Some Basic Concepts of Chemistry - Result Question 54
####53. A sugar syrup of weight $214.2 g$ contains $34.2 g$ of sugar $\left(C _{12} H _{22} O _{11}\right)$. Calculate (i) molal concentration and (ii) mole fraction of sugar in syrup.
(1988, 2M)
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Solution:
- Moles of sugar $=\frac{34.2}{342}=0.1$
Moles of water in syrup $=214.2-34.2=180 g$
Therefore, (i) Molality $=\frac{\text { Moles of solute }}{\text { Weight of Solvent }(g)} \times 1000$
$$ =\frac{0.1}{180} \times 1000=0.55 $$
(ii) Mole fraction of sugar $=\frac{\text { Mole of sugar }}{\text { Mole of sugar }+ \text { Mole of water }}$ $=\frac{0.1}{0.1+10}=9.9 \times 10^{-3}$
