Question

The vapour pressure of a dilute aqueous solution of glucose is $750\text{mm Hg}$ at $373\text{K}$. Calculate:
$\left(i\right)$ Molality
$\left(ii\right)$ Mole fraction of the solute

• A. $\left(i\right)0.013\text{m},\left(ii\right)0.74$
• B. $\left(i\right)0.74\text{m},\left(ii\right)0.0132$
• C. $\left(i\right)0.74\text{m},\left(ii\right)0.98$
• D. $\left(i\right)0.013\text{m},\left(ii\right)0.26$

Answer 1

The correct option is B $\left(i\right)0.74\text{m},\left(ii\right)0.0132$

(ii) We know,
$p_0=\text{Vapour pressure of water at 373 K = 760 mm Hg}$
Given,
$\text{Vapour pressure of solution =}{p}_s=750\text{mm}$
As the solution is dilute solution of glucose, hence
$\text{Molar mass of solvent}{M}_A=18{\text{g mol}}^{-1}$
Relative lowering of vapour pressure in dilute form:
$\frac{p^{\circ }-p_s}{p^{\circ }}=x_B$
$\text{Where},x_B=\text{mole fraction of solute}$
$\Rightarrow x_B=\frac{760-750}{760}{x}_B=0.0132$
$\left(i\right)\text{Molality}=\frac{x_B\times 1000}{(1-x_B)\times M}$
Where,
$x_B=\text{mole fraction of solute}M=\text{Molar mass of solvent}=18{\text{g mol}}^{-1}$

Substituting the values, we get,
$\text{Molality}=\frac{0.013\times 1000}{(1-0.013)\times 18}\text{m}$
$\text{Molality}=0.741\text{m}$