0.1 molal aqueous solution of glucose boils | vedclass

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(A) Given:- For $0.1 \ molal$ solution,$T_b = 100.16^{\circ} C$.- Boiling point of pure water is $100^{\circ} C$.- Boiling point elevation formula: $\

Vedclass · Accepted Answer

$100.80$

Vedclass · Answer

(A) Given:- For $0.1 \ molal$ solution,$T_b = 100.16^{\circ} C$.- Boiling point of pure water is $100^{\circ} C$.- Boiling point elevation formula: $\Delta T_b = K_b \cdot m$ (since glucose is a non-electrolyte,$i = 1$).Step $1$: Calculate elevation for $0.1 \ molal$ solution:$\Delta T_b = 100.16 - 100 = 0.16^{\circ} C$.Step $2$: Use the proportionality $\Delta T_b \propto m$ to find elevation for $0.5 \ molal$ solution:$\Delta T_b^{\prime} = \Delta T_b 	imes \frac{m^{\prime}}{m} = 0.16 	imes \frac{0.5}{0.1} = 0.16 	imes 5 = 0.8^{\circ} C$.Step $3$: Calculate new boiling point:$T_b^{\prime} = 100 + 0.8 = 100.80^{\circ} C$.Thus,the correct option is $A$.

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

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