Chemistry - Chemical Equilibrium Question with Solution | TestHub

According to Graham's law of effusion
$\mathrm{r} \propto \frac{1}{\sqrt{\mathrm{M}}}$

$\frac{{\text{r}}_{\text{mix}}}{{\text{r}}_{{\text{SO}}_2}}=\sqrt{\frac{{\text{M}}_{{\text{SO}}_2}}{{\text{M}}_{\text{mix}}}}$

$\Rightarrow \text{2.56}=\frac{64}{{\text{M}}_{\text{mix}}}$

$\Rightarrow {\text{M}}_{\text{mixture}}=25$

Let mole fraction of${\mathrm{F}}_2$is x. Therefore, mole fraction of$\mathrm{F}$will be$\left(1-\mathrm{x}\right)$

$25=\frac{38\times \text{x}+\left(1-\text{x}\right)\times 19}{1}$

$\text{x}=0.315$;

$\Rightarrow {\text{K}}_{\text{p}}=\frac{{\text{P}}_{\text{F}}^2}{{\text{P}}_{{\text{F}}_2}}=\frac{{\left(\text{0.685P}\right)}^2}{\left(\text{0.315P}\right)}\simeq \text{1.49 atm}$