What is the boiling point of a 0.5 m aqueous | vedclass

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(B) The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m$. For the glucose solution: $T_{b} - T_{b}^{\circ} = K_{b} \cdot m$. $100

Vedclass · Accepted Answer

$100.80$

Vedclass · Answer

(B) The elevation in boiling point is given by $\Delta T_{b} = K_{b} \cdot m$. For the glucose solution: $T_{b} - T_{b}^{\circ} = K_{b} \cdot m$. $100.16^{\circ} C - 100^{\circ} C = K_{b} 	imes 0.1 \ m$. $0.16 = K_{b} 	imes 0.1$,so $K_{b} = 1.6 \ ^{\circ} C \ kg \ mol^{-1}$. Now,for the sucrose solution: $\Delta T_{b} = K_{b} 	imes m = 1.6 	imes 0.5 = 0.80^{\circ} C$. The boiling point of the sucrose solution is $T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100^{\circ} C + 0.80^{\circ} C = 100.80^{\circ} C$.

What is the boiling point of a $0.5 \ m$ aqueous solution of sucrose if a $0.1 \ m$ aqueous solution of glucose boils at $100.16^{\circ} C$ (in $^{\circ} C$)?

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