Question

$V$ ml of $O_2$ effuses through an aperture in $2.733$ minutes, whereas same volume of mixture of $O_2$ and another gas $X$ containing $20$% gas $X$ (by moles) effuses from the same aperture in $3.900$ minutes. What is the molecular mass of gas $X$?

• A. $68.5$
• B. $112.5$
• C. $576$
• D. $98.7$

Answer 1

The correct option is C $576$

Graham's law of diffusion, $\frac{r1}{r2}=\sqrt{\frac{M2}{M2}}$ where r is the rate of diffusion and M is the molar mass

$t_{O2}$=2.733 MIN=164 sec

$t_{mix}$=3.900 min=234 sec

$\frac{t_{mix}}{t_{o2}}=\frac{r_{o2}}{r_{mix}}=\sqrt{\frac{M_{mix}}{32}}$

$\frac{234}{164}=\sqrt{\frac{M_{mix}}{32}}$

$M_{mix}=65.15$

$\frac{1}{\sqrt{M_{mix}}}=\frac{X_{gas}}{\sqrt{M_{gas}}}+\frac{X_{O2}}{\sqrt{M_{O2}}}\frac{1}{\sqrt{65.15}}=\frac{0.2}{\sqrt{M_{gas}}}+\frac{0.8}{\sqrt{32}}{M}_{gas}=576$