# Question #cc7f0

##### 1 Answer

#### Explanation:

Let's start with the vapor pressure.

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_"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

#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

#n_"total" = n_"urea" + n_"water"#

#n_"total" = 0.10 + 5.55 = "5.65 moles"#

The vapor pressure of pure water at

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

The mole fraction of water will be

#chi_"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**!

In your case, you have

#b = n_"urea"/m_"water"#

#b = "0.10 moles"/(100 * 10^(-3)"kg") = "1.0 molal"#

The equation for *freezing-point depression* is

#DeltaT_f = i * K_f * b" "# , where

*freezing-point depression*;

*van't Hoff factor*, equal to

*cryoscopic constant* of the solvent.

For water,

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

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")))#

#DeltaT_f = 1.853^@"C"#

The frezing point of the solution will be

#DeltaT_f = T_f^0 - T_"f sol" implies T_"f sol" = T_f^0 - DeltaT_f#

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

The equation for *boiling-point elevation* is

#DeltaT_b = 8 * K_b * b" "# , where

*boiling-point elevation*.

*ebullioscopic constant* of the solvent.

For water,

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")))#

#DeltaT_b = 0.512^@"C"#

The boiling point of the solution will be

#DeltaT_b = T_"b sol" - T_b^0 implies T_"b sol" = DeltaT_b + T_b^0#

#T_"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.