What is the boiling point of a 6.00 m aqueous solution of a nonvolatile nonelectrolyte? (The boiling point elevation constant for water is 0.512^° C/m) A)100.^° C B)103.^° C C)3.07^° C D)96.9^° C

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Quizplus · Accepted Answer

The boiling point elevation ($\Delta T_b$) can be calculated using the formula $\Delta T_b = i \cdot K_b \cdot m$, where $i$ is the van't Hoff factor (which is 1 for a nonvolatile nonelectrolyte), $K_b$ is the boiling point elevation constant, and $m$ is the molality of the solution. Given $K_b = 0.512^\circ C/m$ and $m = 6.00 m$, $\Delta T_b = 1 \cdot 0.512^\circ C/m \cdot 6.00 m = 3.072^\circ C$. The normal boiling point of water is $100^\circ C$, so the boiling point of the solution is $100^\circ C + 3.072^\circ C = 103.072^\circ C$.

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