Question

Pure $O_2$ diffuses through an aperture in 228 seconds whereas, mixture of $O_2$ and another gas containing 64% $O_2$ by mass diffuses from the same in 190 seconds. The molecular mass of gas (in g) will be:

• A. 51
• B. 28
• C. 18
• D. 36

Answer 1

The correct option is C 18

Let the molecular mass of gas be M.
The mass of $O_2$ in the mixture is $64\text{g}$ and the mass of another gas is $36\text{g}$.
Number of moles of $O_2=\frac{64}{32}=2$
Number of moles of gas$=\frac{36}{M}$
Using Graham's law of effusion,
$\frac{t_{\text{mix}}}{t_{O_2}}=\frac{r_{O_2}}{r_{\text{mix}}}=\sqrt{\frac{M_{\text{mix}}}{36}}$
$\frac{190}{228}=\sqrt{\frac{M_{\text{mix}}}{36}}$
$M_{\text{mix}}=25$
$\Rightarrow M_{\text{mix}}=\frac{\text{Total mass}}{\text{Totle mole}}$
$\Rightarrow M_{\text{mix}}=\frac{100}{2+36/M}$
$M=18g$