Question

A solution containing $30g$ of a non-volatile solute in exactly $90g$ water has a vapour pressure of $21.85mm$ $Hg$ at ${25}^{o}C$. Further $18g$ of water is then added to the solution.The resulting solution has a vapour pressure of $22.15mm$ $Hg$ at ${25}^{o}C$. Calculate the molecular weight of the solute.

• A. $39.9$
• B. $67.9$
• C. $78.8$
• D. $125.3$

Answer 1

Answer: (B)

$67.9$

Acc. to Raoult's law

given $30gm$ volatile $90gm$ water

$21\cdot 85\cdot =\left(P_A\right)\left(\frac{\frac{30}{x}}{\frac{30}{x}+\frac{90}{18}}\right)eq-\left(1\right)$

Case 2 when water is added

$22\cdot 15=\left(P_A\right)\left(\frac{\frac{30}{x}}{\frac{30}{x}+\frac{108}{18}}\right)eq-\left(2\right)$

dividing 2 equations and solve then $\left(x\right)$

we got. $x=67.9$ grams which molecular weight of solute