1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to –4°C before freezing.
1 kg of 0.75 molal aqueous solution of sucrose can be cooled up to –4°C before freezing. The amount of ice (in g) that will be separated out is ________ . (Nearest integer)
$\left[\right.$ Given : $\left.\mathrm{K}_{\mathrm{r}}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right]$
Let mass of water initially present = x gm
$\Rightarrow$ Mass of sucrose $=(1000-\mathrm{x}) \mathrm{gm}$
$\Rightarrow$ moles of sucrose $=\left(\frac{1000-x}{342}\right)$
$\Rightarrow 0.75=\frac{\left(\frac{1000-x}{342}\right)}{\left(\frac{x}{1000}\right)} \Rightarrow \frac{x}{1000}=\frac{1000-x}{342 \times 0.75}$
$\Rightarrow 256.5 x=10^{6}-1000 x$
$\Rightarrow x=795.86 \mathrm{gm}$
$\Rightarrow$ moles of sucrose $=0.5969$
New mass of $\mathrm{H}_{2} \mathrm{O}=\mathrm{a} \mathrm{kg}$
$\Rightarrow 4=\frac{0.5969}{\mathrm{a}} \times 1.86 \Rightarrow \mathrm{a}=0.2775 \mathrm{~kg}$
$\Rightarrow$ ice separated $=(795.86-277.5)=518.3 \mathrm{gm}$
Always double-check your answer by verifying the new molality. After 518 g of ice separates, remaining water = 277.5 g and moles of sucrose = 0.5968 mol. New molality = 0.5968 / 0.2775 ≈ 2.15 mol/kg. Then ΔTf = 1.86 × 2.15 ≈ 4.0°C. ✔ This back-verification in the exam takes 20 seconds and saves marks.
Why Do Students Get This Wrong? Common Mistakes
Mistake 1 — Treating 1 kg as the Mass of Water Alone
The most frequent error: students assume 1 kg = mass of solvent (water). In reality, 1 kg is the total mass of the solution (water + sucrose). Always define variables for the water mass separately when the total solution mass is given.
Mistake 2 — Forgetting That Moles of Solute Stay Constant
When ice separates, only pure water leaves the liquid phase. The sucrose stays dissolved. Students sometimes recalculate moles of sucrose incorrectly for the second part.
Mistake 3 — Using ΔTf = 4°C vs ΔTf = –4°C
ΔTf is always a positive depression value. The solution freezes at –4°C, so the depression from 0°C is +4 K. Never plug in –4 into the formula.
Mistake 4 — Unit Error in Molality Denominator
Kf is in K kg mol⁻¹. The mass of water in the denominator of molality must be in kilograms, not grams. Forgetting to convert grams to kilograms is a very common slip.
Colligative Properties Formula Comparison Table
| Property | Formula | Key Constant | Depends On |
|---|---|---|---|
| Depression in Freezing Point | ΔTf = Kf × m | Kf (Cryoscopic Constant) | Molality of Solute |
| Elevation in Boiling Point | ΔTb = Kb × m | Kb (Ebullioscopic Constant) | Molality of Solute |
| Relative Lowering of Vapour Pressure | (P° − P) / P° = xsolute | — | Mole Fraction of Solute |
| Osmotic Pressure | π = MRT | R (Gas Constant) | Molarity of Solute |
For water: Kf = 1.86 K kg mol⁻¹, Kb = 0.512 K kg mol⁻¹
All four colligative properties depend only on the number of solute particles, not on their chemical identity. This is why sucrose (a non-electrolyte) uses i = 1 in these calculations.
For a complete chapter-wise breakdown and solved exercises, explore the NCERT Solutions for Class 12 Chemistry and the NCERT Books for Class 12 available on eSaral.
Frequently Asked Questions
Find answers to common questions.
Why does ice separate out when a solution is cooled below 0°C?
When a solution is cooled below 0°C, pure water molecules freeze out as ice because the dissolved solute lowers the chemical potential of liquid water, not the solid. As ice forms, the remaining solution becomes more concentrated, further depressing its freezing point until equilibrium is reached at the applied temperature.
What is the value of Kf for water and what are its units?
The cryoscopic constant Kf for water is 1.86 K kg mol⁻¹ (also written as 1.86 °C kg mol⁻¹). It represents the depression in freezing point caused by dissolving 1 mole of a non-electrolyte solute in 1 kg of water. Its value is a fixed property of the solvent and does not depend on the solute.
How much ice separates when 1 kg of 0.75 molal sucrose solution is cooled to –4°C?
Approximately 518 g of ice separates out. The calculation uses ΔTf = Kf × m in two stages: first to find the original water content (≈795.86 g) from the 0.75 molal condition, then to find the water mass remaining in solution (≈277.5 g) at –4°C depression, and finally subtracting to get the ice.
What is the difference between molality and molarity in colligative property problems?
Molality (m) = moles of solute / kg of solvent — it is temperature-independent and used in ΔTf, ΔTb calculations. Molarity (M) = moles of solute / litre of solution — it changes with temperature as volume changes. For freezing/boiling point problems, always use molality. For osmotic pressure (π = MRT), use molarity.
Is this type of question asked in JEE Main?
Yes. Freezing point depression with ice separation is a standard JEE Main integer-type question from the chapter Solutions (Class 12 Chemistry). NTA has included this exact pattern multiple times. Students are expected to know the two-step approach: find initial water content from the given molality, then find new water content from the given ΔTf.
What is the molar mass of sucrose used in this calculation?
The molar mass of sucrose (C₁₂H₂₂O₁₁) is 342 g mol⁻¹. This is calculated as: (12 × 12) + (22 × 1) + (11 × 16) = 144 + 22 + 176 = 342 g mol⁻¹. This value must be memorised for JEE as sucrose numericals are a recurring question type.