# Question 918cb

May 31, 2015

Start by writing the three equations that you're going to use to solve for the boiling point, freezing point, and vapor pressure of the solution.

• Freezing point depression

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

${K}_{f}$ - the cryoscopic constant - depends on the solvent;
$b$ - the molality of the solution;
$i$ - the van't Hoff factor - the number of ions per individual molecule of solute.
$\Delta {T}_{\text{f}}$ - the freezing point depression - defined as
${T}_{\text{f"^@ - T_"f sol}}$.

In your case, the van't Hoff factor will be equal to 1 because you're dealing with a non-electrolyte.

Plug in your values and solve for $\Delta {T}_{\text{f}}$

$\Delta {T}_{\text{f" = 1 * 1.86^@"C"/cancel("molal") * 0.750cancel("molal") = 1.40^@"C}}$

The freezing point of the solution will be

T_"f sol" = T_"f"^@ - DeltaT_"f" = 0^@"C" - 1.40^@"C" = color(green)("-1.40"^@"C")

• Boiling point elevation

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

$i$ - the van't Hoff factor;
${K}_{b}$ - the ebullioscopic constant;
$b$ - the molality of the solution.
$\Delta {T}_{b}$ - the poiling point elevation - defined as ${T}_{\text{b sol" - T_"b}}^{\circ}$

Once again, plug in your values and solve for $\Delta {T}_{\text{b}}$

$\Delta {T}_{\text{b" = 1 * 0.51^@"C"/cancel("molal") * 0.750cancel("molal") = 0.383^@"C}}$

The boiling point of the solution will be

T_"b sol" = T_"b"^@ + DeltaT_"b" = 100^@"C" + 0.383^@"C" = color(green)(100.4^@"C")

• Vapor pressure lowering

${P}_{\text{solution" = chi_"[solvent](http://socratic.org/chemistry/solutions-and-their-behavior/solvent)" * P_"solvent}}^{\circ}$, where

${P}_{\text{solution}}$ - the vapor pressure of the solution;
${\chi}_{\text{solvent}}$ - the mole fraction of the solvent;
${P}_{\text{solvent}}^{\circ}$ - the vapor pressure of the pure solvent.

Since you didn't provide the mass of water used to make your solution, I'll assume that you have a 1-kg sample.

The number of moles of water present in 1000 g will be

1000cancel("g") * "1 mole water"/(18.02cancel("g")) = "55.5 moles"

The mole fraction of water, which is defined as the number of moles of water divided by the total number of moles in the solution, will be

chi_"water" = (55.5cancel("moles"))/((55.5 + 0.750)cancel("moles")) = 0.9867

SIDE NOTE Since molality is defined as moles of solute per kg of solvent, the 1 kg of water implies that you have 0.750 moles of sucrose.

Therefore, the vapor pressure of the solution will be

P_"solution" = 0.9867 * "23.76 torr" = color(green)("23.44 torr")#