Question

Pure $O_2$ diffuses through an aperture in 224 s, whereas a mixture of $O_2$ and another gas containing 80% $O_2$ diffuses from the same in 234s under similar conditions of pressure and temperature.What is molecular weight of gas?

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

The gaseous mixture contains 80% $O_2$ and 20% gas.The average molecular weight of mixture is

$M_{mix}=M_i{x}_i$ ----------------------1

$As{M}_{mix}=M_i{x}_{i}where{x}_i=molefraction.Nowfordiffusionofgaseousmixtureandpure{O}_2$

$\frac{r_{O_2}}{r_{mix}}=\sqrt{\frac{M_{mix}}{M_{O_2}}}\Rightarrow ⟮\frac{V_{O_2}}{V_{mix}}⟯\times ⟮\frac{t_{mix}}{t_{O_2}}⟯$

= $\sqrt{\frac{M_{mix}}{32}}\Rightarrow \frac{1}{224}\times \frac{234}{1}=\frac{M_{mix}}{32}$ -------------------2

On solving , we get $M_{mix}=34.92gmo{l}^{-1}.$ Thus, substituting in eq(1), we get molecular weight of gas(m)

as $M=\frac{⟮34.92\times 100⟯-⟮32\times 80⟯}{20}=46.6gmo{l}^{-1}$