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Kohlrausch Law

10 g of a non...

Question

$10 g$ of a non-electrolyte solute $(M = 60 g/mol)$ is added to $250 g$ of water. The boiling point of the solution is ( $K_{b}$ for water is $0.52 {K kg mol}^{- 1}$ and boiling point of water $= 373.15 K$ )
[0.77 Mark]

373.5 K

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383.15 K

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372.82 K

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364.48 K

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Solution

The correct option is A $373.5 K$
$W_{s o l u t e} = 10 g$
$M_{s o l u t e} = 60 g/mol$
$W_{s o l v e n t} = 250 g$
$K_{b} = 0.52 {K kg mol}^{- 1}$
By using the relation,
$Δ T_{b} = K_{b} m$
$= 0.52 \times \frac{10 \times 1000}{60 \times 250} = 0.35$
So, $T_{b} = T_{b}^{0} + Δ T_{b}$
$= 373.15 + 0.35 = 373.5 K$

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