Chemistry - Liquid Solution Question with Solution | TestHub

Step I : Calculation of vapour pressure of pure liquid A (p°_A)
​Number of moles of liquid A
$\left({\text{n}}_{\text{A}}\right)=\frac{{\text{W}}_{\text{A}}}{{\text{M}}_{\text{A}}}=\frac{\text{(100 g)}}{\left({\text{140 g mol}}^{-1}\right)}=\text{0.7143 mol}$
Number of moles of liquid B
$\left({\text{n}}_{\text{B}}\right)=\frac{{\text{W}}_{\text{B}}}{{\text{M}}_{\text{B}}}=\frac{\text{(1000 g)}}{\left({\text{180 g mol}}^{-1}\right)}=\text{5.5556 mol}$
Mole fraction of A
$\left({\text{x}}_{\text{A}}\right)=\frac{{\text{n}}_{\text{A}}}{{\text{n}}_{\text{A}}+{\text{n}}_{\text{B}}}$
$=\frac{\text{(0.7143 mol)}}{\text{(0.7143 + 5.5556) mol}}=\frac{\text{0.7143}}{\text{6.2699}}=\text{0.1139}$
Mole fraction of${\text{B(x}}_{\text{B}}\text{)}\text{=}1-0.1139=0.8861$
Vapour pressure of pure liquid B,${\text{B(p}}_{\text{B}}^{\text{o}}\text{)}\text{=}\text{500}\text{torr}$
Total vapour pressure of solution$\text{(p)}=475\text{torr}$
According to the Raoult's law
$\text{p}\text{=}{\text{p}}_{\text{A}}^{\text{o}}{\text{x}}_{\text{A}}\text{+}{\text{p}}_{\text{B}}^{\text{o}}{\text{x}}_{\text{B}}$
$\text{475}\text{torr}\text{=}{\text{p}}_{\text{A}}^{\text{o}}\text{×}\text{(0}\text{.1139)}\text{+}\text{500}\text{torr}\text{×}\text{(0}\text{.8861)}$
$\text{475}\text{torr}\text{=}{\text{p}}_{\text{A}}^{\text{o}}\text{×}\text{(0}\text{.1139)}\text{+}\text{443}\text{.05}\text{torr}$
${\text{p}}_{\text{A}}^{\text{o}}=\frac{\left(\text{475}-\text{443.05}\right)\text{torr}}{\text{(0.1139)}}=\frac{\text{31.95}}{\text{0.1139}}\text{torr}=\text{280.5 torr}$

Step II : Calculation of vapour pressure of A in the solution (p_A)
According to Raoult's law,
${\text{p}}_{\text{A}}\text{=}{\text{p}}_{\text{A}}^{\text{o}}{\text{x}}_{\text{A}}\text{=}\text{(280}\text{.5}\text{torr)}\text{×}\text{(0}\text{.1139)}$
${\text{p}}_{\text{A}}\text{=}\text{32}\text{.0}\text{torr}$