For an aqueous solution, freezing point is $- 0.186℃$. Elevation of the boiling point of the same solution is ($k_{f} = 1.86{^\circ}\ mol^{- 1}\$kg and $k_{b} = 0.512{^\circ}\ mol^{- 1}\ kg)$

Question

Dr. Abubacker · Accepted Answer

(b)
$\mathrm{\Delta}T_{b} = mk_{b}$
$\ \ \ \ \ \ \ \ \ \ \ \mathrm{\Delta}T_{f} = mk_{f}\$
$\frac{\mathrm{\Delta}T_{b}}{\mathrm{\Delta}T_{f}} = \frac{k_{b}}{k_{f}} = \frac{0.512}{1.86}$
$\mathrm{\Delta}T_{b} = \frac{0.512}{1.86} 	imes 0.186$
=0.0512${^\circ}$

Q. For an aqueous solution, freezing point is $- 0.186℃$. Elevation of the boiling point of the same solution is ($k_{f} = 1.86{^\circ}\ mol^{- 1}\$kg and $k_{b} = 0.512{^\circ}\ mol^{- 1}\ kg)$

Explanation

Reviewed by Dr. Abubacker

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