# Question e16a6

Sep 3, 2017

Here's what I got.

#### Explanation:

The key here is the difference between the ebullioscopic constant, ${K}_{b}$, and the cryoscopic constant, ${K}_{f}$, of water. the solvent of the solution.

You know that you have--see here for the source

• ${K}_{b} = {0.512}^{\circ} {\text{C kg mol}}^{- 1}$
• ${K}_{f} = {1.86}^{\circ} {\text{C kg mol}}^{- 1}$

Now, the boiling-point elevation is calculated using

$\Delta {T}_{\text{b}} = i \cdot {K}_{b} \cdot b$

Here

• $i$ is the van't Hoff factor
• $b$ is the molality of the solution

Similarly, the freezing-point depression is calculated using

$\Delta {T}_{\text{f}} = i \cdot {K}_{f} \cdot b$

Since you're dealing with a single solution here, i.e. the same molality and van't Hoff factor, you can divide the two equations to get

$\left(\Delta {T}_{\text{b")/(DeltaT_"f}}\right) = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{i}}} \cdot {K}_{b} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{b}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{i}}} \cdot {K}_{f} \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{b}}}}$

This means that you have

$\Delta {T}_{\text{f" = K_"f"/K_"b" * DeltaT_"b}}$

Now, water has a normal boiling point of ${100}^{\circ} \text{C}$. In your case, the boiling point of the solution is said to be equal to ${100.52}^{\circ} \text{C}$, which means that the boiling-point elevation is equal to

$\Delta {T}_{\text{b" = 100.52^@"C" - 100^@"C" = 0.52^@"C}}$

Plug in your values to find the freezing-point depression

$\Delta {T}_{\text{f" = (1.86 color(red)(cancel(color(black)(""^@"C kg mol"^(-1)))))/(0.512color(red)(cancel(color(black)(""^@"C kg mol"^(-1))))) * 0.52^@"C}}$

$\Delta {T}_{\text{f" = 1.89^@"C}}$

Water has a normal freezing point of ${0}^{\circ} \text{C}$. The freezing-point depression is equal to ${1.89}^{\circ} \text{C}$, which means that the freezing point of the solution will be ${1.89}^{\circ} \text{C}$ lower than the normal freezing point of pure water.

You thus have

T_"f sol" = 0^@"C" - 1.89^@"C" = color(darkgreen)(ul(color(black)(-1.89^@"C")))#

SIDE NOTE The difference between this value and the value given to you in (a) is most likely caused by different values used for ${K}_{\text{b}}$ and/or ${K}_{\text{f}}$.