Some Basic Concepts of Chemistry - Result Question 40
53. A sugar syrup of weight 214.2 g contains 34.2 g of sugar $\left(\mathrm{C}{12} \mathrm{H}{22} \mathrm{O}_{11}\right)$. Calculate (i) molal concentration and (ii) mole fraction of sugar in syrup.
(1988, 2M)
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Answer:
Correct Answer: 53. $\left(9.9 \times 10^{-3}\right)$
Solution:
- Moles of sugar=34.2/342=0.1 Moles of water in syrup=214.2-34.2=180 g Therefore, (i) Molality=Moles of solute/Weight of Solvent (kg) × 1000 =0.1/0.18 × 1000=5.56
$\text { (ii) Mole fraction of sugar } \begin{aligned} & =\frac{\text { Mole of sugar }}{\text { Mole of sugar }+ \text { Mole of water }} \\ & =\frac{0.1}{0.1+10}=9.9 \times 10^{-3} \end{aligned}$
BETA
