Depression of Freezing Point Numericals

CHEColligative Properties

A non-volatile solute lowers the freezing point in proportion to molality, the basis for antifreeze use and for cryoscopic molar-mass determination. ISC numericals apply $\Delta T_f = K_f m$ to find molar mass, the freezing point, the mass of solute, or $K_f$ itself.

Freezing-point depression

\Delta T_f = i K_f m = T_f(\text{solvent}) - T_f(\text{solution})

K_f

cryoscopic constant,

m

molality,

i

van't Hoff factor

Molar mass from depression

M_2 = \dfrac{i \, K_f \, w_2 \times 1000}{\Delta T_f \times w_1}

w_2

solute mass (g),

w_1

solvent mass (g)

Cryoscopic constant from data

K_f = \dfrac{\Delta T_f \times M_2 \times w_1}{w_2 \times 1000}

use when the solute molar mass

M_2

is known (non-electrolyte,

i = 1

)

Depression of freezing point: $\Delta T_f = i K_f m$ , where $K_f$ is the molal (cryoscopic) constant.
$\Delta T_f = T_f(\text{pure solvent}) - T_f(\text{solution})$ ; for water $K_f = 1.86\,\text{K kg mol}^{-1}$ .
Molar mass from depression: $M = \dfrac{i K_f \times w_2 \times 1000}{\Delta T_f \times w_1}$ , with masses $w_2$ (solute) and $w_1$ (solvent) in grams.
Rearranged to find $K_f$ for a known non-electrolyte: $K_f = \dfrac{\Delta T_f \times M \times w_1}{w_2 \times 1000}$ .
Mass of solute for a target freezing point: get $m$ from $\Delta T_f = K_f m$ , then $w = m \times (\text{kg solvent}) \times M_r$ .
Ethylene glycol is the common antifreeze example — a non-electrolyte ( $i = 1$ ) added to water to lower its freezing point.
A solute lowers the freezing point because it reduces the solvent's vapour pressure, so the solid-liquid equilibrium is reached at a lower temperature.
$K_f$ is the depression for a $1\,\text{molal}$ non-electrolyte solution and is a property of the solvent alone, with units $\text{K kg mol}^{-1}$ .
For benzene $K_f = 5.12\,\text{K kg mol}^{-1}$ , larger than water's $1.86$ , so benzene gives bigger, more precisely measured depressions — useful in molar-mass work.
Order of solving: find $\Delta T_f$ from the two freezing points, then molality, then moles and molar mass (including $i$ for electrolytes).
Freezing-point depression is the basis of antifreeze (ethylene glycol in radiators) and of de-icing roads with salt, which gives a large $\Delta T_f$ because of its $i$ .
Trap: a freezing point of $-0.372^\circ\text{C}$ means $\Delta T_f = 0.372\,\text{K}$ (magnitude below $0^\circ\text{C}$ ), not the temperature itself.

Where the marks go

Substituting the negative freezing-point temperature for $\Delta T_f$ ; the depression is its magnitude below the pure-solvent freezing point.
Computing $\Delta T_f$ as solution minus solvent; depression is solvent minus solution (a positive value).
Forgetting $i$ for an electrolyte (e.g. de-icing salt), which underestimates the depression.
Using water's $K_f = 1.86$ when the solvent is benzene ( $5.12$ ) or another liquid — $K_f$ is solvent-specific.
Unit slip: molality requires solvent mass in kg, while the molar-mass formula keeps $w_1$ in grams with the $\times 1000$ factor.

How the board asks it

Numericalmolar mass from depression
On dissolving $1.9\,\text{g}$ of a non-volatile, non-electrolyte solute in $75\,\text{g}$ of water, the freezing point of the solution is found to be $-0.465\,^\circ\text{C}$ . Calculate the molar mass of the solute ( $K_f$ for water $= 1.86\,\text{K kg mol}^{-1}$ ).
Numericalfreezing point from $\Delta T_f = K_f m$
Calculate the freezing point of a solution prepared by dissolving $18\,\text{g}$ of glucose ( $M = 180\,\text{g mol}^{-1}$ ) in $250\,\text{g}$ of water, given $K_f = 1.86\,\text{K kg mol}^{-1}$ for water.
Numericalmass of solute for a target freezing point
What mass of ethylene glycol ( $M = 62\,\text{g mol}^{-1}$ ) must be added to $5.5\,\text{kg}$ of water to lower its freezing point to $-10\,^\circ\text{C}$ ? ( $K_f = 1.86\,\text{K kg mol}^{-1}$ )
Numerical $K_f$ of solvent from a known non-electrolyte
A solution of $1.0\,\text{g}$ of a non-electrolyte ( $M = 256\,\text{g mol}^{-1}$ ) in $50\,\text{g}$ of benzene freezes $0.40\,\text{K}$ below the freezing point of pure benzene. Calculate the molal depression constant $K_f$ of benzene.
Numericalvan't Hoff factor and degree of dissociation
$1.0\,\text{g}$ of a strong electrolyte $AB$ ( $M = 100\,\text{g mol}^{-1}$ ) is dissolved in $100\,\text{g}$ of water and the solution freezes at $-0.31\,^\circ\text{C}$ . Calculate the van't Hoff factor $i$ and the degree of dissociation ( $K_f = 1.86\,\text{K kg mol}^{-1}$ ).
Numericaldepression as proportional to molality
Equal masses of urea ( $M = 60\,\text{g mol}^{-1}$ ) and glucose ( $M = 180\,\text{g mol}^{-1}$ ) are each dissolved in $100\,\text{g}$ of water. Calculate the ratio of the freezing-point depressions of the two solutions.

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.