Calculating Molar Mass Using Colligative Properties — Examples

We've been learning about colligative properties. They are extensive properties. So, they only depend on the amount of solute. Each of the colligative properties relates to the moles of solute. That means if we start with a solute's mass, we can calculate its molar mass. We'll look at an example for each colligative property.

Steps for Calculating Molar Mass

The units for molar mass are grams per mole. So, for each colligative property, our strategy is to find the moles of solute. The problem will always tell you the solute's mass.

Freezing Point Depression and Boiling Point Elevation

Solve for molality. You will need the temperature change and constant.
Convert molality to moles of solute. The problem will tell you the solvent's volume.
Calculate the molar mass.

Osmotic Pressure

Solve the osmotic pressure formula for moles, $n=\frac{\Pi\times V}{i\times R\times T}$ .
Divide the grams of solute by moles.

Vapor Pressure Lowering

First we need to take advantage of Raoult's law. It connects the mole fraction and the partial pressures. Typically, the problem tells you the partial pressures. So, we can solve for the mole fraction. $P_{A}=X \times P^{o}_{A}\rightarrow X=\frac{P_{A}}{P^{o}_{A}}=\frac{n_{solvent}}{n_{solvent}+n_{solute}}$
$n_{solvent}=X(n_{solvent}+n_{solute})\rightarrow \frac{1-X}{X}n_{solvent}=n_{solute}$
$MM=\frac{g_{solvent}}{n_{solvent}}$

Freezing Point Depression Example

Let's say that we added 46.0 grams of an unknown solute to 1000.0 mL of water. If the freezing point lessened by $2.1^{o}C$ , what is the molar mass of the solute?

$K_{f}\times m=\Delta T\rightarrow m=\frac{\Delta T}{K_{f}}=\frac{2.1^{o}C}{1.86\frac{^{o}C}{m}}=1.13m$

Remember, water's $K_{f}$ is $1.86\frac{^{o}C}{m}$ .

$m=\frac{mol_{solute}}{kg_solvent}=1.3\frac{mol_{solute}}{kg_solvent}\times 1.0kg_{water}=1.3 mols_{solvent}$

Remember water's density is $\frac{1.0g}{ml}$ which equals $\frac{1.0kg}{L}$ .

$MM=\frac{g}{mol}=\frac{46.0g}{1.3mols}=35.4\frac{g}{mol}$

Boiling Point Elevation Example

The boiling point elevation is similar to freezing point depression. Again, let's say that we added 46.0 grams of the same unknown solute to 1000.0 mL of water. If the boiling point elevated by $0.58^{o}C$ , what is the molar mass of the solute?

$K_{B}\times m=\Delta T\rightarrow m=\frac{\Delta T}{K_{B}}=\frac{0.58^{o}C}{0.51\frac{^{o}C}{m}}=1.13m$

Remember, water's $K_{B}$ is $0.51\frac{^{o}C}{m}$ .

$m=\frac{mol_{solute}}{kg_solvent}=1.3\frac{mol_{solute}}{kg_solvent}\times 1.0kg_{water}=1.3 mols_{solvent}$

$MM=\frac{g}{mol}=\frac{46.0g}{1.3mols}=35.4\frac{g}{mol}$

Because it's the same solute, we get the same answer. We see that freezing point depression is much larger than boiling point elevation.

Osmotic Pressure Example

When we studied osmotic pressure, we looked at blood. At $37^{o}C$ , its osmotic pressure was $7.6atm$ . The average person's blood volume is $5.0L$ . One liter of blood contains 8.7 grams of solute. For the solute, the Van't Hoff factor is 2. Calculate its molar mass.

$n=\frac{\Pi\times V}{i\times R\times T}=\frac{7.6atm\times 5.0L}{2\times 0.0821\frac{L\times atm}{mols\times K}\times 310.15K}=0.75mols$

$8.7\frac{g}{L}\times5.0L=43.5g_{solute}$

$\frac{43.5g}{0.75mols}=58.0\frac{g}{mol}$

In this case, we were looking at sodium chloride ( $NaCl$ ).

Vapor Pressure Lowering Example

Let's say we added $237.0g$ of an unknown solute to 1000.0 milliliters of water. The partial pressure is $2.80kPa$ . Calculate the molar mass of the solute.

$X=\frac{P_{A}}{P^{o}_{A}}=\frac{2.80kPa}{3.17kPa}=0.88$

The vapor pressure of pure water is $3.17kPa$ .

$n_{solute}=\frac{1.00-X}{X}n_{solvent} = \frac{1.00-0.88}{.88}\times 55.5\frac{mols}{L}\times 1.0L=13.88 mols$

$MM=\frac{237.0g}{13.88mols}=17.07\frac{g}{mol}$