# Question e695c

Nov 11, 2015

${T}_{\text{f" = -1.28^@"C}}$

#### Explanation:

I'm not entirely sure why the problem provides the density of the water, since that is not needed to find the boiling point and the freezing point of the solution.

I can only assume that your solution contains $\text{57.8 g}$ glucose dissolved in $\text{465 mL}$ of water. Since the density of water is to be taken as $\text{1.00 g/mL}$, this will not make any difference for the final result.

So, the first thing would be to determine the mass of water

465color(red)(cancel(color(black)("mL"))) * "1.00 g"/(1color(red)(cancel(color(black)("mL")))) = "465 g"

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

$\Delta {T}_{f}$ - the freezing-point depression;
$i$ - the van't Hoff factor, equal to $1$ for non-electrolytes;
${K}_{f}$ - the cryoscopic constant of the solvent;
$b$ - the molality of the solution.

In your case, glucose is a non-electrolyte, which means that it does not dissociate to form ions in solution. As a result, you have $i = 1$.

In order to calculate the molality of the solution, you need to know two things

• the number of moles of solute, in your case sucrose
• the mass of the solvent, in your case water, expressed in kilograms

Use glucose's molar mass to determine how many moles you have in that sample

57.8color(red)(cancel(color(black)("g"))) * " 1 mole glucose"/(180.156color(red)(cancel(color(black)("mL")))) = "0.3208 moles glucose"

The molality of the solution will thus be - do not forget to convert the mass of the water from grams to kilograms

$\textcolor{b l u e}{b = {n}_{\text{solute"/m_"solvent}}}$

$b = \text{0.3028 moles"/(465 * 10^(-3)"kg") = "0.6899 moles/kg" = "0.6899 molal}$

You can find the values of water's cryoscopic and ebullioscopic constants here

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

So, plug in your values and determine the freezing-point depression

$\Delta {T}_{\text{f" = 1 * 1.86^@"C" color(red)(cancel(color(black)("kg")))/color(red)(cancel(color(black)("mol"))) * 0.6899color(red)(cancel(color(black)("moles")))/color(red)(cancel(color(black)("kg"))) = 1.283^@"C}}$

The freezing-point depression is defined as

$\Delta {T}_{\text{f" = T_"f"^@ - T_"f"" }}$, where

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

This means that the freezing point of the solution will be

${T}_{\text{f" = T_"f"^@ - DeltaT_"f}}$

T_"f" = 0^@"C" - 1.283^@"C" = color(green)(-1.28^@"C")#

I'll leave the calculation of the solution's boiling point to you as practice. The equation for boiling-point elevation is

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

$\Delta {T}_{b}$ - the boiling-point elevation;
$i$ - the van't Hoff factor
${K}_{b}$ - the ebullioscopic constant of the solvent;
$b$ - the molality of the solution.

Plug in your values and solve for $\Delta {T}_{b}$, then use

$\Delta {T}_{\text{b" = T_"b" - T_"b}}^{\circ}$