Chemistry - Liquid Solution Question with Solution | TestHub

Liquid A & Liquid $\mathrm{B}\to$ both Volatile

$$\begin{gathered} \text{displaylines}{n}_A=2 \\ {n}_B=3 \end{gathered}$$

$$\begin{gathered} \text{displaylines}{P}_A^o=100mmHg \\ {P}_B^o=200mmHg \end{gathered}$$

$\to itmeansBismorevolatileatsametemperature$

$$\begin{gathered} \text{displaylines}{X}_A=\frac{2}{5} \\ {X}_B=\frac{3}{5} \end{gathered}$$

$P_T=P_A^o{X}_A+P_B^o{X}_B$

$$\begin{gathered} \text{displaylines}=100\times \frac{2}{5}+200\times \frac{3}{5}=100\left\{\frac{2}{5}+\frac{6}{5}\right\}=\frac{800}{5} \\ {P}_T=160mmHg \end{gathered}$$

(A) on ${\uparrow }_{\text{ing }}$ Temp. → V.P. $\uparrow$

(B) On ${\uparrow }_{\text{ing }}$ Temp. → V.p. of $\mathrm{A}\uparrow$

(C) On ${\mathrm{X}}_{\mathrm{B}}{\uparrow }_{\text{ing }}\to$ V.P. $\uparrow$ because B is move volatile ( ${\mathrm{P}}^o{}_{\mathrm{B}}>{\mathrm{P}}^o{}_{\mathrm{A}}$ )

(D) On ${\uparrow }_{\text{ing }}{\mathrm{X}}_{\mathrm{A}}\to$ V.P. $\uparrow$ Because A is less volatile. ( ${\mathrm{P}}^{\mathrm{o}}{}_{\mathrm{A}}<{\mathrm{P}}^{\mathrm{o}}{}_{\mathrm{B}}$ )