Question

chapter 12 boiling point elevation practice worksheet

1. what is the boiling point of a solution prepared by adding 29.3 g of methanol (c₁₀h₂₀o) to 0.0590 kg of chloroform? (normal boiling point = 61.2 °c and kb = 3.85 °c/m)

Explanation:

Step1: Calculate moles of methanol

$n=\frac{m}{M}$, where $m = 29.3\ g$ and $M=156.26\ g/mol$. So $n=\frac{29.3\ g}{156.26\ g/mol}\approx0.1875\ mol$.

Step2: Calculate molality of the solution

$m=\frac{n}{m_{solvent}(kg)}$, with $n = 0.1875\ mol$ and $m_{solvent}=0.0590\ kg$. Then $m=\frac{0.1875\ mol}{0.0590\ kg}\approx3.18\ m$.

Step3: Calculate boiling - point elevation

$\Delta T_b = K_b\times m$, where $K_b = 3.85^{\circ}C/m$ and $m = 3.18\ m$. So $\Delta T_b=3.85^{\circ}C/m\times3.18\ m\approx12.2^{\circ}C$.

Step4: Calculate the boiling point of the solution

$T_{solution}=T_{solvent}+\Delta T_b$, with $T_{solvent}=61.2^{\circ}C$ and $\Delta T_b = 12.2^{\circ}C$. Then $T_{solution}=61.2^{\circ}C + 12.2^{\circ}C=73.4^{\circ}C$.

Answer:

$73.4^{\circ}C$