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NCERT Exemplar · Q49

Q.Match the laws given in Column I with expressions given in Column II.
Column I:

(i) Raoult's law;
(ii) Henry's law;
(iii) Elevation of boiling point;
(iv) Depression in freezing point;
(v) Osmotic pressure.
Column II:
(a) ΔTf=Kfm\Delta T_f = K_f m.
(b) Π=CRT\Pi = CRT.
(c) p=x1p1o+x2p2op = x_1 p_1^{o} + x_2 p_2^{o}.
(d) ΔTb=Kbm\Delta T_b = K_b m.
(e) p=KH⋅xp = K_H \cdot x.
Himachal HpboseShort· 2mImportance★★★★★
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This is a matching problem linking five solution-phase laws (Raoult, Henry, boiling-point elevation, freezing-point depression, osmotic pressure) to their standard mathematical expressions. The correct matches are: (i)→(c), (ii)→(e), (iii)→(d), (iv)→(a), (v)→(b).

The key to solving this cleanly is to recall what each law says — not just the formula, but the physical idea behind it. Once you know the concept, the expression follows naturally.


1. Raoult’s law describes the vapour pressure of an ideal liquid-liquid solution. It says: the partial pressure of each component above the solution equals its mole fraction times its pure vapour pressure. For a two-component mixture, the total pressure is

p=x1p1∘+x2p2∘p = x_1 p_1^\circ + x_2 p_2^\circ.

That matches option (c).

Tip

Raoult’s law is for volatile solutes. If the solute is non-volatile, the formula reduces to p=xsolventpsolvent∘p = x_{\text{solvent}} p^\circ_{\text{solvent}}, but the general form given here is the one listed.

2. Henry’s law governs the solubility of a gas in a liquid. It states: at constant temperature, the partial pressure of the gas above the liquid is directly proportional to its mole fraction in the liquid.

p=KH⋅xp = K_H \cdot x

Here KHK_H is Henry’s constant. That’s option (e).

Watch out

A common confusion: Henry’s law looks similar to Raoult’s law (p=p∘xp = p^\circ x), but the constant is different — KHK_H is not the pure vapour pressure of the solute. Henry’s law applies to dilute solutions of gases, not to ideal liquid mixtures.

3. Elevation of boiling point — when a non-volatile solute is added to a solvent, the boiling point rises. The change is proportional to the molality of the solution:

ΔTb=Kbm\Delta T_b = K_b m

where KbK_b is the ebullioscopic constant. That’s option (d).

4. Depression in freezing point — similarly, adding a solute lowers the freezing point. The depression is:

ΔTf=Kfm\Delta T_f = K_f m

where KfK_f is the cryoscopic constant. That’s option (a). …

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