum .When the same vessel was filled with another gas, the pressure dropped from 2000 mm to 1500 mm in 74 minute.What is the molecular weight of the gas ?

• A. 73
• B. 75
• C. 79
• D. 80

Answer 1

The correct option is C 79

Given,

pressure dropped in vessel containing oxygen from $2000mm$ to $1500mm$ in $47$ minutes

pressure dropped in vessel containing $x$ gas from $2000mm$ to $1500mm$ in $74$ minutes

Set rate of diffusion of $O_2=r_1$

& rate of diffusion of $X$ gas=$r_x$

According to Graham's law of diffusion,

$r\propto \sqrt{\frac{1}{\alpha }}\propto \sqrt{\frac{1}{M}}$ where $d$=density of gas, $M$= Molar mass of gas

$\Rightarrow \frac{r_1}{r_x}=\frac{t_x}{t_1}=\sqrt{\frac{M_x}{M_{O_2}}}$ where $M_x$=Mass of $x$, $M_{O_2}$=Mass of $O_2$, $T_x$=time required by $x$, $t_{O_2}$=time required by $O_2$ $[\because O_2$ Molar mass=$32]$

$\Rightarrow \frac{t_x}{t_1}=\sqrt{\frac{M_x}{M_{O_2}}}$

$=\frac{74}{47}=\sqrt{\frac{M_x}{32}}$

$\therefore$ $M_x=\frac{{74}^2}{{47}^2}\times 32$

$=79.3$

$=79$

$\therefore$ Molar mass of $x$ gas is $79g/mol$