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An aqueous solution containing ‘m’ moles of non-volatile non-electrolyte solute freezes at $- 0 . 186^{0} C .$ The elevation in the boiling point of the same aqueous solution $(K = 1.86 K kg {mol}^{- 1}, K = 0.512 K kg {mol}^{- 1})$ would be

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0.186

0.512

0.0512

0.512/1.86

answer is C.

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Detailed Solution

$ΔTb = i \times k_{b} \times M$

$\frac{∆ T_{b}}{∆ T_{f}} = \frac{K_{b}}{K_{f}}$

$∆ T_{b} = \frac{0.512 \times 0.186}{1.86} = 0.0512$

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