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Elevation in Boiling Point

The boiling p...

Question

The boiling point of pure solvent is is $27^{\circ} C$ and its heat of vapourisation is $84 c a l / g$ . Determine the boiling point elevation constant $(K_{b})$ for the solvent

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Solution

We know,
Boiling point elevation constant,
$K_{b} = \frac{R T_{b}^{2}}{1000 L_{v}}$
where,
$T_{b}$ is boiling point of pure solvent in $K$
$R in c a l / (m o l . K)$
$L_{v}$ is latent heat of vaourisation of solvent in $c a l / g$
Thus,
$K_{b} = \frac{2 \times (27 + 273)^{2}}{1000 \times 84} = 2.14 K k g m o l^{- 1}$

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