# Colligative Properties

**Colligative properties**are properties that depend only on the number (concentration) of solute particles present in a solution. These properties do not depend on the identity of the solute. As an example let's talk about two aqueous solutions; one containing glucose and the other containing urea. As far as the colligative properties are concerned, the two solutions are not different as long as the number of glucose and urea particles are the same. In other words, if the concentration of the given samples of urea and glucose are the same, the colligative properties of both solutions will be the same.

The colligative properties that you have to be familiar with for the MCAT are:

- vapor-pressure lowering
- boiling-point elevation
- freezing-point depression
- osmotic pressure generating

A simple **solution** is a mixture of a solute and a solvent.

SOLUTE + SOLVENT = SOLUTION

# Vapor-pressure Lowering

P_{A} = X_{A}x P_{A}

Here, P_{A}is the vapor pressure of the solution (impure solvent),

X_{A} is the mole fraction of the solvent, and

P_{A} is the vapor pressure of the pure solvent.

Calculate the mole fractions of sodium chloride and water in a solution containing 11.7 g of sodium chloride and 9 g of water.

In this solution, we have 11.7 grams of NaCl and 9 grams of H_{2}O. First, you have to find the number of moles of NaCl and H_{2}O.

The total number of moles in the solution = 0.2 + 0.5 = 0.7 mol

_{}

# Boiling-point Elevation

**boiling point**of a liquid is defined as the temperature at which its vapor pressure equals the prevailing pressure. The prevailing pressure is normally the atmospheric pressure, provided that the container in which the liquid is present is kept open while boiling. When a nonvolatile solute is added to a pure solvent, the boiling point increases. The increase is proportional to the number of moles of the solute added to the solvent. The relationship is mathematically represented as shown below:

Î”T_{b} = i K_{b} C_{m}

Here, Î”T_{b}is the boiling-point elevation,

i is the ionization factor,

K_{b}is the boiling-point elevation constant, and

C_{m}is the molal concentration of the solution.

Calculate the boiling point of 0.2 *m* aqueous solution of glucose. (*K _{b}*of water is 0.512

^{o}C/

*m.*)

The formula for finding the boiling-point elevation is:

Î”*T _{b} = i K_{b} C_{m}*

In this problem, we have all the necessary values to find the boiling-point elevation. Here, the ionization factor is 1, since glucose doesn't ionize. Let's substitute the values into the formula.

Î”*T _{b} = K_{b} C_{m}*= 0.512

^{o}C/

*m*x 0.2

*m*Ã¢â€°Ë† 0.1

^{o}C

The boiling point of 0.2 *m* solution of glucose is 100 + 0.1 = 100.1^{o}C

# Freezing-point Depression

Î”T_{f} = i K_{f} C_{m}

Here, Î”T_{f}is the freezing-point depression,

i is the ionization factor,

K_{f}is the freezing-point depression constant, and

C_{m} is the molal concentration of the solution.

Calculate the freezing point of 2 m aqueous solution of glucose. (*K _{f}* of water is 1.86

^{o}C/m)

The formula for finding the freezing-point depression is:

Î”*T _{f }*=

*i K*

_{f}C_{m}

In this problem, we have all the necessary values to find the freezing-point depression. The ionization factor is 1.

Î”*T _{f}* =

*K*= 1.86

_{f}C_{m}^{o}C/

*m*x 2

*m*â‰ˆ 3.7

^{o}C

So the freezing point of 2*m* solution of glucose is 0.0 â€“ 3.7 = â€“ 3.7^{o}C

# Osmotic Pressure

**osmosis**is defined as the flow of solvent through a semipermeable membrane resulting in the equilibrium of concentrations on both sides of the semipermeable membrane.

**Osmosis experiment**

_{2}O).

**osmotic pressure**is the pressure required or applied to the solution to stop the flow of the solvent, or in other words, to stop the process of osmosis. The osmotic pressure and concentration are related by the following equation:

Osmotic pressure, Ï€ = M R T

Here, M is the molar concentration of the solute,

R is the gas constant, and

T is the absolute temperature.