Q.45 g of ethylene glycol ($C_2H_6O_2$) is mixed with 600 g of water. Calculate

Concept: Freezing Point Depression

When a non-volatile solute dissolves in a solvent, the freezing point of the solution drops below that of the pure solvent. The depression is given by $\Delta T_f = K_f \cdot m$, where $K_f$ is the cryoscopic constant and $m$ is molality.

Step 1. Find the molar mass of ethylene glycol: $M = 2(12) + 6(1) + 2(16) = 62 \text{ g/mol}$.

Step 2. Calculate moles of solute: $n = \frac{45}{62} = 0.7258 \text{ mol}$.

Step 3. Calculate molality: $m = \frac{0.7258}{0.6} = 1.2097 \approx 1.2 \text{ mol/kg}$. NCERT rounds the molality to $1.2 \text{ mol kg}^{-1}$ before the final multiplication, and we follow the book.

Step 4. Apply the formula with $K_f = 1.86 \text{ K kg/mol}$ for water:

Step 5. The freezing point of the solution is $273.15 \text{ K} - 2.2 \text{ K} = 270.95 \text{ K}$.

When you dissolve a non-volatile solute like ethylene glycol in water, the solute particles disrupt the orderly arrangement water molecules need to form ice. This interference means the solution must be cooled below the normal freezing point before it can freeze. The extent of this freezing point depression depends on how many solute particles are present per kilogram of solvent—the molality—and a solvent-specific constant $K_f$ that captures how "sensitive" the solvent is to dissolved particles.

The relationship is beautifully simple:

where $\Delta T_f$ is the freezing point depression (always positive), $K_f$ is the cryoscopic constant (for water, $K_f = 1.86 \, \text{K kg mol}^{-1}$), and $m$ is the molality in mol/kg.

Step-by-step solution

1. Find the molar mass of ethylene glycol

Ethylene glycol is $\text{C}_2\text{H}_6\text{O}_2$. Adding up atomic masses:

2. Calculate moles of ethylene glycol

We have 45 g of solute:

3. Convert mass of water to kilograms

The solvent mass is 600 g:

4. Calculate molality

Molality is moles of solute per kilogram of solvent:

NCERT's Solution rounds the molality to $1.2 \, \text{mol kg}^{-1}$ at this point and carries that rounded value forward; we do the same so our final answers match the book's.

5. Apply the freezing point depression formula

Using $K_f = 1.86 \, \text{K kg mol}^{-1}$ for water:

6. Determine the new freezing point

Pure water freezes at 273.15 K. The solution freezes at:

Method: Freezing Point Depression (Cryoscopy)

This is a colligative property problem — the solute particles (ethylene glycol) lower the freezing point based only on how many particles are present, not their identity.

Step 1 — Identify the formula

The freezing point depression is given by:

Where:

• $\Delta T_f$ = freezing point depression (in K)
• $i$ = van't Hoff factor (for non-electrolytes like ethylene glycol, $i = 1$)
• $K_f$ = cryoscopic constant of the solvent (for water, $K_f = 1.86 \, \text{K kg mol}^{-1}$)
• $m$ = molality of the solution (mol solute per kg solvent)

Step 2 — Calculate moles of solute

Molar mass of ethylene glycol ($C_2H_6O_2$):

Moles of solute:

Step 3 — Calculate molality

Mass of solvent (water) = 600 g = 0.600 kg

NCERT rounds the molality to $1.2 \, \text{mol kg}^{-1}$ before the next step — do the same so your answer matches the book's.

Step 4 — Calculate freezing point depression

(Carrying the unrounded 1.2097 instead gives 2.25 K — a rounding choice only; quote the book's 2.2 K.)

Step 5 — Calculate the freezing point of the solution

Pure water freezes at $273.15 \, \text{K}$. The solution freezes lower by $\Delta T_f$:

Final Answer

Key concept check: Ethylene glycol is a non-electrolyte ($i=1$), so each molecule contributes one particle. If it were an electrolyte (like NaCl, $i=2$), the depression would double for the same molality.

Here are the most common mistakes students make when solving freezing point depression problems, using this question as the example.

1. Using the wrong formula (Freezing vs. Boiling)

Mistake:

Students often confuse the formulas for freezing point depression and boiling point elevation. They might use $K_b$ when they need $K_f$, or forget the sign convention.

Correct approach:

• Freezing point depression:

• Boiling point elevation:

For water:

• $K_f = 1.86 \, \text{K kg mol}^{-1}$
• $K_b = 0.52 \, \text{K kg mol}^{-1}$

How to avoid:

Write the correct formula before plugging in numbers. Circle whether the problem asks for freezing or boiling.

2. Forgetting the van't Hoff factor ($i$)

Mistake:

Assuming $i = 1$ for all solutes. Ethylene glycol is a non-electrolyte, so $i = 1$ is correct here — but students often forget to check.

Correct approach:

• Non-electrolyte (sugar, glycol, urea): $i = 1$
• Electrolyte (NaCl, CaCl$_2$): $i$ = number of ions produced

How to avoid:

Always ask: Does this solute dissociate in water? If yes, find $i$ from the dissociation equation.

3. Incorrect molality calculation

Mistake:

Using mass of solvent in grams instead of kilograms, or using mass of solution instead of solvent.

Correct calculation for this example:

• Molar mass of ethylene glycol ($C_2H_6O_2$):

• Moles of solute:

• Mass of solvent (water) = 600 g = 0.600 kg
• Molality:

How to avoid:

Convert solvent mass to kg before dividing. Write units at every step.

4. Sign error in final freezing point

Mistake:

Adding $\Delta T_f$ to the pure solvent's freezing point instead of subtracting.

Correct:

Freezing point of pure water = $273.15 \, \text{K}$

Depression means the solution freezes lower:

How to avoid:

Remember: Depression = decrease. The new freezing point is always lower than the pure solvent's freezing point.

5. Rounding differently from the intended answer

Mistake:

Mixing rounding conventions mid-way and then wondering why the answer differs from the textbook's.

What happens here:

NCERT rounds the molality to $1.2 \, \text{mol kg}^{-1}$ before multiplying, giving $\Delta T_f = 1.86 \times 1.2 = 2.2 \, \text{K}$. If you instead carry the unrounded 1.2097 all the way, you get 2.25 K. Both computations are honest arithmetic — but the book's printed answer is 2.2 K, so state where you rounded.

How to avoid:

Keep 3–4 significant figures in intermediate steps, state clearly when you round, and match the expected answer's precision at the end.

6. Forgetting to state the final answer clearly

Mistake:

Writing only $\Delta T_f$ when the question asks for both (a) depression and (b) freezing point.

Correct final answer for this problem:

• (a) Freezing point depression:

• (b) Freezing point of solution:

How to avoid:

Read the question twice. Label each part of your answer clearly.

Quick checklist to avoid all these mistakes