Calculate the molal elevation constant (K_b) | vedclass

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Question

(C) The elevation in boiling point is given by the formula: $\Delta T_b = T_b - T_b^\circ$ Given,$T_b = 319.8 \ K$ and $T_b^\circ = 319.5 \ K$. So,$\D

Vedclass · Accepted Answer

$2.5 \ K \ kg \ mol^{-1}$

Vedclass · Answer

(C) The elevation in boiling point is given by the formula: $\Delta T_b = T_b - T_b^\circ$ Given,$T_b = 319.8 \ K$ and $T_b^\circ = 319.5 \ K$. So,$\Delta T_b = 319.8 \ K - 319.5 \ K = 0.3 \ K$. The formula for elevation in boiling point is $\Delta T_b = K_b 	imes m$,where $m$ is the molality. Given $m = 0.12 \ m$. Therefore,$K_b = \frac{\Delta T_b}{m} = \frac{0.3 \ K}{0.12 \ mol \ kg^{-1}} = 2.5 \ K \ kg \ mol^{-1}$. Thus,the correct option is $C$.

Calculate the molal elevation constant $(K_b)$ of a solvent if the boiling point of a $0.12 \ m$ solution is $319.8 \ K$ (Boiling point of pure solvent = $319.5 \ K$).

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