Chemistry · General chemistry II · Worked example

Freezing and boiling points of antifreeze

$25.0 g$ of ethylene glycol, $C_{2} H_{6} O_{2}$ , is dissolved in $100.0 g$ of water. Find the freezing and boiling points of the solution. For water, $K_{f} = 1.86 ° C \cdot kg / mol$ and $K_{b} = 0.512 ° C \cdot kg / mol$ .

Δ T_{f} = i K_{f} m, Δ T_{b} = i K_{b} m

Moles of solute

The freezing point counts particles, so start with moles. Ethylene glycol’s molar mass is $2 (12.01)$ $+ 6 (1.008)$ $+ 2 (16.00)$ $= 62.07 g / mol$ . It is a molecular compound that does not ionize, so $i = 1$ .

n = \frac{25.0 g}{62.07 g/mol} = 0.4028 mol

Molality

Divide by kilograms of solvent, not by the solution’s mass or volume. Keep the extra digit, $4.028$ , for the next two steps and round only the reported temperatures.

m = \frac{0.4028 mol}{0.1000 kg} = 4.028 mol/kg

Freezing-point depression

Multiply by $K_{f}$ for water. With $i = 1$ , $Δ T_{f} = K_{f} m$ . The freezing point drops below $0.00 ° C$ by $Δ T_{f}$ .

Δ T_{f} = (1) (1.86) (4.028) = 7.49 \circ C \Rightarrow T_{f} = - 7.49 \circ C

\begin{array}{c} Δ T_{f} = (1) (1.86) (4.028) \\ = 7.49 \circ C \\ T_{f} = - 7.49 \circ C \end{array}

Boiling-point elevation

The same molality with the smaller $K_{b}$ . The boiling point rises above $100.00 ° C$ by $Δ T_{b}$ .

Δ T_{b} = (1) (0.512) (4.028) = 2.06 \circ C \Rightarrow T_{b} = 102.06 \circ C

\begin{array}{c} Δ T_{b} = (1) (0.512) (4.028) \\ = 2.06 \circ C \\ T_{b} = 102.06 \circ C \end{array}

Result

The solution freezes at $- 7.49 ° C$ and boils at $102.06 ° C$ (at $1 atm$ , treating it as ideal).

Your turn

1. One like the example

$15.0 g$ of urea, $CO {({NH}_{2})}_{2}$ $(60.06 g / mol)$ , is dissolved in $250.0 g$ of water. At what temperature does the solution freeze?

Show the answer and explanation

$- 1.86 ° C$ .

Urea is molecular, so $i = 1$ . $15.0 g \div 60.06 g / mol$ $= 0.2498 mol$ , and $0.2498 mol \div 0.2500 kg$ $= 0.9992 mol / kg$ . $Δ T_{f} = (1.86) (0.9992)$ $= 1.86 ° C$ , so the solution freezes at $0.00 - 1.86 = - 1.86 ° C$ .

\begin{array}{l} 15.0 g CO {({NH}_{2})}_{2} \times \frac{1 mol CO {({NH}_{2})}_{2}}{60.06 g CO {({NH}_{2})}_{2}} = 0.2498 mol CO {({NH}_{2})}_{2} \\ m = \frac{0.2498 mol}{0.2500 kg} = 0.9992 mol/kg \\ (1.86^{\circ} C kg/mol) (0.9992 mol/kg) = 1.86^{\circ} C \\ 0.00^{\circ} C - 1.86^{\circ} C = - 1.86^{\circ} C \end{array}

\begin{array}{c} 15.0 g CO {({NH}_{2})}_{2} \\ \times \frac{1 mol}{60.06 g} \\ = 0.2498 mol CO {({NH}_{2})}_{2} \\ m = \frac{0.2498 mol}{0.2500 kg} = 0.9992 mol/kg \\ (1.86^{\circ} C kg/mol) \\ \times (0.9992 mol/kg) \\ = 1.86^{\circ} C \\ 0.00^{\circ} C - 1.86^{\circ} C = - 1.86^{\circ} C \end{array}

2. A variation

$0.0500 mol$ of ${CaCl}_{2}$ is dissolved in $500.0 g$ of water. Assuming ideal dissociation, at what temperature does it freeze?

Show the answer and explanation

About $- 0.558 ° C$ .

The molality is $0.0500 mol \div 0.5000 kg$ $= 0.100 mol / kg$ . ${CaCl}_{2}$ gives one ${Ca}^{2 +}$ and two ${Cl}^{-}$ , so $i = 3$ and $Δ T_{f} = 3 \times 1.86 \times 0.100$ $= 0.558 ° C$ . The solution freezes at $- 0.558 ° C$ . A measured value is a little closer to zero, because some ions pair.

\begin{array}{l} m = \frac{0.0500 mol}{0.5000 kg} = 0.100 mol/kg \\ Δ T_{f} = i K_{f} m \\ i = 3 \\ K_{f} = 1.86^{\circ} C kg/mol \\ Δ T_{f} = (3) (1.86^{\circ} C kg/mol) (0.100 mol/kg) = 0.558^{\circ} C \\ 0.00^{\circ} C - 0.558^{\circ} C = - 0.558^{\circ} C \end{array}

\begin{array}{c} m = \frac{0.0500 mol}{0.5000 kg} \\ = 0.100 mol/kg \\ Δ T_{f} = i K_{f} m \\ i = 3 \\ K_{f} = 1.86^{\circ} C kg/mol \\ Δ T_{f} = (3) (1.86^{\circ} C kg/mol) \\ \times (0.100 mol/kg) \\ = 0.558^{\circ} C \\ 0.00^{\circ} C - 0.558^{\circ} C = - 0.558^{\circ} C \end{array}

3. Somewhere new

$2.50 g$ of an unknown molecular compound dissolves in $50.0 g$ of water, and the solution freezes at $- 1.55 ° C$ . What is the compound’s molar mass?

Show the answer and explanation

$60.0 g / mol$ .

Run the equation backward. With $i = 1$ , $m = Δ T_{f} \div K_{f}$ $= 1.55 \div 1.86$ $= 0.833 mol / kg$ . The $0.0500 kg$ of water holds $0.833 \times 0.0500$ $= 0.0417 mol$ of solute. $2.50 g \div 0.0417 mol$ $= 60.0 g / mol$ , the molar mass of urea.

\begin{array}{l} \frac{1.55^{\circ} C}{1.86^{\circ} C kg/mol} = 0.833 mol/kg \\ (0.833 mol/kg) (0.0500 kg) = 0.0417 mol \\ \frac{2.50 g}{0.0417 mol} = 60.0 g/mol \end{array}

\begin{array}{c} \frac{1.55^{\circ} C}{1.86^{\circ} C kg/mol} \\ = 0.833 mol/kg \\ (0.833 mol/kg) (0.0500 kg) \\ = 0.0417 mol \\ \frac{2.50 g}{0.0417 mol} = 60.0 g/mol \end{array}

Keep exploring

Open the Colligative studio with this solution, switch to boiling-point elevation, then try $i = 2$ to see what an ionic solute would do.

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Moles of solute

Molality

Freezing-point depression

Boiling-point elevation

Result

Your turn

1. One like the example

−1.86 °⁢C−1.86 ∘C.

2. A variation

About −0.558 °⁢C−0.558 ∘C.

3. Somewhere new

60.0 g/⁢mol60.0 g/mol.

Keep exploring

Study it, then try it

$- 1.86 ° C$ .

About $- 0.558 ° C$ .

$60.0 g / mol$ .