# Question 0c4c5

Jun 13, 2016

${T}_{\text{f sol" = -1.16^@"C}}$

#### Explanation:

When dealing with the freezing point of a solution, your first goal is to determine its molality by

• determining the number of moles of solute present in the solution
• determining the mass of the solvent in kilograms

Once you know the molality of the solution, you can find the freezing-point depression by using the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \Delta {T}_{f} = i \cdot {K}_{f} \cdot b \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$\Delta {T}_{f}$ - the freezing-point depression;
$i$ - the van't Hoff factor
${K}_{f}$ - the cryoscopic constant of the solvent;
$b$ - the molality of the solution.

In your case, the cryoscopic constant of water is said to be

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

Glucose is a non-electrolyte, i.e. it remains undissociated in aqueous solution, which means that its van't Hoff factor will be $i = 1$.

So, you know that your solution contains $\text{53.4 g}$ of glucose. To determine how many moles of glucose it contains, use the compound's molar mass

53.4 color(red)(cancel(color(black)("g"))) * "1 mole glucose"/(180.16color(red)(cancel(color(black)("g")))) = "0.2964 moles glucose"

The solution contains $\text{475 g}$ of water, which is the solvent. Convert this to kilograms

475color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.475 kg"

The molaity of the solution, which is defined as

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{molality" = "moles of solute"/"kilogram of solvent} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

will thus be

$b = {\text{0.2964 moles"/"0.475 kg" = "0.624 mol kg}}^{- 1}$

Now plug in your values and solve for the freezing-point depression

DeltaT_f = 1 * 1.86^@"C" color(red)(cancel(color(black)("kg")))color(red)(cancel(color(black)("mol"^(-1)))) * 0.624 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))

$\Delta {T}_{f} = {1.161}^{\circ} \text{C}$

The freezing-point depression is defined as

color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = T_f^@ - T_"f sol"color(white)(a/a)|)))" ", where

${T}_{f}^{\circ}$ - the freezing point of the pure solvent
${T}_{\text{f sol}}$ - the freezing point of the solution

This means that the freezing point of the solution will be

${T}_{\text{f sol}} = {T}_{f}^{\circ} - \Delta {T}_{f}$

T_"f sol" = 0^@"C" - 1.161^@"C" = color(green)(|bar(ul(color(white)(a/a)color(black)(-1.16^@"C")color(white)(a/a)|)))#

The answer is rounded to three sig figs.