Question

A sugar syrup of weight $214.2g$ contains $34.2g$ of sugar $(C_{12}{H}_{22}{O}_{11})$ . Calculate (i) molal concentration and (ii) mole fraction of sugar in syrup.

Answer 1

Moles of sugar $=\frac{34.2}{342}=0.1$
Moles of water in syrup $=214.2-34.2$
$=180g$
Therefore, (i) Molality $=\frac{\text{Moles of solute}}{\text{Weight of Solvent }(g)}\times 1000$
$=\frac{0.1}{1800}\times 1000=0.55$
(ii) Mole fraction of sugar $=\frac{\text{Mole of sugar}}{\text{Mole of sugar + Mole of water}}$
$=\frac{0.1}{0.1+10}=9.9\times {10}^{-3}$