# 6.4 Colligative properties  (Page 9/30)

 Page 9 / 30

## The freezing point of a solution of an electrolyte

The concentration of ions in seawater is approximately the same as that in a solution containing 4.2 g of NaCl dissolved in 125 g of water. Assume that each of the ions in the NaCl solution has the same effect on the freezing point of water as a nonelectrolyte molecule, and determine the freezing temperature the solution (which is approximately equal to the freezing temperature of seawater).

## Solution

We can solve this problem using the following series of steps.

1. Convert from grams to moles of NaCl using the molar mass of NaCl in the unit conversion factor.
Result: 0.072 mol NaCl
2. Determine the number of moles of ions present in the solution using the number of moles of ions in 1 mole of NaCl as the conversion factor (2 mol ions/1 mol NaCl).
Result: 0.14 mol ions
3. Determine the molality of the ions in the solution from the number of moles of ions and the mass of solvent, in kilograms.
Result: 1.1 m
4. Use the direct proportionality between the change in freezing point and molal concentration to determine how much the freezing point changes.
Result: 2.0 °C
5. Determine the new freezing point from the freezing point of the pure solvent and the change.
Result: −2.0 °C
Check each result as a self-assessment.

## Check your learning

Assume that each of the ions in calcium chloride, CaCl 2 , has the same effect on the freezing point of water as a nonelectrolyte molecule. Calculate the freezing point of a solution of 0.724 g of CaCl 2 in 175 g of water.

−0.208 °C

Assuming complete dissociation, a 1.0 m aqueous solution of NaCl contains 1.0 mole of ions (1.0 mol Na + and 1.0 mol Cl ) per each kilogram of water, and its freezing point depression is expected to be

$\text{Δ}{T}_{\text{f}}=2.0\phantom{\rule{0.2em}{0ex}}\text{mol ions/kg water}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}1.86\phantom{\rule{0.2em}{0ex}}\text{°}\text{C kg water/mol ion}=3.7\phantom{\rule{0.2em}{0ex}}\text{°}\text{C}.$

When this solution is actually prepared and its freezing point depression measured, however, a value of 3.4 °C is obtained. Similar discrepancies are observed for other ionic compounds, and the differences between the measured and expected colligative property values typically become more significant as solute concentrations increase. These observations suggest that the ions of sodium chloride (and other strong electrolytes) are not completely dissociated in solution.

To account for this and avoid the errors accompanying the assumption of total dissociation, an experimentally measured parameter named in honor of Nobel Prize-winning German chemist Jacobus Henricus van’t Hoff is used. The van’t Hoff factor ( i )    is defined as the ratio of solute particles in solution to the number of formula units dissolved:

$i\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{moles of particles in solution}}{\text{moles of formula units dissolved}}$

Values for measured van’t Hoff factors for several solutes, along with predicted values assuming complete dissociation, are shown in [link] .

Expected and Observed van’t Hoff Factors for Several 0.050 m Aqueous Electrolyte Solutions
Electrolyte Particles in Solution i (Predicted) i (Measured)
HCl H + , Cl 2 1.9
NaCl Na + , Cl 2 1.9
MgSO 4 Mg 2+ , ${\text{SO}}_{4}{}^{2-}$ 2 1.3
MgCl 2 Mg 2+ , 2Cl 3 2.7
FeCl 3 Fe 3+ , 3Cl 4 3.4
glucose A nonelectrolyte shown for comparison. C 12 H 22 O 11 1 1.0

#### Questions & Answers

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A teaspoon of the carbohydrate sucrose contains 16 calories, what is the mass of one teaspoo of sucrose if the average number of calories for carbohydrate is 4.1 calories/g?
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