Question

The vapour pressure of $5$% aqueous solution of a non-volatile organic substance at $373$ K is $750$ mm. Calculate the molar mass of the solute. Given vapour pressure of pure water at $373$ K is $760$ mm of $Hg$.

• A. $50$ g
• B. $29$ g
• C. $36$ g
• D. $72$ g

Answer 1

The correct option is D $72$ g

$5$% aqueous solution of a non-volatile organic substance means, $100g$ of solution contains $5g$ organic solute and $95g$ water.

Thus,

$W_B=5g$, that is mass of solute
$W_A=95g$, that is mass of solvent
$M_A=18g$, that is a molecular mass of water.
$M_B$ that is a molecular mass of solute is unknown.

Vapor pressure of pure water $p_A^o=760$ mm of $Hg$

Vapor pressure of the solution after addition of solute that is $p_A=50$ mm of $Hg$

Lowering of vapor pressure of a solution states that $\frac{p_A^o-p_A}{p_A^o}=X_B$

$\frac{p_A^o-p_A}{p_A^o}=\frac{n_B}{n_A}$

$\frac{p_A^o-p_A}{p_A^o}=\frac{W_B\times M_A}{W_A\times M_B}$

Putting all the values in the equation we get-

$\frac{760-750}{760}=\frac{5\times 18}{95\times M_B}$

From this the value of $M_B$ is obtained as $72g$.