Question

An ideal aqueous solution containing liquid A ($M.wt=128$) $64\%$ by weight has a vapor pressure of 145 mm Hg. If the vapor pressure of A is x mm of Hg and that of water is 155 mm Hg at the same temperature. Then find $\dfrac{x}{5}$. The solution is ideal.

Answer 1

Hint: Question given above based on the raoult's law which defines that a solvents partial vapor in a solution or in any mixture is equal to the vapor pressure of the pure solvent multiplied by its mole fraction in the solution.

Complete step by step solution:
According to roult’s law the total pressure of the given solvent can be calculated with the formula:
${{P}_{t}}={{P}_{A}}+{{P}_{B}}$; where ${{P}_{t}}$= total pressure, ${{P}_{A}}$= pressure exerted by liquid A and ${{P}_{B}}$pressure exerted by liquid B.
While ${{P}_{A}}=P{{{}^\circ }_{A}}+{{x}_{A}}$ and ${{P}_{B}}=P{{{}^\circ }_{B}}+{{x}_{A}}$; Where $P{{{}^\circ }_{A}}$ and $P{{{}^\circ }_{B}}$ are partial vapour pressure exerted by liquids A and B, respectively and ${{x}_{A}}$ and ${{x}_{B}}$ showing mole fractions of A and B.
\[\therefore {{P}_{T}}=P{{{}^\circ }_{A}}{{x}_{A}}+P{{{}^\circ }_{B}}{{x}_{B}}\]
Total pressure given is = 145 mm
Molecular weight of A = 64gm, B = 100-64 = 36 gm
Mole fraction of A is given by $\dfrac{moles\, of\, A}{Total \,moles}=\dfrac{\dfrac{64}{128}}{\dfrac{64}{128}+\dfrac{36}{18}}=0.2$; $P{{{}^\circ }_{A}}$= x
$P{{{}^\circ }_{B}}$=155; Mole fraction of B = \[\dfrac{\text{moles of B}}{\text {Total moles}}=(1-{{x}_{A}})\]
Put all the values in equation
$145=x(0.2)+155(0.8)$; $x = 105$ $nm$
Then $\dfrac{x}{5}=\dfrac{105}{5}=21mm$

Note: Raoult’s law is valid only in the case of ideal solutions. Ideal solutions are those solutions in which the solvent-solute interaction is the same as a solvent – solvent or solute – solute interaction. Solute is the substance which is present in less amount while solvent is one which is present in larger amount.