Question

A sugar syrup of weight $214.2 \,g$ contains $34.2 \,g$ 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 $= 214.2 - 34.2$
$= 180\, g$
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 $= \frac{\text{Mole of sugar}}{\text{Mole of sugar + Mole of water}}$
$= \frac{0.1}{0.1+10} =9.9 \times {10}^{-3}$