Ranking Four Solutions Without a Thermometer
In a winter practical class at a Patna college, four solutions are prepared and left in the same ice–salt bath: (A) 1.80 g of glucose (M = 180 g mol⁻¹) dissolved in 100 g of water; (B) 0.585 g of sodium chloride (M = 58.5 g mol⁻¹) in 100 g of water; (C) 0.600 g of urea (M = 60 g mol⁻¹) in 100 g of water; (D) 1.22 g of benzoic acid (M = 122 g mol⁻¹) in 100 g of benzene. Kf for water is 1.86 K kg mol⁻¹ and for benzene 5.12 K kg mol⁻¹. The students find that A and C freeze at the same temperature even though different masses were weighed out, while D freezes at a temperature corresponding to only half the expected number of particles. One student concludes that D is contaminated with a low-molar-mass impurity and wants to recrystallise it; another disagrees, saying the benzoic acid is simply behaving normally for that solvent. The instructor asks for a quantitative verdict before any recrystallisation is attempted.
1. Calculate the expected freezing-point depression for A, B and C, and explain why A and C behave identically despite different masses.
Read the worked answer
Convert every mass into moles first, then divide by the mass of solvent in kg (0.100 kg in each case).
- A: 1.80/180 = 0.0100 mol → molality 0.100 m; ΔTf = 1.86 × 0.100 = 0.186 K.
- C: 0.600/60 = 0.0100 mol → molality 0.100 m; ΔTf = 1.86 × 0.100 = 0.186 K.
- B: 0.585/58.5 = 0.0100 mol → 0.100 m of formula units, but NaCl is a strong electrolyte, i ≈ 2, so the effective particle molality is 0.200 m; ΔTf = 1.86 × 0.200 = 0.372 K.
A and C coincide because freezing-point depression depends on the number of particles, not on the mass weighed out. Glucose and urea are both non-electrolytes that neither dissociate nor associate, so each gram-mole contributes one mole of particles. The masses 1.80 g and 0.600 g were deliberately chosen so that each gives 0.0100 mol; the difference in mass merely reflects the threefold difference in molar mass. B, though weighing least, depresses the freezing point most because each NaCl formula unit releases two ions. Strictly, the measured i for 0.1 m NaCl is about 1.87 rather than 2.00 because of ion pairing, so 0.372 K is an upper estimate - a point worth stating before comparing with measured values.
2. Analyse solution D quantitatively. Is the impurity explanation tenable? What is actually happening?
Read the worked answer
Calculate what the expected depression would be if benzoic acid behaved as a normal solute: molality = 1.22/122 = 0.0100 mol in 0.100 kg = 0.100 m, so ΔTf(expected) = 5.12 × 0.100 = 0.512 K. The observed depression is about 0.256 K.
Use ΔTf(observed) = i·Kf·m: i = 0.256/(5.12 × 0.100) = 0.50, i.e. the number of particles is half the number of formula units. An impurity would do the opposite - extra particles would make the depression larger than 0.512 K, not smaller. The impurity hypothesis is therefore untenable, and recrystallisation would be a wasted step.
The only process that halves the particle count is association: two benzoic acid molecules join through a pair of hydrogen bonds between their –COOH groups to give a cyclic dimer, 2 C6H5COOH ⇌ (C6H5COOH)2. Benzene is non-polar and cannot solvate or disrupt these hydrogen bonds, so association is favourable; in water the same acid would ionise instead and give i > 1. The apparent molar mass is 122/0.5 = 244 g mol⁻¹, matching the dimer. This is a classic abnormal molar mass, and it shows the limits of using colligative properties for molar mass unless the solvent and the association behaviour are known.
3. A radiator holding 1.0 kg of water must be protected down to −5.0 °C. Calculate the mass of glucose needed and comment on whether glucose is the sensible industrial choice.
Read the worked answer
Required depression = 5.0 K. For a non-electrolyte, ΔTf = Kf·m, so m = 5.0/1.86 = 2.69 mol kg⁻¹. For 1.0 kg of water this is 2.69 mol of glucose: 2.69 × 180 = about 484 g, i.e. roughly half the mass of the water itself.
Judgement, not just arithmetic: glucose would work on paper but is a poor engineering choice. First, it is inefficient by mass - the same 2.69 mol kg⁻¹ of ethylene glycol (M = 62 g mol⁻¹) needs only about 167 g, roughly a third as much. Second, an aqueous glucose solution in a hot radiator is a nutrient medium for micro-organisms, will caramelise and foul the passages at engine temperatures, and offers no protection against boiling. Ethylene glycol is preferred because it is non-volatile, raises the boiling point as well as lowering the freezing point, does not support microbial growth and is stable over repeated heating cycles. The calculation also illustrates a concept point: colligative protection depends on moles of particles, so a low molar mass per particle is a practical advantage - the opposite of what the sodium chloride result in solution B showed about ion count, where the benefit came from dissociation rather than from low molar mass.