Calculate the boiling point of a 4.0 m solution of sucrose in water? Sucrose is a nonelectrolyte.The normal boiling point of pure water is 100.0 $\degree$C.K_b (water) = 0.52 $\degree$C/m A)2.1 $\degree$C B)97.9 $\degree$C C)96.0 $\degree$C D)94.0 $\degree$C E)102.1 $\degree$C

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Quizplus · Accepted Answer

The boiling point elevation is calculated using the formula $\Delta T_b = i \cdot K_b \cdot m$. For sucrose, $i = 1$ (nonelectrolyte), $K_b = 0.52 \degree C/m$, and $m = 4.0$. Thus, $\Delta T_b = 1 \cdot 0.52 \cdot 4.0 = 2.08 \degree C$. The new boiling point is $100.0 + 2.08 = 102.08 \degree C$, approximately 102.1 $\degree$C.

Calculate the Boiling Point of a 4 °\degree° CKb (Water) = 0

Calculate the Boiling Point of a 4 $\degree$ CK_b (Water) = 0