Q.A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Concept: Freezing Point Depression — the lowering of freezing point is proportional to the molality of the solution ($\Delta T_f = K_f \cdot m$). For two solutes in the same solvent, $K_f$ is constant, so $\Delta T_f \propto m$.

Step 1: Find $\Delta T_f$ for sugar solution.

Pure water freezes at 273.15 K, observed at 271 K.

$\Delta T_f = 273.15 - 271 = 2.15\ \text{K}$.

Step 2: Relate $\Delta T_f$ to molality.

For sugar (molar mass $M_s = 342\ \text{g/mol}$), 5% by mass means 5 g sugar in 100 g solution → 5 g in 95 g water.

Molality of sugar:

$m_s = \frac{5/342}{0.095} \approx 0.154\ \text{mol/kg}$.

Step 3: Find $\Delta T_f$ for glucose.

Glucose molar mass $M_g = 180\ \text{g/mol}$. Same 5% by mass → 5 g glucose in 95 g water.

$m_g = \frac{5/180}{0.095} \approx 0.292\ \text{mol/kg}$.

Since $\Delta T_f \propto m$, …

1. The core concept: Freezing point depression

When a non-volatile solute is added to a solvent, the freezing point of the solution is lower than that of the pure solvent. This is a colligative property — it depends only on the number of solute particles, not on their chemical identity.

The relationship is given by:

Where:

• $\Delta T_f$ = depression in freezing point = $T_f^\circ - T_f$
• $K_f$ = cryoscopic constant (freezing point depression constant) of the solvent
• $m$ = molality of the solution (moles of solute per kg of solvent)

For water, $K_f$ is a fixed value (1.86 K kg mol⁻¹), but we don't need its numerical value here — we can work by ratio.

2. What we know from the cane sugar data

Cane sugar is sucrose, $\text{C}_{12}\text{H}_{22}\text{O}_{11}$, molar mass = 342 g/mol.

A 5% solution by mass means: 5 g of sugar in 100 g of solution. That means 5 g of solute and 95 g of solvent (water).

Step 1: Find molality of the sugar solution

Moles of sugar = $\frac{5}{342}$ mol

Mass of solvent = 95 g = 0.095 kg

So:

Let's compute:

Step 2: Find the depression for sugar

Pure water freezes at 273.15 K. The sugar solution freezes at 271 K.

So:

Step 3: Find $K_f$ for water

From $\Delta T_f = K_f \cdot m$:

3. Now for glucose

Glucose is $\text{C}_6\text{H}_{12}\text{O}_6$, molar mass = 180 g/mol.

A 5% solution by mass means: 5 g glucose in 95 g water (same solvent mass as before). …

Method: Freezing Point Depression (Cryoscopy) — Using Colligative Property Relation

This problem uses the colligative property of freezing point depression. Since both solutions have the same mass percentage (5% w/w) but different solutes (cane sugar vs glucose), we compare their molalities and use the fact that $\Delta T_f \propto m$ for the same solvent.

Step 1: Recall the formula

For a non-electrolyte solute:

where:

• $\Delta T_f = T_f^\circ - T_f$ (depression in freezing point)
• $K_f$ = cryoscopic constant of water (same for both)
• $m$ = molality of solution (moles of solute per kg of solvent)

Step 2: Find molality of cane sugar solution

• 5% by mass means 5 g cane sugar in 100 g solution → 5 g solute + 95 g solvent
• Molar mass of cane sugar (sucrose, $\text{C}_{12}\text{H}_{22}\text{O}_{11}$) = 342 g/mol
• Moles of cane sugar = $\frac{5}{342} = 0.01462$ mol
• Mass of solvent = 95 g = 0.095 kg
• Molality of cane sugar:

Step 3: Find $\Delta T_f$ for cane sugar

Given:

• $T_f^\circ$ (pure water) = 273.15 K
• $T_f$ (cane sugar solution) = 271 K

Step 4: Find molality of glucose solution

• 5% glucose → 5 g glucose in 95 g water
• Molar mass of glucose ($\text{C}_6\text{H}_{12}\text{O}_6$) = 180 g/mol
• Moles of glucose = $\frac{5}{180} = 0.02778$ mol
• Mass of solvent = 0.095 kg
• Molality of glucose: …

Here are the common mistakes students make on this exact type of boiling/freezing point elevation problem, and how to avoid each.

✗ Mistake 1: Confusing Freezing Point Depression with Boiling Point Elevation

The error:

Students see “boiling point elevation” in the title and try to apply the boiling point formula ($\Delta T_b = K_b \cdot m$), even though the problem gives freezing point data.

Why it happens:

Both concepts use colligative properties, but the constants ($K_f$ vs $K_b$) and the direction of temperature change are different.

How to avoid:

• Read the problem carefully: freezing point means use $\Delta T_f = K_f \cdot m$.
• Remember:
  • Freezing point decreases → $\Delta T_f = T_f^\circ - T_f$ (positive).
  • Boiling point increases → $\Delta T_b = T_b - T_b^\circ$.

✗ Mistake 2: Forgetting to Convert Mass Percent to Molality

The error:

Students treat “5% solution” as 5 g solute in 100 g solution, but then incorrectly use 100 g as the solvent mass.

Why it happens:

Mass percent is mass of solute per 100 g of solution, not per 100 g of solvent. Molality requires kg of solvent.

How to avoid:

• For a 5% solution:
  • Mass of solute = 5 g
  • Mass of solution = 100 g
  • Mass of solvent = $100 - 5 = 95 \, \text{g} = 0.095 \, \text{kg}$
• Always calculate solvent mass by subtracting solute mass from total solution mass.

✗ Mistake 3: Using Molar Mass of Sucrose Incorrectly

The error:

Students use the molar mass of glucose (180 g/mol) for sucrose, or vice versa, or forget to calculate moles at all.

Why it happens:

Both are sugars, but their molar masses are different:

• Sucrose ($C_{12}H_{22}O_{11}$) = 342 g/mol
• Glucose ($C_6H_{12}O_6$) = 180 g/mol

How to avoid:

• Write the molecular formula before calculating molar mass.
• Double-check: sucrose has 12 carbons, glucose has 6.

✗ Mistake 4: Assuming $K_f$ is the Same for Both Without Calculation

The error: …