Question

Calculate the molality of an aqueous glucose solution in which mole fraction of glucose is 0.25.

• A. 20.64 m
• B. 13..88 m
• C. 18.26 m
• D. 18.52 m

Answer 1

The correct option is D 18.52 m

Given: mole fraction of glucose = 0.25
Let moles of glucose be $\left(n_g\right)$ and moles of water be $\left(n_w\right)$
Mole fraction = $\frac{\text{moles of glucose}\left(n_g\right)}{\text{moles of glucose}\left(n_g\right)+\text{moles of water}\left(n_w\right)}\Rightarrow 0.25=\frac{n_g}{n_g+n_w}$
solving we get $n_w$ = $3n_g$ ---- (1)
$\Rightarrow Molality=\frac{\text{moles of solute}}{\text{mass of solvent (g)}}\times 1000=\frac{n_g}{n_w\times 18}\times 1000$
By using the relation (1) we get,
Molality = $\frac{n_g}{3n_g\times 18}\times 1000=18.52$ m