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Elevation in Boiling Point

The elevation...

Question

The elevation in boiling point of a solution of $9.43 g$ of $M g {C l}_{2}$ in $1 k g$ of water is : ( $K_{b} = 0.52 K$ $k g$ ${m o l}^{- 1}$ , Molar mass of $M g {C l}_{2} = 94.3 g$ ${m o l}^{- 1}$ )

0.156

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0.52

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0.17

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0.94

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Solution

The correct option is A $0.156$
${Δ T}_{b} = i \times K_{b} \times$ molarity
$= 3 \times 0.52 \times \frac{9.43}{95 \times 1}$ { $i$ for ${M g C l}_{2} = 3$ }
${Δ T}_{b} = 0.156 K$

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