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Question

At certain hill station pure water boils at ${99.725}^{o} C$ . If $K_{b}$ for water is ${0.513}^{o} C$ Kg $m o 1^{- 1}$ . The boiling point of 0.69 molar solution of Urea will be:

{100.079}^{o} C

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103^{o} C

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{100.359}^{o} C

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{10.9}^{o} C

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Solution

The correct option is A ${100.079}^{o} C$
$Δ T_{b} = K_{b} m$
$= 0.513 \times 0.69$
$= {0.35397}^{0} C$
$T - T_{b} = 0.35397$
$T = 99.725 + .35397$
$= {100.079}^{0} C$

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