How many moles of sucrose should be dissolved it 500 g of water so as to get a solution which has 3 difference of $104 ° C$ between boiling point and freezing point? $(K_{f} = 1.86 K kg {mol}^{- 1}, K_{b} = 0.52 K kg m o l^{- 1})$

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1.68

1.492

8.40

0.840

answer is D.

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Detailed Solution

We know that the boiling point and freezing point of water are $100 ° C$ and $0 ° C$ respectively, So

$104^{0} C = {ΔT}_{b} - {ΔT}_{f} + 100^{\circ} C$

$\begin{array}{l} (Δ T_{b} - Δ T_{f}) = 4 \\ (1.86 + 0.52) \times \frac{n}{500} \times 1000 = 4 \Rightarrow n = 0.84 0 \end{array}$

$=$ 0.840 moles of sucrose.

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