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:
$\begin{aligned} & \text { 53. } \text { Moles of sugar }=\frac{34.2}{342}=0.1 \\ & \text { Moles of water in syrup }=214.2-34.2=180 \mathrm{~g} \\ & \text { Therefore, (i) Molality }=\frac{\text { Moles of solute }}{\text { Weight of Solvent (g) }} \times 1000 \\ & =\frac{0.1}{180} \times 1000=0.55 \\ & \end{aligned}$
$\text { (ii) Mole fraction of sugar } \begin{aligned} & =\frac{\text { Mole of sugar }}{\text { Mole of sugar }+ \text { Mole of water }} \\ & =\frac{0.1}{0.1+10}=9.9 \times 10^{-3} \end{aligned}$
BETA
