## General Chemistry

Have you ever wondered what they are sparing on airplanes on cold days before they take off?

What they are spraying is a solution typically containing ethylene glycol, CHO which lowers the freezing point of water. It is essentially the same solution that is used as antifreeze in cars.
This feature is called the freezing point depression and it is also a type of colligative properties. The freezing point of the solution is generally lower than that of the solvent.

We can also see this in the phase diagram of the solvent and the solution. It includes two regions of interest for the colligative properties. On the left side, is the freezing/melting point region, and on the right side, toward higher temperatures, we have the boiling point region. We have already discussed the boiling point elevation of the solution, so let’s now see how the freezing point depression is explained.

On the x axis, we have the temperature and as it increases, the phase is transitioning from solid-liquid-gas. Notice, on the left side, that the freezing point of the solution is lower than that of the solvent.

To understand this property, let’s again recall that the entropy of a solution is higher than the entropy of the solvent because the mixing of components increases the randomness of the system.

Freezing, on the other hand, decreases the randomness, and thus the process requires energy to be removed from the system to create a more ordered state. Now, because the solution has higher entropy, it requires more energy to be removed to transition to a solid phase. Therefore, a solution has a lower freezing point than the solvent.

# Calculating the Freezing Point Depression

The Freezing point depression is calculated with the following formula:

Where Kf is the molal freezing point depression constant characteristic of a given solvent. The values of Kf for some common solvents are given below:

and m is the molality of the solute which is calculated using this formula:

For example, using the appropriate data in the table, determine the freezing point depression of the solution that contains 24.1 g urea (NH2)2CO) in 485 mL of water.

The freezing point depression is calculated by the following formula:

ΔTf = m x Kf

Kf is the freezing point depression constant for the solvent, and from the table, we find that it is 1.86 oC/m

m is the molality of the solute which is equal to:

${\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{mol}}\;{\rm{(solute)}}}}{{{\rm{kg}}\,{\rm{(solvent)}}}}$

The moles of urea are calculated from the mass:

${\rm{n}}\,{\left( {{\rm{N}}{{\rm{H}}_{\rm{2}}}} \right)_{\rm{2}}}{\rm{CO)}}\;{\rm{ = }}\,{\rm{24}}{\rm{.1}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{60}}{\rm{.1}}\;\cancel{{\rm{g}}}}}\;{\rm{ = }}\,{\rm{0}}{\rm{.401}}\;{\rm{mol}}$

Convert the mL to kg for water:

${\rm{m}}\,{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{O)}}\;{\rm{ = }}\,{\rm{485}}\,\cancel{{{\rm{mL}}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}{\rm{.00}}\,\cancel{{\rm{g}}}}}{{{\rm{1}}\,\cancel{{{\rm{mL}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\;{\rm{kg}}}}{{{\rm{1000}}\;\cancel{{\rm{g}}}}}{\rm{ = }}\,{\rm{0}}{\rm{.485}}\,{\rm{kg}}$

The molality of the solute is:

${\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.401}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.485}}\,{\rm{kg}}}}\; = \;0.827\,m$

The freezing point depression would then be:

ΔTf = Kfm = 1.86 oC /m x 0.827 m = 0.902 oC

# The Freezing Point Depression of Electrolyte Solutions

As in the case of vapor pressure lowering, when a strong electrolyte is dissolved in water, the concentration of the solute particles is given by the ions rather than the formula of the compound because strong electrolytes dissociate into ions in aqueous solutions.

So, if we dissolve 1 mole of NaCl in water it will dissociate into ions and two moles of ions will be formed:

NaCl(aq)  →  Na+(aq) + Cl(aq)

In general, we can calculate the number of ions based on the formula of the salt. For example, 1 mole of MgBr2 is expected to produce 3 moles of ions because each formula unit contains one Mg2+ and 2 Br ions. However, the dissociation of most ionic compounds does not occur at 100%,  and the solution of an ionic compound usually contains fewer particles than what its formula suggests. The actual extent of dissociation can be expressed as a van’t Hoff factor (i).

For most ionic compounds, the van’t Hoff’s constant is determined experimentally, and will likely be given to you in the test. If it is not, and there is nothing mentioned about, you can go based on the formula of the compound. For nonelectrolytes, it is assumed to be 1 as we do not worry about their negligible dissociation.

For example, calculate the freezing point of the solution prepared by dissolving 4.80 g NaCl in 25.0 g of water.

The freezing point depression is calculated by the following formula:

ΔTf = m x Kf

Kf is the freezing point depression constant for the solvent, and from the table, we find that it is 1.86 oC/m

m is the molality of the solute which is equal to:

${\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{mol}}\;{\rm{(solute)}}}}{{{\rm{kg}}\,{\rm{(solvent)}}}}$

The moles of particles from 4.80 g NaCl are calculated by multiplying the moles of NaCl by two since it dissociates into two ions:

${\rm{n}}\,({\rm{particles)}}\;{\rm{ = }}\,2\,{\rm{ \times }}\;{\rm{4}}{\rm{.80}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{58}}{\rm{.44}}\;\cancel{{\rm{g}}}}}\;{\rm{ = }}\,{\rm{0}}{\rm{.164}}\;{\rm{mol}}$

The mass of water is 0.0250 kg, therefore, the molality of the solute is:

${\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.164}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.0250}}\,{\rm{kg}}}}\; = \;6.56\,m$

The freezing point depression would then be:

ΔTf = Kfm = 1.86 oC /m x 6.56 m = 12.2 oC

The normal freezing point of water is 0oC, therefore, the freeing point of the solution will be:

0 – 12.2 = -12.2 oC

Check Also

#### Practice

1.

Determine the freezing point of a solution containing 1.80 g of glycerol, C3H8O3, in 26.0 g of water.

-1.40 oC

Solution

The freezing point depression is calculated by the following formula:

ΔTf = m x Kf

Kf is the freezing point depression constant for the solvent, and from the table, we find that it is 1.86 oC/m

m is the molality of the solute which is equal to:

${\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{mol}}\;{\rm{(solute)}}}}{{{\rm{kg}}\,{\rm{(solvent)}}}}$

The moles of glycerol are calculated from the mass:

${\rm{n}}\,{\rm{(}}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\,{\rm{0}}{\rm{.180}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{92}}{\rm{.1}}\;\cancel{{\rm{g}}}}}\;{\rm{ = }}\,{\rm{0}}{\rm{.00195}}\;{\rm{mol}}$

The mass of water in kg is:

${\rm{m}}\,{\rm{(}}{{\rm{H}}_{\rm{2}}}{\rm{O)}}\;{\rm{ = }}\,{\rm{26}}{\rm{.0}}\,\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\;{\rm{kg}}}}{{{\rm{1000}}\;\cancel{{\rm{g}}}}}{\rm{ = }}\,{\rm{0}}{\rm{.0260}}\,{\rm{kg}}$

The molality of the solute is:

${\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.0195}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.0260}}\,{\rm{kg}}}}\; = \;0.750\,m$

The freezing point depression and the freezing point of the solution would then be:

ΔTf = Kfm = 1.86 oC /m x 0.750 m = 1.40 oC

Tf = 0 – 1.40 = -1.40 oC

2.

A solution is prepared by dissolving 75.0 g glycerin (C3H8O3) in 240. g water. Calculate the boiling point of the solution. Glycerin is a nonelectrolyte.

101.74 oC

Solution

The normal boiling point of water is 100 oC. So, to find the boiling point of the solution, we first calculate the boiling point elevation and add that to 100 oC.

For the boiling point elevation, we use a similar formula with a different constant:

ΔTb = Kbm

To find the molality, m, we need to moles of the solute and mass of the solvent in kg. 240. g of water is 0.240 kg.

The moles of glycerin are calculated from the mass:

${\rm{n}}\,{\rm{(}}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\;{\rm{ = }}\,{\rm{75}}{\rm{.0}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{92}}{\rm{.1}}\;\cancel{{\rm{g}}}}}\;{\rm{ = }}\,{\rm{0}}{\rm{.814}}\;{\rm{mol}}$

The molality of the solute is:

${\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.814}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.240}}\,{\rm{kg}}}}\; = \;3.393\,m$

ΔTb = Kbm = 0.512 oC /m x 3.393 m = 1.74 oC

The boiling point of the solution would then be 100 + 1.74 = 101.74 oC

3.

A 1.60 g sample of a newly synthesized supplement was dissolved in 12.0 g carbon tetrachloride, CCl4. The boiling point of this solution was determined to be 79.2 oC while the boiling point of pure carbon tetrachloride is 76.5 oC. Calculate the molar mass of the supplement considering that is a nonelectrolyte.

244 g/mol

Solution

We are first going to determine the molality of the solution, from which the moles of the solute can be calculated based on the mass of the solvent. Once we know the moles, we can calculate the molar mass since the mass is also given.

ΔTb= 79.2 – 76.5 = 2.70 oC

The mass of carbon tetrachloride is 0.0120 kg, and the boiling point elevation constant for CCl4 is 4.95 oC/m.

The molality is derived from the expression of ΔTb:

$m\, = \,\frac{{\Delta T}}{{{K_b}}}\; = \;\frac{{{\rm{2}}{\rm{.70}}}}{{{\rm{4}}{\rm{.95}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.5455}}\,{\rm{mol/kg}}$

So, there are 0.5455 moles of solute in one kg of solvent. We need to find how many moles of solute we have in 0.0120 kg of solvent.

${\rm{n}}\,{\rm{(solute)}}\,{\rm{0}}{\rm{.0120}}\;\cancel{{{\rm{kg}}\;{\rm{solvent}}}}\;{\rm{ \times }}\;\frac{{{\rm{0}}{\rm{.5455}}\,{\rm{mol}}}}{{{\rm{1}}\;\cancel{{{\rm{kg}}\;{\rm{solvent}}}}}}{\rm{ = }}\,0.00655\,{\rm{mol}}$

The given mass of the supplement was 1.60 g, therefore, the molar mass of it would be:

1.60 g ÷ 0.00655 mol = 244 g/mol

4.

A 12.4 g of an unknown nonelectrolyte was dissolved in 100.0 g of water to prepare a solution with a freezing point of -1.7 °C. Calculate the molar mass of the unknown compound.

136 g/mol

Solution

We are going to first determine the molality of the solution, from which the moles of the solute can be calculated based on the mass of the solvent. Once we know the moles, we can calculate the molar mass since the mass is also given.

ΔTf= 0 – (-1.7) = 1.70 oC

The mass of the water is 0.1000 kg, and the freezing point depression constant for water is 1.86 oC/m.

The molality is derived from the expression of ΔTf:

$m\, = \,\frac{{\Delta T}}{{{K_f}}}\; = \;\frac{{{\rm{1}}{\rm{.70}}}}{{{\rm{1}}{\rm{.86}}}}\,{\rm{ = }}\,{\rm{0}}{\rm{.914}}\,{\rm{mol/kg}}$

So, there are 0.914 moles of solute in one kg of water. We need to find how many moles of solute we have in 0.1000 kg of water.

${\rm{n}}\,{\rm{(solute)}}\,{\rm{0}}{\rm{.1000}}\;\cancel{{{\rm{kg}}\;{\rm{solvent}}}}\;{\rm{ \times }}\;\frac{{{\rm{0}}{\rm{.914}}\,{\rm{mol}}}}{{{\rm{1}}\;\cancel{{{\rm{kg}}\;{\rm{solvent}}}}}}{\rm{ = }}\,0.0914\,{\rm{mol}}$

The given mass of the unknown was 12.4 g, therefore, the molar mass of it would be:

12.4 g ÷ 0.0914 mol = 136 g/mol

5.

How many grams of ethylene glycol (C2H6O2), a nonelectrolyte, must be added to 5.40 L water to prepare an antifreeze solution with a freezing point of -30.0 oC?

5.40 x 103 g

Solution

The mass of ethylene glycol can be determined from its moles which are calculated from the molality. The molality can be found from the expression for the freezing point depression. The normal freezing point of water is 0 oC, therefore, ΔTf= 30.0 oC. Based on these, the molality is:

$m\, = \,\frac{{\Delta T}}{{{K_f}}}\; = \;\frac{{{\rm{30}}{\rm{.0}}}}{{{\rm{1}}{\rm{.86}}}}\,{\rm{ = }}\,{\rm{16}}{\rm{.1}}\,{\rm{mol/kg}}$

The mass of water is 5.40 kg because the density is 1 g/mL, which means there are 16.1 mol of solute in one kg of water.

We need to find how many moles of solute we have in 5.40 kg of water.

${\rm{n}}\,{\rm{(solute)}}\,{\rm{5}}{\rm{.40}}\;\cancel{{{\rm{kg}}\;{\rm{solvent}}}}\;{\rm{ \times }}\;\frac{{{\rm{16}}{\rm{.1}}\,{\rm{mol}}}}{{{\rm{1}}\;\cancel{{{\rm{kg}}\;{\rm{solvent}}}}}}{\rm{ = }}\,86.94\,{\rm{mol}}$

The mass of ethylene glycol would be:

${\rm{m}}\,{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\;{\rm{86}}{\rm{.94}}\;{\rm{mol}}\;{\rm{ \times }}\;\frac{{{\rm{62}}{\rm{.1}}\;{\rm{g}}}}{{{\rm{1}}\,{\rm{mol}}}}\;{\rm{ = }}\,{\rm{5,399}}\,{\rm{g}}\,{\rm{ = }}\,{\rm{5}}{\rm{.40}}\;{\rm{ \times }}\;{\rm{1}}{{\rm{0}}^{\rm{3}}}\,{\rm{g}}$

So, to prepare an antifreeze with a freezing point of -30 oC, ethylene glycol and water are mixed in a 1:1 mass ratio.

6.

A sample of 23.0 g naphthalene (C10H8) was added to benzene (C6H6) and the resulting solution had a boiling point of 83.7 oC. How many liters of benzene were used to prepare the solution if the normal boiling point of benzene is 80.1 oC and the density is 0.877 g/mL?

0.144 L

Solution

The plan is to determine the mass of benzene from the molality and convert it to L.

ΔTb= 83.7 – 80.1 = 3.60 oC

The moles of naphthalene are calculated from the mass and the molar mass:

${\rm{n}}\,{\rm{(}}{{\rm{C}}_{{\rm{10}}}}{{\rm{H}}_{\rm{8}}}{\rm{)}}\;{\rm{ = }}\,{\rm{23}}{\rm{.0}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{128}}{\rm{.2}}\;\cancel{{\rm{g}}}}}\;{\rm{ = }}\,{\rm{0}}{\rm{.1794}}\;{\rm{mol}}$

Let’s set the mass of benzene as x kg, and write the expression for the molality:

$m\, = \,\frac{{\Delta T}}{{{K_b}}}\;{\rm{ = }}\;\frac{{{\rm{3}}{\rm{.60}}}}{{{\rm{2}}{\rm{.53}}}}\,{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.1794}}}}{{\rm{x}}}\;{\rm{ = }}\frac{{{\rm{n}}\,{\rm{(solute)}}}}{{{\rm{kg}}\;{\rm{(solvent)}}}}$

3.60x = 2.53 x 0.1794

3.60 x = 0.45389

x = 0.126 kg

This corresponds to 126 g, and the volume is:

${\rm{v}}\,{\rm{(benzene)}}\;{\rm{ = }}\,{\rm{126}}\;{\rm{g}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mL}}}}{{{\rm{0}}{\rm{.877}}\;{\rm{g}}}}\;{\rm{ = }}\;{\rm{144}}\;{\rm{mL}}\;{\rm{ = }}\;{\rm{0}}{\rm{.144}}\;{\rm{L}}\,$

7.

You are planning a winter trip and you have an antifreeze solution that is 40.0 % by mass of ethylene glycol (C2H6O2) in water. Would this antifreeze be good enough for the trip considering that the temperature goes down as far as -30.0 °C? Ethylene glycol is a nonelectrolyte.

The antifreeze wouldn’t be suitable.

Solution

The aim here is to find the ΔTf and see if it is greater than 30.0 oC. If it is, then the antifreeze would be suitable.

ΔTf = mKf

The Kf for water is 1.86 oC/m, so we need to determine the molality.

Suppose we take 1.00 kg of water, which means there would be 0.400 kg of ethylene glycol. The moles of ethylene glycol then would be:

${\rm{n}}\,{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}{{\rm{O}}_{\rm{2}}}{\rm{)}}\;{\rm{ = }}\,{\rm{400}}{\rm{.}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{62}}{\rm{.1}}\;\cancel{{\rm{g}}}}}\;{\rm{ = }}\,{\rm{6}}{\rm{.44}}\;{\rm{mol}}$

The molality of the solution is:

${\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{6}}{\rm{.44}}\;{\rm{mol}}}}{{{\rm{1}}{\rm{.00}}\,{\rm{kg}}}}\; = \;6.44\,m$

ΔTf = mKf = 6.44 m x 1.86 = 12.0 oC

The freezing point will only drop by 12.0 oC, so this ratio doesn’t yield a suitable antifreeze to drive at -30 oC.

8.

Calculate the boiling point of the solution prepared by dissolving 5.4 g FeCl3 in 36.0 g of water.

102 oC

Solution

The boiling point elevation is calculated by the following formula:

ΔTb = m x Kb

The moles of particles from 5.40 g FeCl3 are calculated by multiplying the moles of FeCl3 by four since it dissociates into four ions:

FeCl3(aq) →Fe3+(aq) + 3Cl(aq)

${\rm{n}}\,({\rm{particles)}}\;{\rm{ = }}\,4\,{\rm{ \times }}\;{\rm{5}}{\rm{.40}}\;\cancel{{\rm{g}}}\;{\rm{ \times }}\;\frac{{{\rm{1}}\;{\rm{mol}}}}{{{\rm{162}}{\rm{.2}}\;\cancel{{\rm{g}}}}}\;{\rm{ = }}\,{\rm{0}}{\rm{.133}}\;{\rm{mol}}$

The mass of water is 0.0360 kg, therefore, the molality of the solute is:

${\rm{m}}\;{\rm{ = }}\,\frac{{{\rm{0}}{\rm{.133}}\;{\rm{mol}}}}{{{\rm{0}}{\rm{.0360}}\,{\rm{kg}}}}\; = \;3.69\,m$

The boiling point elevation would then be:

ΔTb = Kbm = 0.512 oC /m x 3.69 m = 1.89 oC

The normal boiling point of water is 100 oC, therefore, the boiling point of the solution will be 101.89 oC ≈ 102 oC.

9.

How many grams of NaCl were added to 1.00 L of water if the resulting solution has a freezing point of -7.40 °C? The density of water is 1.00 g/mL.

116 g

Solution

The ΔTf = 7.40 oC, and we need to use this to determine the molality of the solution. Once we know the molality, we can determine the moles of the solute, and hence the given mass.

The molality is derived from the expression of ΔTf:

ΔTf = Kfm

$m\, = \,\frac{{\Delta T}}{{{K_f}}}\; = \;\frac{{{\rm{7}}{\rm{.40}}}}{{{\rm{1}}{\rm{.86}}}}\,{\rm{ = }}\,{\rm{3}}{\rm{.98}}\,{\rm{mol/kg}}$

This means there are 3.98 mol of particles in 1 kg of water, and because we have 1 kg of water (1L = 1 kg), this is how many moles of particles there are in the solution.

Importantly, the number of NaCl moles is half of this because each NaCl dissociates into two ions. Therefore, we had 1.99 mol NaCl dissolved in water.

m (NaCl) = 1.99 mol x 58.44 g/mol = 116 g