Chemistry - Liquid Solution Question with Solution | TestHub

4.00
${\mathrm{X}}_{{\mathrm{N}}_2}=\frac{{\mathrm{P}}_{{\mathrm{N}}_2}^1}{{\mathrm{K}}_{\mathrm{H}}}=\frac{5\times 0·8}{1\times {10}^{-5}}$
$\left({\mathrm{P}}_{{\mathrm{N}}_2}^1={\mathrm{P}}_{\mathrm{a}\mathrm{i}\mathrm{r }}{\text{mole fraction of N}}_2\right)$
$=4\times {10}^{-5}$
$\therefore \frac{{\text{n}}_{{\text{N}}_2}}{{\text{n}}_{{\text{N}}_2}+{\text{n}}_{{\text{H}}_2\text{O}}}=4\times {10}^{–5}=\frac{{\text{n}}_{{\text{N}}_2}}{{\text{n}}_{{\text{N}}_2}+10}\simeq \frac{{\text{n}}_{{\text{N}}_{\text{2}}}}{10}$
$\therefore \text{n}=4\times {10}^{–4}$
Hence$\mathrm{x}=4$