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MHT-CET Chemistry · Solutions and Colligative Properties

Depression of Freezing Point

A non-volatile solute lowers the freezing point by ΔTf = Kf · m (times i for an electrolyte); the molar-mass form is M₂ = 1000 Kf W₂/(ΔTf W₁), with Kf = 1.86 K kg mol⁻¹ for water.

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Why this matters

22 PYQs, none HARD — the boiling-point page with a different constant, and the exam sets it the same way: molality from ΔTf (often given as a freezing point of −0.36 °C or −0.95 °C), Kf from a measured depression, molar mass from masses, and a ranking of electrolytes by i·m. The one new habit is reading a negative Celsius freezing point as a positive ΔTf.

Concept 1 of 3: ΔTf = Kf · m: Molality, Kf, or ΔTf From a Freezing Point

Freezing point falls by KfK_f per molal. Water's KfK_f is 1.861.86 K kg mol−1^{-1}; a solution freezing at −0.93∘-0.93^\circC has ΔTf=0.93\Delta T_f = 0.93 K and molality 0.50.5.

Definition

  • m=ΔTfKfm = \dfrac{\Delta T_f}{K_f}: 0.931.86=0.5\dfrac{0.93}{1.86} = 0.5, 0.181.6=0.113\dfrac{0.18}{1.6} = 0.113, 0.951.86=0.51\dfrac{0.95}{1.86} = 0.51, 0.361.86=0.193\dfrac{0.36}{1.86} = 0.193 mol kg−1^{-1}.
  • ΔTf=Tf∘−Tf\Delta T_f = T_f^\circ - T_f: water freezing at −0.95∘-0.95^\circC gives ΔTf=0.95\Delta T_f = 0.95 K.
  • Kf=ΔTfmK_f = \dfrac{\Delta T_f}{m}: 0.20.18=1.11\dfrac{0.2}{0.18} = 1.11; from masses, 2.52.5 g of a 117117 g mol−1^{-1} solute in 3535 g solvent is m=0.611m = 0.611, so Kf=30.611=4.91K_f = \dfrac{3}{0.611} = 4.91; 11 g of a 6060 g mol−1^{-1} solute in 100100 g with ΔTf=0.3\Delta T_f = 0.3 gives Kf=1.8K_f = 1.8.
  • ΔTf\Delta T_f from masses: 3.23.2 g of 128128 g mol−1^{-1} in 8080 g with Kf=4.8K_f = 4.8: m=0.3125m = 0.3125, ΔTf=1.5\Delta T_f = 1.5 K; 44 g of 126126 g mol−1^{-1} in 8080 mL water: m=0.397m = 0.397, ΔTf=0.74\Delta T_f = 0.74 K.
  • A 11 molal solution has ΔTf=Kf\Delta T_f = K_f numerically — 'which concentration makes ΔTf\Delta T_f and KfK_f equal' is 11 m.

Freezing point depression

ΔTf=Kf m,ΔTf=Tf∘−Tf(Kfwater=1.86 K kg mol−1)\Delta T_f = K_f\,m,\qquad \Delta T_f = T_f^\circ - T_f \qquad (K_f^{\text{water}} = 1.86\ \text{K kg mol}^{-1})

Worked example

An aqueous solution freezes at −1.24∘-1.24^\circC. Find its molality, and the mass of glucose (180180 g mol−1^{-1}) in 500500 g of water that would produce it.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q82Easy

Example 1 · Solutions and Colligative Properties · Depression of Freezing Point

Calculate the molality of solution of non-volatile solute having depression in freezing point 0.93 K and cryoscopic constant of solvent 1.861.86 K kg mol−1^{-1}.

Reading a freezing point as the depression

−0.95∘-0.95^\circC is the freezing POINT; the depression is +0.95+0.95 K. A stem in kelvin that says 'freezes at −0.7-0.7 K' is using the same convention — take the magnitude.
Drill 8 more on δtf = kf · m: molality, kf, or δtf from a freezing point

Concept 2 of 3: Molar Mass From ΔTf: M₂ = 1000 Kf W₂ / (ΔTf W₁)

The same rearrangement as for boiling: solute mass and KfK_f on top, depression and solvent mass below, with 10001000 for the gram-to-kilogram conversion.

Definition

  • 11 g in 100100 g, ΔTf=0.2\Delta T_f = 0.2, Kf=1.2K_f = 1.2: 1000×1.2×10.2×100=60\dfrac{1000 \times 1.2 \times 1}{0.2 \times 100} = 60. 1.51.5 g in 9090 g, 0.250.25 K: 8080. 1515 g in 200200 mL water, 0.750.75 K: 186186. 55 g in 5050 g water, 0.20.2 K: 930930.
  • Formula recognition: the correct option has Kf×W2K_f \times W_2 over ΔTf×W1\Delta T_f \times W_1.
  • Water's mass in grams equals its volume in millilitres (density 11), so 200200 mL means W1=200W_1 = 200 g.

Molar mass

M2=1000 Kf W2ΔTf W1M_2 = \frac{1000\,K_f\,W_2}{\Delta T_f\,W_1}

Worked example

2.52.5 g of a solute in 125125 g of water lowers the freezing point by 0.3720.372 K. Find the molar mass.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q66Easy

Example 2 · Solutions and Colligative Properties · Depression of Freezing Point

Calculate the molar mass of non-volatile solute when 1 g of it is dissolved in 100 g solvent decreases its freezing point by 0.2 K. Kf=1.2 K kg mol−1K_f = 1.2\text{ K kg mol}^{-1}

Choosing the formula with W₁ on top

On the formula-recognition stem the four options permute W1W_1, W2W_2, KfK_f and ΔTf\Delta T_f. Check with units: KfK_f (K kg mol−1^{-1}) ×\times g ×1000\times 1000, over K ×\times g, leaves g mol−1^{-1}.
Drill 4 more on molar mass from δtf: m₂ = 1000 kf w₂ / (δtf w₁)

Concept 3 of 3: Highest and Lowest Depression: Compare i × m

ΔTf=iKfm\Delta T_f = i K_f m, so at the same solvent the largest i×mi \times m freezes lowest. Count the ions per formula unit for ii and multiply by the molality.

Definition

  • Highest among 0.10.1 m NaCl (0.20.2), 0.050.05 m MgSO4_4 (0.10.1), 11 m AlPO4_4 (22), 0.050.05 m Al2_2(SO4_4)3_3 (0.250.25): AlPO4_4.
  • Lowest among 0.10.1 m NaCl (0.20.2), 0.050.05 m MgSO4_4 (0.10.1), 0.080.08 m AlPO4_4 (0.160.16), 0.060.06 m Al2_2(SO4_4)3_3 (0.30.3): MgSO4_4.
  • Both salts of a 2+2+/2−2- or 3+3+/3−3- pairing (MgSO4_4, AlPO4_4) give only two ions; the charge does not add particles.

Electrolyte depression

ΔTf=i Kf mrank by i×m\Delta T_f = i\,K_f\,m \qquad \text{rank by } i \times m

Worked example

Which freezes lowest: 0.10.1 m urea, 0.060.06 m KCl, 0.040.04 m CaCl2_2, 0.030.03 m AlCl3_3?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q79Moderate

Example 3 · Solutions and Colligative Properties · Depression of Freezing Point

Which of the following solutions exhibits highest freezing point depression?

Ranking by molality alone

11 m AlPO4_4 beats 0.050.05 m Al2_2(SO4_4)3_3 only after the ion count: 22 against 0.250.25. Without ii, the highest molality still wins here — but the LOWEST-depression stem is decided by ii.
Drill 1 more on highest and lowest depression: compare i × m

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