# 6.4 Colligative properties  (Page 8/30)

 Page 8 / 30

## Determination of molar masses

Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements.

## Determination of a molar mass from a freezing point depression

A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. What is the molar mass of this compound?

## Solution

We can solve this problem using the following steps.

1. Determine the change in freezing point from the observed freezing point and the freezing point of pure benzene ( [link] ).
$\text{Δ}{T}_{\text{f}}=5.5\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}-2.32\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}=3.2\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}$
2. Determine the molal concentration from K f , the freezing point depression constant for benzene ( [link] ), and Δ T f .
$\begin{array}{l}\hfill \text{Δ}{T}_{\text{f}}={K}_{\text{f}}m\hfill \\ \\ m\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{Δ}{T}_{\text{f}}}{{K}_{\text{f}}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{3.2\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}}{5.12\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}\phantom{\rule{0.2em}{0ex}}{m}^{-1}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}0.63\phantom{\rule{0.2em}{0ex}}m\end{array}$
3. Determine the number of moles of compound in the solution from the molal concentration and the mass of solvent used to make the solution.
$\text{Moles of solute}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{0.62\phantom{\rule{0.2em}{0ex}}\text{mol solute}}{1.00\phantom{\rule{0.2em}{0ex}}\overline{)\text{kg solvent}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0.0550\phantom{\rule{0.2em}{0ex}}\overline{)\text{kg solvent}}=0.035\phantom{\rule{0.2em}{0ex}}\text{mol}$
4. Determine the molar mass from the mass of the solute and the number of moles in that mass.
$\text{Molar mass}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{4.00\phantom{\rule{0.2em}{0ex}}\text{g}}{0.034\phantom{\rule{0.2em}{0ex}}\text{mol}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\text{g/mol}$

## Check your learning

A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. What is the molar mass of this compound?

1.8 $×$ 10 2 g/mol

## Determination of a molar mass from osmotic pressure

A 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at 22 °C. What is the molar mass of hemoglobin?

## Solution

Here is one set of steps that can be used to solve the problem:

1. Convert the osmotic pressure to atmospheres, then determine the molar concentration from the osmotic pressure.
$\begin{array}{}\\ \Pi \phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{5.9\phantom{\rule{0.2em}{0ex}}\text{torr}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}\text{atm}}{760\phantom{\rule{0.2em}{0ex}}\text{torr}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}7.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\text{atm}\\ \Pi =\mathit{\text{MRT}}\\ \\ M\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\Pi }{RT}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{7.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-3}\phantom{\rule{0.2em}{0ex}}\text{atm}}{\left(0.08206\phantom{\rule{0.2em}{0ex}}\text{L atm/mol K}\right)\left(295\phantom{\rule{0.2em}{0ex}}\text{K}\right)}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}3.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\text{M}\end{array}$
2. Determine the number of moles of hemoglobin in the solution from the concentration and the volume of the solution.
$\text{moles of hemoglobin}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{3.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\text{mol}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{L solution}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}0.500\phantom{\rule{0.2em}{0ex}}\overline{)\text{L solution}}=1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\text{mol}$
3. Determine the molar mass from the mass of hemoglobin and the number of moles in that mass.
$\text{molar mass}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{10.0\phantom{\rule{0.2em}{0ex}}\text{g}}{1.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\text{mol}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}6.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\text{g/mol}$

## Check your learning

What is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C?

2.7 $×$ 10 4 g/mol

## Colligative properties of electrolytes

As noted previously in this module, the colligative properties of a solution depend only on the number, not on the kind, of solute species dissolved. For example, 1 mole of any nonelectrolyte dissolved in 1 kilogram of solvent produces the same lowering of the freezing point as does 1 mole of any other nonelectrolyte. However, 1 mole of sodium chloride (an electrolyte) forms 2 moles of ions when dissolved in solution. Each individual ion produces the same effect on the freezing point as a single molecule does.

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