A solution is made by dissolving 100. g of ethylene glycol (C2H6O2) in 200. g of water. What is the freezing point of this solution?

Question

A solution is made by dissolving 100. g of ethylene glycol [latex](C_2H_6O_2)[/latex] in 200. g of water. What is the freezing point of this solution?

Accepted Answer

To calculate the freezing point of the solution, we first need to calculate [latex]\Delta t_f [/latex],the change in freezing point. Use the equation

[latex]\Delta t_f= mK_f = \frac{mol solute}{kg solvent} imes K_f[/latex]

[latex]K_f (for water): \frac{1.86^\circ C kg solvent}{mol solute}[/latex] (from Table 14.5)

mol solute : [latex](100. \cancel{g C_2H_6O_2} ) (\frac{1 mol C_2H_6O_2}{62.07 \cancel{g C_2H_6O_2} } )[/latex]

= 1.61 mol [latex]C_2H_6O_2 [/latex]

kg solvent: [latex](200. \cancel{g} H_2O) (\frac{1 kg}{1000 \cancel{g}} ) =0.200 kg H_2O[/latex]

[latex]\Delta t_f= (\frac{1.61 \cancel{mol C_2H_6O_2}}{0.200 \cancel{kg H_2O}} ) (\frac{1.86^\circ C \cancel{kg H_2O}}{1 \cancel{mol C_2H_6O_2}} ) = 15.0^\circ C[/latex]

The freezing point depression, 15.0°C, must be subtracted from 0°C, the freezing point of the pure solvent (water):

freezing point of solution = freezing point of solvent - [latex]\Delta t_f [/latex]

=0.0°C - 15.0°C = -15.0°C

Therefore, the freezing point of the solution is -15.0°C .This solution will protect an automobile radiator down to -15.0°C (5°F).

TABLE 14.5 Freezing Point Depression and Boiling Point Elevation Constants of Selected Solvents
Solvent	Freezing point of pure solvent (°C)	Freezing point depression constant, [latex]K_f[/latex] [latex]\frac{^\circ ext{C kg solvent}}{ ext{mol solute}} [/latex]	Boiling point of pure solvent of pure solven (°C)	Boiling point elevation constant [latex]K_b[/latex] [latex]\frac{^\circ ext{C kg solvent}}{ ext{mol solute}} [/latex]
Water	0.00	1.86	100.0	0.512
Acetic acid	16.6	3.90	118.5	3.07
Benzene	5.5	5.1	80.1	2.53
Camphor	178	40	208.2	5.95

Question 14.15: A solution is made by dissolving 100. g of ethylene glycol (...

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