Chemistry - Liquid Solution Question with Solution | TestHub

(P) ${\mathrm{P}}_{\mathrm{A}}={\mathrm{X}}_{\mathrm{A}}\cdot {\mathrm{P}}_{\mathrm{A}}^0=0.50\times 40=20\mathrm{cm}\mathrm{Hg}$ and $P_{\text{total }}=X_A\cdot P_A^0+X_B\cdot P_B^0=0.50\times 40+0.5\times 80=60\mathrm{cm}\mathrm{Hg}$

(Q) $P_A=X_A\cdot P_A^0=0.40\times 40=16\mathrm{cm}\mathrm{Hg}$ and $P_{\text{total }}=X_A\cdot P_A^0+X_B\cdot P_B^0=0.4\times 40+0.6\times 80=64\mathrm{cm}\mathrm{Hg}$

(R) $\frac{1}{P_{\text{total }}}=\frac{Y_A}{P_A^0}+\frac{Y_B}{P_B^0}=\frac{0.5}{40}+\frac{0.5}{80}\Rightarrow P_{\text{total }}=53.33\mathrm{cm}\mathrm{Hg}$ and ${\mathrm{P}}_{\mathrm{A}}={\mathrm{Y}}_{\mathrm{A}}\cdot {\mathrm{P}}_{\text{total }}=0.50\times 53.33=26.67\mathrm{cm}\mathrm{Hg}$

(S) $\frac{1}{P_{\text{total }}}=\frac{Y_A}{P_A^0}+\frac{Y_B}{P_B^0}=\frac{0.40}{40}+\frac{0.60}{80}\Rightarrow P_{\text{total }}=57.14\mathrm{cm}\mathrm{Hg}$ and ${\mathrm{P}}_{\mathrm{A}}={\mathrm{Y}}_{\mathrm{A}}\cdot {\mathrm{P}}_{\text{total }}=0.40\times 57.14=22.86\mathrm{cm}\mathrm{Hg}$