Question

Pure $O_2$ diffuses through an aperture in $224$ second, whereas of $O_2$ and another gas containing $80$% $O_2$ diffuses from the same in $234$ second. The molecular mass of gas will be :

• A. $51.5$
• B. $48.6$
• C. $55$
• D. $46.6$

Answer 1

The correct option is D $46.6$

Let the weight of other gas in the mixture be x.

For gaseous mixture of $80\%O_2$ and $20\%$ another gas.

Average molecular weight of mixture $=\frac{32\times 80+20\times x}{100}⟶\left(1\right)$

Now, for diffusion of gaseous mixture (m) and pure $O_2$,

$\frac{r_{O_2}}{r_{mixture}}=\sqrt{\frac{M_{mixture}}{M_{O_2}}}=\frac{V_{O_2}}{t_{O_2}}\times \frac{t_{mixture}}{V_{mixture}}$

$\Rightarrow \sqrt{\frac{M_{mixture}}{32}}=\frac{1}{224}\times \frac{234}{1}$

$\Rightarrow M_{mixture}=32\times 1.0912$

$\Rightarrow M_{mixture}=34.92⟶\left(2\right)$

Now, from ${eq}^n\left(1\right)\&\left(2\right)$, we have

$\frac{32\times 80+20\times x}{100}=34.92$

$\Rightarrow 20x=3492-2560$

$\Rightarrow x=\frac{932}{20}=46.6$

Hence, the molecular weight of another gas is $46.6$.