# Question cc7f0

Oct 7, 2015

${P}_{\text{sol" = "0.031 atm}}$

#### Explanation:

You can calculate the vapor pressure of the urea solution by using the vapor pressure of the pure solvent, which in your case would be water, and the mole fraction of the solvent - this is known as Raoult's Law.

${P}_{\text{sol" = chi_"water" * P_"water}}^{0}$

The mole fraction of water is the ratio between the number of moles of water and the total number of moles present in solution.

To figure out how many moles of water you have in $\text{100 g}$ of water, use its molar mass

100color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.02color(red)(cancel(color(black)("g")))) = "5.55 moles H"_2"O"

SInce you know that you have $\text{0.10 moles}$ of urea, the total number of moles will be

${n}_{\text{total" = n_"urea" + n_"water}}$

${n}_{\text{total" = 0.10 + 5.55 = "5.65 moles}}$

The vapor pressure of pure water at ${26}^{\circ} \text{C}$ is equal to $\text{0.0312 atm}$

http://www.endmemo.com/chem/vaporpressurewater.php

The mole fraction of water will be

${\chi}_{\text{water" = n_"water"/n_"total}}$

chi_"water" = (5.55color(red)(cancel(color(black)("g"))))/(5.65color(red)(cancel(color(black)("g")))) = 0.9823

The vapor pressure of the solution will thus be

P_"sol" = 0.9823 * "0.0312 atm" = "0.0306 atm" = color(green)("0.031 atm")

For the boiling point and freezing point of the solution, you need to find its molality.

Molality is defined as moles of solute, in your case urea, per mass of solvent - expressed in kilograms!

$b = {n}_{\text{urea"/m_"water}}$

$b = \text{0.10 moles"/(100 * 10^(-3)"kg") = "1.0 molal}$

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

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

For water, ${K}_{f} = 1.853 \text{^@"C kg/mol}$.

Urea is a non-electrolyte, so the van't Hoff factor will be equal to $1$.

The freezing-point depression* for this solution will thus be

DeltaT_f = 1 * 1.853""^@"C"color(red)(cancel(color(black)("kg")))/color(red)(cancel(color(black)("mol"))) * 1.0color(red)(cancel(color(black)("mol")))/color(red)(cancel(color(black)("kg")))

$\Delta {T}_{f} = {1.853}^{\circ} \text{C}$

The frezing point of the solution will be

$\Delta {T}_{f} = {T}_{f}^{0} - {T}_{\text{f sol" implies T_"f sol}} = {T}_{f}^{0} - \Delta {T}_{f}$

T_"f sol" = 0^@"C" - 1.853^@"C" = color(green)(-1.9^@"C")

The equation for boiling-point elevation is

$\Delta {T}_{b} = 8 \cdot {K}_{b} \cdot b \text{ }$, where

$\Delta {T}_{b}$ - the boiling-point elevation.
${K}_{b}$ - the ebullioscopic constant of the solvent.

For water, ${K}_{b} = 0.512 \text{^@"C kg/mol}$.

Once again, plug in your values to get

DeltaT_b = 1 * 0.512""^@"C"color(red)(cancel(color(black)("kg")))/color(red)(cancel(color(black)("mol"))) * 1.0color(red)(cancel(color(black)("mol")))/color(red)(cancel(color(black)("kg")))#

$\Delta {T}_{b} = {0.512}^{\circ} \text{C}$

The boiling point of the solution will be

$\Delta {T}_{b} = {T}_{\text{b sol" - T_b^0 implies T_"b sol}} = \Delta {T}_{b} + {T}_{b}^{0}$

${T}_{\text{b sol" = 0.512^@"C" + 100^@"C" = 100.5^@"C}}$

I'll leave this answer rounded to four sig figs to emphasize the small increase in the boiling point of the solution when compared with that of pure water.

When it comes to colligative properties problems, you need to know the formulas for boiling-point elevation, freezing-point depression, and the vapor pressure of the solution.

When it comes to molality, keep in mind that everything revolves around moles of solute and kilograms of solvent.