Question

Pure $O_2$ diffuses through an aperture in $224$ second, whereas mixture of $O_2$ and another gas containing $10$% $O_2$ diffuses from the same in $336$ second. What is the molecular mass of the other gas?

• A. 36
• B. 76.4
• C. 53
• D. 30.8

Answer 1

The correct option is B 76.4

To find the molecular mass of gas,

for gaseous mixture $10\%O_2,90\%$ gas:

The average molecular weight of mixture-

$M=\frac{32\times 10+90\times x}{100}$ ...(1)

For the diffusion of gaseous mixture and pure $O_2$

$\frac{r_{O_2}}{r_x}=\sqrt{\frac{M}{M_{O_2}}}$:

$\frac{V_{O_2}}{t_{O_2}}\times \frac{t_x}{V_x}$$=\sqrt{\frac{M}{M_{O_2}}}$

(Since volume diffuses)

$\frac{1}{224}\times \frac{336}{1}=\sqrt{\frac{M}{32}}$

$M=72$

Subsituting $M=72$ in equation (1)

$72=\frac{320+90\times x}{100}$

$x=76.4$

The molecular mass of gas is $76.4$.

Hence, option B is correct.