Abnormal molar masses

Almost twice as far

Dissolve $0.10\ \text{mol}$ of glucose in $1\ \text{kg}$ of water. The formula gives a fall of the freezing point of $0.186\ \text{K}$. Now dissolve $0.10\ \text{mol}$ of sodium chloride in the same mass of water. The fall that we measure is about $0.35\ \text{K}$, nearly twice as much.

A colligative property counts particles. A glucose molecule stays in one piece. A formula unit of sodium chloride splits into two ions, $\text{Na}^+$ and $\text{Cl}^-$. So $0.10\ \text{mol}$ of salt makes nearly $0.20\ \text{mol}$ of particles.

The picture shows the idea. The same number of formula units makes one, two or three times as many particles. The effect grows with the number of particles.

Some solutes do the opposite. Ethanoic acid in benzene, which is not a polar solvent, has molecules that join in pairs through hydrogen bonds. A pair is one particle, so there are fewer particles than we expect, and the effect is smaller. A molar mass worked out from such a solution is too large. Chemists call these results abnormal molar masses.

The van’t Hoff factor

We measure how far a solute splits or joins with the van’t Hoff factor, $i$:

The calculated effect is the value for a solute that stays whole. The factor can also be written with the moles of particles, or with the molar masses:

• $i = \dfrac{\text{moles of particles}}{\text{moles of solute dissolved}}$
• $i = \dfrac{\text{normal molar mass}}{\text{observed molar mass}}$

Fewer particles per mole of solute means a larger molar mass, which is why the last form is upside down. Look at the three cases:

• $i = 1$: the solute stays whole. Glucose is an example.
• $i > 1$: the solute splits. Sodium chloride gives a value near $2$.
• $i < 1$: the solute joins. Ethanoic acid in benzene gives a value near $0.5$ when all of it has paired.

Every colligative formula gets a factor $i$:

• Relative lowering of the vapour pressure: $\dfrac{p_1^0 - p_1}{p_1^0} = i\,x_2$
• Boiling point: $\Delta T_b = i\,K_b\,m$
• Freezing point: $\Delta T_f = i\,K_f\,m$
• Osmotic pressure: $\pi = i\,C\,R\,T$

A solute that splits

Let the formula unit give $\nu$ ions (the Greek letter $\nu$ is read "nu"), and let $\alpha$ be the degree of dissociation, the fraction of the solute that has split. Start with $1\ \text{mol}$:

• $(1 - \alpha)\ \text{mol}$ stays whole.
• $\alpha\ \text{mol}$ splits and gives $\nu\alpha\ \text{mol}$ of ions.

The total is $1 - \alpha + \nu\alpha$, so

When all of it splits, $\alpha = 1$ and $i = \nu$. Here are some values of $\nu$:

• $\text{NaCl}$ gives $\text{Na}^+$ and $\text{Cl}^-$: $\nu = 2$.
• $\text{K}_2\text{SO}_4$ gives two $\text{K}^+$ and one $\text{SO}_4^{2-}$: $\nu = 3$.
• $\text{Al}_2(\text{SO}_4)_3$ gives two $\text{Al}^{3+}$ and three $\text{SO}_4^{2-}$: $\nu = 5$.
• A weak acid $\text{HA}$ gives $\text{H}^+$ and $\text{A}^-$: $\nu = 2$, with a small $\alpha$.

For strong electrolytes the measured $i$ is a little less than $\nu$, even though the salt has split completely. The reason is that oppositely charged ions attract one another and move a little like pairs.

Worked example: calcium chloride. A solution has $1.11\ \text{g}$ of $\text{CaCl}_2$, of molar mass $111\ \text{g/mol}$, in $400\ \text{g}$ of water. Find the fall of the freezing point (a) if it splits completely, and (b) if only $80\%$ of it splits.

1. Molality. $\dfrac{1.11\ \text{g}}{111\ \text{g/mol}} = 0.0100\ \text{mol}$, and $\dfrac{0.0100\ \text{mol}}{0.400\ \text{kg}} = 0.0250\ \text{mol/kg}$.
2. (a) $\nu = 3$ and $i = 3$, so $\Delta T_f = 3 \times 1.86\ \text{K kg/mol} \times 0.0250\ \text{mol/kg} = 0.1395\ \text{K}$.
3. (b) $i = 1 + 0.80 \times (3 - 1) = 2.6$, so $\Delta T_f = 2.6 \times 1.86\ \text{K kg/mol} \times 0.0250\ \text{mol/kg} = 0.121\ \text{K}$.

A solute that joins

Let $n$ molecules join into one, and let $\alpha$ be the fraction that has joined. Start with $1\ \text{mol}$:

• $(1 - \alpha)\ \text{mol}$ stays single.
• $\alpha\ \text{mol}$ joins, and makes $\dfrac{\alpha}{n}\ \text{mol}$ of larger particles.

The total is $1 - \alpha + \dfrac{\alpha}{n}$, so

For pairs, $n = 2$ and $i = 1 - \dfrac{\alpha}{2}$. If every molecule has paired, $\alpha = 1$ and $i = 0.5$.

Worked example: ethanoic acid in benzene. A solution of $1.5\ \text{g}$ of ethanoic acid, of molar mass $60\ \text{g/mol}$, in $40\ \text{g}$ of benzene freezes $1.79\ \text{K}$ below pure benzene. How much of the acid is in pairs?

1. Molality. $\dfrac{1.5/60\ \text{mol}}{0.040\ \text{kg}} = 0.625\ \text{mol/kg}$.
2. Fall for single molecules. $5.12\ \text{K kg/mol} \times 0.625\ \text{mol/kg} = 3.20\ \text{K}$.
3. The factor. $i = \dfrac{1.79\ \text{K}}{3.20\ \text{K}} = 0.559$.
4. The share in pairs. From $i = 1 - \dfrac{\alpha}{2}$ we want $\alpha$, so $\alpha = 2(1 - i) = 2 \times 0.441 = 0.88$.

About $88\%$ of the molecules are in pairs. The molar mass that we would have found is $\dfrac{60}{0.559} = 107\ \text{g/mol}$. That lies between $60$ and $120$, the molar mass of a pair.

Salt on the road and salt in the blood

Road salt. Each formula unit of $\text{NaCl}$ gives two ions, so salt lowers the freezing point twice as much as sugar does for the same number of moles. One kilogram of $\text{NaCl}$ gives $\dfrac{2 \times 1000\ \text{g}}{58.5\ \text{g/mol}} = 34\ \text{mol}$ of ions. One kilogram of $\text{CaCl}_2$ gives $\dfrac{3 \times 1000\ \text{g}}{111\ \text{g/mol}} = 27\ \text{mol}$. So for the same mass, $\text{NaCl}$ makes more particles. The formulas hold only for dilute solutions, and a mixture of salt and water cannot go below about $-21\ ^\circ\text{C}$.

Saline. The saline in a hospital drip has $9.0\ \text{g}$ of $\text{NaCl}$ in each litre. That is $\dfrac{9.0}{58.5} = 0.154\ \text{mol/L}$. If every unit splits, the particles are $0.308\ \text{mol/L}$, and $\pi = 0.308\ \text{mol/L} \times 0.0821\ \text{L atm/(K mol)} \times 310\ \text{K} = 7.83\ \text{atm}$. This is close to the pressure of blood, which is why such saline is isotonic with it.

Practise with new numbers

Try it in Python

The program finds $i$ and then $\alpha$ for three salts, from the fall of the freezing point.

Common mistakes

• Taking $i = \nu$ when the question says that only part of the salt splits. Use $i = 1 + \alpha(\nu - 1)$.
• Counting molecules and not ions for $\nu$. One unit of $\text{CaCl}_2$ gives three ions.
• Writing $\alpha$ as a percentage in the formulas. Use $0.88$, not $88$.
• Using the formula for splitting when the solute joins. For joining, $i$ is less than $1$.
• Giving $i \neq 1$ to a solute such as glucose. A solute that stays whole has $i = 1$.

Remember these

• $i = \dfrac{\text{observed effect}}{\text{calculated effect}} = \dfrac{\text{normal molar mass}}{\text{observed molar mass}}$.
• Splitting: $i = 1 + \alpha(\nu - 1)$, so $i > 1$. Joining: $i = 1 - \alpha\left(1 - \dfrac{1}{n}\right)$, so $i < 1$.
• With the factor: $\Delta T_b = i\,K_b\,m$, $\Delta T_f = i\,K_f\,m$ and $\pi = i\,C\,R\,T$.
• Strong electrolytes show an $i$ a little below $\nu$ because the ions attract each other.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. What is the largest van’t Hoff factor that aluminium chloride, $\text{AlCl}_3$, can have in water?

2. A salt that gives $2$ ions is $90\%$ split. What is $i$?

3. The molar mass of a solute that joins in pairs is found to be $110\ \text{g/mol}$, and its normal molar mass is $60\ \text{g/mol}$. What is $i$?

4. Why does a solution of $0.10\ \text{mol/kg}$ sodium chloride boil at a higher temperature than one of $0.10\ \text{mol/kg}$ glucose?