Calculation of the Freezing-Point Depression of Water

Determine the freezing-point depression of water as a result of the addition of 0.01 $g / cm ^{3}$ of (a) methanol and (b) a protein whose molecular weight is 60 000.

Question

Calculation of the Freezing-Point Depression of Water

Determine the freezing-point depression of water as a result of the addition of 0.01 [latex]g / cm ^{3}[/latex] of (a) methanol and (b) a protein whose molecular weight is 60 000.

Accepted Answer

Since such a small amount of solute is involved, we will assume that the density of the solution is the same as that of pure water, 1 [latex]g / cm ^{3}[/latex]. Thus the solute mole fraction is

[latex]\begin{aligned}x_{ ext {solute }} &=\frac{ ext { Moles of solute in } 1 cm ^{3} ext { of solution }}{( ext { Moles of solute }+ ext { moles of water }) ext { in } 1 cm ^{3} ext { of solution }} \&=\frac{\frac{0.01}{m}}{\frac{0.99}{18}+\frac{0.01}{m}}\end{aligned}[/latex]

where m is the molecular weight of the solute. [latex]\Delta_{ ext {fus }} \underline{H}\left(T_{m} ight)[/latex] for water is 6025 J/mol.

a. The molecular weight of methanol is 32. Thus, [latex]x_{ ext {solute }}=0.00565[/latex] and

[latex]\Delta T=\frac{8.314 \frac{ J }{ mol K } imes(273.15 K )^{2} imes 5.65 imes 10^{-3}}{6025 J / mol }=0.58 K[/latex]

b. The molecular weight of the protein is 60 000, so [latex]x_{ ext {solute }} \sim 3 imes 10^{-6}[/latex] and

[latex]\Delta T=\frac{8.314 \frac{ J }{ mol K } imes(273.15 K )^{2} imes 3 imes 10^{-6}}{6025 J / mol }=3.1 imes 10^{-4} K[/latex]

This small depression of the freezing point is barely measurable. Compare this result with the osmotic pressure difference found in Illustration 11.5.1.

Question 12.3.1: Calculation of the Freezing-Point Depression of Water Determ...

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