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Molality

The boiling p...

Question

The boiling point elevation constant $(K_{b})$ of water is $0.52 {K kg mol}^{- 1}$ . The boiling point of $0.1 m a q .$ solution of urea is:

{105.2}^{\circ} C

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{1002.5}^{\circ} C

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{100.52}^{\circ} C

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{100.052}^{\circ} C

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Solution

The correct option is D ${100.052}^{\circ} C$
Given : $K_{b} = 0.52 K {kg mol}^{- 1}$ , molality = $0.1 m$
$Δ T_{b} = K_{b} m = 0.52 \times 0.1 = 0.052 K$
Since, difference on kelvin scale and celsius scale is same, so $Δ T_{b} = 0.052 K = {0.052}^{\circ} C$
Hence, boiling point of solution $= (T_{b})_{H_{2} O} + Δ T_{b}$
$= 100^{\circ} C + {0.052}^{\circ} C$
$= {100.052}^{\circ} C$

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