Question

Pure $O_2$ diffuses through an aperture in 224 sec, whereas mixture of $O_2$ and another gas containing
80 % $O_2$ takes 234 sec to effuse out same volume. What is molecular weight of the gas ?

(Answer should be the closest integer to the molecular weight).

Answer 1

From the Graham's law of effusion, the time taken for effusion is directly proportional to the square root of the molar mass of a gas.
$\frac{t_{mix}}{t_{O_2}}=\sqrt{\frac{M_{mix}}{M_{O_2}}}$

Substitute values in the above expression.

$\frac{234}{224}=\sqrt{\frac{M_{mix}}{32}}$

$M_{mix}=34.92$ g/mol

Hence, the molar mass of the mixture is 34.92 g/mol.
Let $M$ be the molar mass of another gas.
Hence, $\frac{32\times 80+M\times 20}{100}=34.92$

$M=46.6$ g/mol

Thus, the molar mass of the gas is 46.6 g/mol.