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(B) The formula for elevation in boiling point is $\Delta T_{b} = K_{b} 	imes m$. First,calculate the molality $(m)$: $m = \frac{	ext{moles of solut

Vedclass · Accepted Answer

$100.256$

Vedclass · Answer

(B) The formula for elevation in boiling point is $\Delta T_{b} = K_{b} 	imes m$. First,calculate the molality $(m)$: $m = \frac{	ext{moles of solute}}{	ext{mass of solvent in kg}} = \frac{0.1 \ mol}{0.2 \ kg} = 0.5 \ mol \ kg^{-1}$. Now,calculate the elevation in boiling point: $\Delta T_{b} = 0.513 \ ^{\circ}C \ kg \ mol^{-1} 	imes 0.5 \ mol \ kg^{-1} = 0.2565 \ ^{\circ}C$. The boiling point of the solution $(T)$ is $T_{b} + \Delta T_{b} = 100 \ ^{\circ}C + 0.2565 \ ^{\circ}C = 100.2565 \ ^{\circ}C$,which is approximately $100.256 \ ^{\circ}C$.

The molal elevation boiling point constant for water is $0.513 \ ^{\circ}C \ kg \ mol^{-1}$. Calculate the boiling point of the solution if $0.1 \ mol$ of sugar is dissolved in $200 \ g$ of water. (in $^{\circ}C$)

Similar Questions

If $K_b$ denotes the molal elevation constant of water,then the boiling point of an aqueous solution containing $36 \ g$ of glucose (molar mass $= 180 \ g \ mol^{-1}$) per $dm^3$ is:

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