Q.1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point of benzene by 0.40 K. The freezing point depression constant of benzene is 5.12 K kg mol$^{-1}$. Find the molar mass of the solute.

Concept: Freezing Point Depression (colligative property).

Step 1 — Write the formula

For a non-electrolyte, $\Delta T_f = K_f \cdot m$, where $m$ is molality (mol solute per kg solvent).

Step 2 — Express molality in terms of molar mass

$m = \frac{\text{moles of solute}}{\text{kg of solvent}} = \frac{w / M}{W_{\text{kg}}}$,

with $w = 1.00\ \text{g}$, $W = 50\ \text{g} = 0.050\ \text{kg}$, and $M$ = molar mass (g/mol).

Step 3 — Substitute and solve

$\Delta T_f = K_f \cdot \frac{w}{M \cdot W_{\text{kg}}}$ …

Why this works: The idea behind boiling point elevation (and freezing point depression)

When you dissolve a non-volatile solute in a solvent, the solvent's freezing point drops. This happens because the solute particles disrupt the orderly arrangement needed for the solvent to freeze — the liquid has to be cooled further before solid forms. The key relationship is beautifully simple: the depression $\Delta T_f$ is directly proportional to the molality of the solution, not the concentration by mass or volume. That's why we use the formula:

where $K_f$ is the cryoscopic constant (freezing point depression constant) of the solvent, and $m$ is the molality of the solution in mol/kg.

The problem gives us $\Delta T_f$, $K_f$, the mass of solute, and the mass of solvent. Our job is to find the molar mass $M$ of the solute. Since molality itself depends on molar mass, we can set up an equation and solve.

Step-by-step solution

1. Write down what we know

• Mass of solute, $w_B = 1.00\ \text{g}$
• Mass of solvent (benzene), $w_A = 50\ \text{g} = 0.050\ \text{kg}$ (always convert to kg for molality)
• Freezing point depression, $\Delta T_f = 0.40\ \text{K}$
• Cryoscopic constant of benzene, $K_f = 5.12\ \text{K kg mol}^{-1}$

We need the molar mass of the solute, $M_B$ (in g/mol).

2. Express molality in terms of molar mass

Molality $m$ is moles of solute per kg of solvent:

So:

3. Plug into the freezing point depression equation

Substitute: …

Method: Freezing Point Depression (Cryoscopy)

This is a colligative property problem — the freezing point depression depends only on the number of solute particles, not their identity.

Step 1: Recall the formula

The freezing point depression is given by:

Where:

• $\Delta T_f$ = depression in freezing point (given: 0.40 K)
• $K_f$ = cryoscopic constant (given: 5.12 K kg mol$^{-1}$)
• $m$ = molality of the solution (mol solute per kg solvent)

Step 2: Find the molality

Rearrange to solve for molality:

Step 3: Relate molality to molar mass

Molality is defined as:

Let $M$ = molar mass of solute (g mol$^{-1}$).

Moles of solute = $\frac{1.00 \text{ g}}{M}$

Mass of solvent = $50 \text{ g} = 0.050 \text{ kg}$ …

Here are the common mistakes students make on this exact problem, along with how to avoid each one.

1. Forgetting to Convert Mass of Solvent from Grams to Kilograms

The Mistake:

Using $50 \, \text{g}$ directly in the formula instead of converting to $0.050 \, \text{kg}$.

Why it happens:

The cryoscopic constant $K_f$ has units of $\text{K kg mol}^{-1}$, so the solvent mass must be in kilograms.

How to avoid:

Always write the formula with units first:

Then substitute:

Key check: If your answer is off by a factor of 1000, this is likely the error.

2. Confusing Freezing Point Depression with Boiling Point Elevation

The Mistake:

Using $K_b$ (boiling point elevation constant) instead of $K_f$ (freezing point depression constant).

Why it happens:

The problem mentions "lowered the freezing point" but students sometimes misread and apply the boiling point formula.

How to avoid:

• Freezing point depression: $\Delta T_f = K_f \cdot m$
• Boiling point elevation: $\Delta T_b = K_b \cdot m$

The problem gives $K_f = 5.12 \, \text{K kg mol}^{-1}$, so use that.

3. Using the Wrong Formula for Molar Mass

The Mistake:

Plugging numbers into $M = \frac{K_f \cdot w_{\text{solute}}}{\Delta T_f \cdot w_{\text{solvent}}}$ without converting solvent mass to kg.

How to avoid:

Use the derived formula carefully:

Here, $w_{\text{solvent}}$ is in grams, and the factor 1000 converts it to kg.

Correct substitution:

4. Forgetting the "Non-Electrolyte" Condition

The Mistake:

Adding a van't Hoff factor $i$ (like $i=2$ for NaCl) when the solute is non-electrolyte.

Why it happens:

Students overcomplicate — they see "solute" and assume dissociation. …