Chemistry - Liquid Solution Question with Solution | TestHub

By raoult's law
$\frac{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}}=\frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}+{\mathrm{n}}_{\mathrm{B}}}$
Or$\frac{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{A}}}=\frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}}$
${\mathrm{P}}_{\mathrm{A}}$:= the partial vapour pressure of the component {display style$\text{i}$} in the gaseous mixture (above the solution),
${\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}$= the vapour pressure of the pure component
${\mathrm{n}}_{\mathrm{B}}$: moles of solute
${\mathrm{n}}_{\mathrm{A}}$: moles of solvent
$\frac{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{o}}-{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{A}}}=\frac{{\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}}}$
$\frac{25.5-{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{A}}}=\frac{\frac{20}{180}}{\frac{100}{18}}$
$\frac{25.5-{\mathrm{P}}_{\mathrm{A}}}{{\mathrm{P}}_{\mathrm{A}}}=\frac{20\times 18}{100\times 180}=0.02$
$25.5-{\text{P}}_{\text{A}}=0.02{\text{P}}_{\text{A}}$
$1.02{\text{P}}_{\text{A}}=25.5$
${\mathrm{P}}_{\mathrm{A}}=25.0$