Q.Two elements A and B form compounds having formula AB2 and AB4. When dissolved in 20g of benzene (C6H6), 1g of AB2 lowers the freezing point by 2.3K whereas 1g of AB4 lowers it by 1.3K. The molar depression constant for benzene is 5.1 K kg mol^-1. Calculate atomic masses of A and B.

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:

$\Delta T_f = K_f \times m$

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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.

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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.

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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$_4^{3-}$: $i = 4$ …