Q.Calculate the depression in the freezing point of water when 20 g of CH3CH2CHClCOOH is added to 500 g of water.
Ka = 1.4 x 10^-3, Kf = 1.86 K kg mol-1
(Atomic mass C = 12u, H = 1u, O = 16u, Cl = 35.5u)

Ethylene glycol is added as an antifreeze; it lowers the freezing point of water so the coolant doesn't freeze in cold winters.

Adding a non-volatile solute like ethylene glycol to water is a classic application of the colligative property of depression in freezing point, $\Delta T_f = K_f \times m$. This lowers the freezing point of the radiator coolant well below $0°C$, preventing it from freezing and expanding (which could crack the engine block) in cold climates. The s …

Freezing-point depression is a colligative property, proportional to i x m. MgCl2 has the highest van't Hoff factor (i = 3), so lowest freezing point. Correct option (c).

Depression in freezing point: dTf = i x Kf x m. At equal molality (1 m) and same Kf, the solution giving the most dissolved particles depresses the freezing point most.

• Sucrose (i = 1) -> 1 particle
• Glucose (i = 1) -> 1 particle
• KCl -> K+ + Cl- (i = 2) -> 2 particles …

Kf (cryoscopic constant) links depression in freezing point directly to molality.

The molal depression constant (freezing point depression constant, or cryoscopic constant, $K_f$) is defined as the depression in freezing point produced when 1 mole of a solute is dissolved in 1 kg (1000 g) of a solvent.

It relates the observed depression in freezing point ($\Delta T_f$) to molality ($m$) via:

$\Delta T_f = K_f \times m$

…

For equimolar solutions, freezing-point depression ΔTf = i·Kf·m increases with the van't Hoff factor i (number of particles produced per formula unit); K2SO4 gives i≈3, the highest of the four, so it has the lowest (most depressed) freezing point.

ΔTf = i × Kf × m, where m is molality (same for all four since they are equimolar) and Kf is the same solvent constant. So the compound with the largest i shows the largest ΔTf, i.e. the lowest freezing point.

• K2SO4 → 2K+ + SO4^2-: i ≈ 3
• NaCl → Na+ + Cl-: i ≈ 2
• Glucose: a non-electrolyte, does not dissociate: i = 1 …

The freezing point depression constant Kf has SI unit K kg mol^-1, obtained from the defining relation ΔTf = Kf x m.

The depression in freezing point is related to molality by:

ΔTf = Kf x m

where ΔTf is in kelvin (K) and molality m is in mol/kg (mol kg^-1). Rearranging:

Kf = ΔTf / m

Units: Kf = K / (mol kg^-1) = K kg mol^-1

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Ethylene glycol does lower water's freezing point (colligative property), but the reason is wrong: ethylene glycol is highly soluble in water because it forms hydrogen bonds. Assertion true, Reason false → (C).

Colligative properties depend only on the number of solute particles. Judging the options exactly as printed, (c) 'Decrease in freezing point' (depression of freezing point) is the genuine colligative property; option (a) as printed reads 'fluid pressure', which is not a standard colligative term.

Checking each option as printed:

• (a) Relative lowering of fluid pressure - the standard colligative property is relative lowering of vapour pressure. As actually printed (both the English and Hindi versions use 'fluid pressure' / 'taral daab') this is not a recognised colligative property, so it cannot be taken as the intended correct term.
• (b) Decrease in boiling point - wrong direction: a non-volatile solute raises (elevates) the boiling point, it does not decrease it. …

Adding NaCl lowers water's vapour pressure, and this lowering of vapour pressure is exactly what causes the solution's freezing point to fall below that of pure water.

[A] Adding a non-volatile solute like NaCl to water does depress its freezing point — this is a well-established colligative property, and is TRUE.

…

Depression in freezing point (a colligative property) increases with the number of particles: van't Hoff factor i. K2SO4 (i = 3) gives the most.

Depression in freezing point: delta Tf = i x Kf x m.

For equal molality, the electrolyte producing the most particles wins:

• K2SO4 -> 2K+ + SO4^2- , i = 3
• NaCl -> Na+ + Cl- , i = 2
• Urea, non-electrolyte, i = 1 …

Freezing point depression depends on the number of particles a solute produces in solution (colligative property). NaCl and KCl both dissociate into 2 ions per formula unit, so equimolal solutions of the two give the same depression.

Freezing point depression is given by ΔTf = i × Kf × m, where i is the van't Hoff factor (number of particles produced per formula unit), Kf is the molal freezing point depression constant of water, and m is the molality.

NaCl → Na+ + Cl- gives i = 2.

KCl → K+ + Cl- gives i = 2.

…

Depression in freezing point is a colligative property proportional to the van't Hoff factor i; MgCl2 gives i = 3 versus i = 1 for glucose.

Depression in freezing point is given by delta Tf = i x Kf x m, where i is the van't Hoff factor (number of particles produced per formula unit) and m is the molality (here comparable at equal molarity for dilute aqueous solutions).

• Glucose is a non-electrolyte; it does not dissociate, so i = 1. …

Depression in freezing point, $\Delta T_f = i K_f m$, so at the same molal concentration the electrolyte that dissociates into the most ions (highest $i$) gives the largest depression.

For a fixed concentration (0.1 M) and the same solvent (same $K_f$), $\Delta T_f \propto i$, the van't Hoff factor, which equals the number of particles (ions) each formula unit dissociates into (assuming complete dissociation):

• Glucose: non-electrolyte, $i = 1$
• NaCl $\to$ Na$^+$ + Cl$^-$: $i = 2$
• Na$_2$SO$_4$ $\to$ 2Na$^+$ + SO$_4^{2-}$: $i = 3$
• Na$_3$PO$_4$ $\to$ 3Na$^+$ + PO$_4^{3-}$: $i = 4$ …