Question

The vapour pressure of a dilute solution of glucose $\left({\text{C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}\right)$ is $750mm$ of mercury at $373K$. Calculate (i) molality, and (ii) mole fraction of the solution.

Answer 1

$\to$ we know that,

Molarity $M=\frac{molesofsolote}{solvent\left(kg\right)}$

Mole fraction $;x_A=\frac{n_A}{n_A+n_B}$

& $x_B=\frac{n_B}{n_A+n_B}$

$P_1=x_1{P}_1^0\Rightarrow x_1=\frac{p_1}{p_1^0}=\frac{750}{760}=0.9868$

$x_2\left(solote\right)=1-0.9868=0.0132$

$\therefore$ Molality $=\frac{x_2}{x_1{M}_1}\times 1000=\frac{0.0132}{0.9868\times 18}\times 1000$ [where x of $H_{2}O=18]$

$=0.7503$ mol $k{g}^{-1}$

$\therefore$ Molality = 0.7503 mol$k{g}^{-1}$ & Mole fraction = $x_2=0.0132$