MHT-CET Chemistry · Solutions and Colligative Properties

Types of Solutions, Solubility and Henry's Law

A solution is named by the physical states of its solute and solvent; a solid's solubility follows the enthalpy of solution and Le Chatelier; a gas's solubility is Henry's law, S = K_H · P.

All Solutions and Colligative Properties notes MHT-CET Chemistry strategy

Why this matters

41 PYQs, one HARD — one of the chapter's largest pages and its most recall-heavy. Half the stems are Henry's law arithmetic (solubility from K_H and pressure, or K_H from solubility), and the rest name a solution type from an example, pick the salt whose solubility falls on heating (sodium sulphate, four sittings), state which concentration term depends on temperature, or add a lattice and a hydration enthalpy. Learn the solution-type table and the one salt; the arithmetic is a single multiplication.

Concept 1 of 4: Types of Solutions by the States of Solute and Solvent

The solvent is the component present in the larger amount and in the same physical state as the solution; the solute is what is dissolved in it. Nine pairings are possible, and the exam asks you to name the pairing for a given everyday example.

Definition

Solvent decides the state of the solution; solute is the minor component.
Gas in liquid: carbonated water ( $\text{CO}_2$ in water). Liquid in liquid: ethanol in water, gasoline (liquid hydrocarbons in each other). Solid in liquid: sea water (salt in water), sugar solution.
Solid in gas: iodine vapour in air, camphor in nitrogen. Liquid in gas: chloroform mixed with nitrogen, humidity. Gas in gas: air.
Solid in solid: alloys — brass (zinc in copper), bronze (tin in copper). Gas in solid: hydrogen in palladium. Liquid in solid: amalgam (mercury in a metal).

Solute	Solvent	Example
Gas	Liquid	Carbonated water ( $\text{CO}_2$ in water), oxygen in water The gas is the solute even though it is what the drink is named for.
Liquid	Liquid	Ethanol in water; gasoline (a liquid-in-liquid mixture of hydrocarbons)
Solid	Liquid	Sea water (salt in water), sugar in water
Solid	Gas	Iodine vapour in air; camphor in nitrogen Iodine in air is solid-in-GAS — air is the solvent, whatever the amount of iodine.
Liquid	Gas	Chloroform mixed with nitrogen; water vapour in air (humidity)
Gas	Gas	Air (oxygen in nitrogen)
Solid	Solid	Alloys — brass (zinc in copper), bronze (tin in copper) An alloy is a solid solution; bronze is NOT solid-in-liquid.
Gas	Solid	Hydrogen adsorbed in palladium
Liquid	Solid	Amalgam — mercury in sodium or in silver

Name the solute first, then the solvent: 'solid in gas' means a solid solute in a gaseous solvent.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q54Easy

Example 1 · Solutions and Colligative Properties · Types of Solutions, Solubility and Henry's Law

Identify an example of solution that consists of solid as solute and liquid as solvent.

Naming the solvent first

'Solid in gas' is a solid solute in a gas solvent. Iodine in air is solid-in-gas, not gas-in-solid; the reversed option is always offered.

Drill 5 more on types of solutions by the states of solute and solvent

Concept 2 of 4: Concentration Terms: Which Ones Change With Temperature

Molarity divides by the solution's VOLUME, and volume expands on heating — so molarity is the temperature-dependent term. Molality, mole fraction and mass per cent use masses and mole counts, which do not change with temperature.

Definition

Molarity $M = \dfrac{n_{\text{solute}}}{V_{\text{solution}}\,(\text{L})}$ — temperature dependent.
Molality $m = \dfrac{n_{\text{solute}}}{W_{\text{solvent}}\,(\text{kg})}$ , mole fraction $x = \dfrac{n}{n_{\text{total}}}$ , mass per cent $= \dfrac{W_{\text{solute}}}{W_{\text{solution}}} \times 100$ — all temperature independent.
Colligative-property formulas use molality (or mole fraction), which is why they hold at any temperature.
A hydrated salt in a mass-per-cent problem: the water of crystallisation counts as solvent. $50$ g of $12\%$ $\text{BaCl}_2$ needs $6$ g of $\text{BaCl}_2$ , i.e. $6 \times \dfrac{244}{208} = 7.04$ g of $\text{BaCl}_2\cdot2\text{H}_2\text{O}$ , and $50 - 7.04 \approx 42.9$ g of added water.

Concentration terms

M = \frac{n}{V(\text{L})} \ (\text{T-dependent}),\qquad m = \frac{n}{W_{\text{solvent}}(\text{kg})},\qquad x = \frac{n}{n_{\text{total}}}

Worked example

4

g of NaOH is dissolved in

96

g of water. Find the mass per cent and the molality (NaOH

= 40

g mol

^{-1}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 1 · Q85Easy

Example 2 · Solutions and Colligative Properties · Types of Solutions, Solubility and Henry's Law

Which of the following concentration terms depends on temperature?

Ignoring the water of crystallisation

\text{BaCl}_2\cdot2\text{H}_2\text{O}

brings its own water; the water to ADD is

50 - 7.04

, not

50 - 6

. Option (D)

50.0

g is the version that forgot the hydrate entirely.

Drill 1 more on concentration terms: which ones change with temperature

Concept 3 of 4: Solubility of Solids: Like Dissolves Like, ΔH of Solution, and the Salt That Dissolves Less on Heating

A solid dissolves when solute–solvent attractions are comparable to the solute–solute and solvent–solvent ones. The enthalpy of solution is the lattice enthalpy (to be paid) plus the hydration enthalpy (recovered); if dissolution is exothermic, heating REDUCES solubility — sodium sulphate is the exam's standard example.

Definition

Like dissolves like: a polar solute dissolves in a polar solvent because solute–solute, solute–solvent and solvent–solvent interactions are of similar magnitude.
$\Delta H_{\text{sol}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hyd}}$ : KCl $699 + (-681.8) = 17.2$ kJ mol $^{-1}$ (endothermic); NaCl with $\Delta H_{\text{sol}} = 4$ and lattice $790$ has $\Delta H_{\text{hyd}} = -786$ kJ mol $^{-1}$ .
Most salts (NaCl, KNO $_3$ , NaNO $_3$ , KBr, NaBr) dissolve endothermically and become MORE soluble on heating. $\text{Na}_2\text{SO}_4$ is the exception — its solubility falls above about $32^\circ$ C, when the decahydrate gives way to the anhydrous salt whose dissolution is exothermic.
Le Chatelier: exothermic dissolution + heating → equilibrium shifts back to the solid.

Enthalpy of solution

\Delta H_{\text{sol}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hyd}}

Worked example

The lattice enthalpy of a salt is

760

kJ mol

^{-1}

and its hydration enthalpy is

-740

kJ mol

^{-1}

. Is its dissolution endothermic, and does its solubility rise on heating?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 12th May Shift 1 · Q53Easy

Example 3 · Solutions and Colligative Properties · Types of Solutions, Solubility and Henry's Law

If lattice enthalpy and hydration enthalpy of KCl are

699\,\text{kJ mol}^{-1}

and

-681.8\,\text{kJ mol}^{-1}

respectively. What is the enthalpy of solution of KCl?

Subtracting the hydration enthalpy

\Delta H_{\text{hyd}}

is already negative; ADD it.

699 - (-681.8)

gives

1380

, not on the list, but a sign slip in the NaCl stem gives

+786

, which is option (A).

Drill 6 more on solubility of solids: like dissolves like, δh of solution, and the salt that dissolves less on heating

Concept 4 of 4: Henry's Law: S = K_H · P, and Partial Pressures From Mole Fractions

The amount of gas that dissolves is proportional to its partial pressure above the liquid:

S = K_H\,P

in the CET form (with

K_H

in mol dm

^{-3}

atm

^{-1}

). Partial pressure itself comes from Dalton's law,

P_i = x_i P_{\text{total}}

Definition

$S = K_H\,P$ : $K_H = 6.85 \times 10^{-4}$ mol dm $^{-3}$ atm $^{-1}$ at $0.8$ atm gives $5.48 \times 10^{-4}$ mol dm $^{-3}$ ; $K_H = 0.16$ , $P = 0.15$ bar gives $2.4 \times 10^{-2}$ .
$K_H = \dfrac{S}{P}$ : $\dfrac{5.14 \times 10^{-4}}{0.75} = 6.85 \times 10^{-4}$ ; $\dfrac{0.028}{0.346} = 0.081$ .
Dalton: $x_i = \dfrac{P_i}{P_{\text{total}}}$ . Partial pressures $4.5$ and $5.5$ bar give mole fractions $0.45$ , $0.55$ ; $64$ g O $_2$ ( $2$ mol) with $160$ g Ne ( $8$ mol) at $25$ bar gives $P_{\text{O}_2} = 0.2 \times 25 = 5$ bar.
Gases that REACT with water ( $\text{CO}_2$ , $\text{NH}_3$ , HCl) do not follow Henry's law strictly; $\text{O}_2$ has very low physical solubility. Solubility of a gas falls as temperature rises.
Statement form: 'solubility of a gas in a liquid is directly proportional to the pressure of the gas over the solution' — Henry's law, not Raoult's or Dalton's.

Henry and Dalton

S = K_H\,P \qquad K_H = \frac{S}{P} \qquad P_i = x_i\,P_{\text{total}}

Worked example

The Henry's law constant of a gas is

0.12

mol dm

^{-3}

bar

^{-1}

. Find its solubility when its partial pressure is

0.25

bar, and the pressure needed for a solubility of

0.06

mol dm

^{-3}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2022 · Shift 1 · Q55Easy

Example 4 · Solutions and Colligative Properties · Types of Solutions, Solubility and Henry's Law

Calculate the solubility of a gas in water at 0.8 atm and 25°C. [Henry's law constant is

6.85 \times 10^{-4}

mol dm

^{-3}

atm

^{-1}

]

Dividing by the pressure

In the CET form

S = K_H P

; solubility is

K_H

TIMES

P

. Dividing gives

8.56 \times 10^{-4}

, and the un-multiplied

K_H

itself is offered as option (C).

Drill 11 more on henry's law: s = k_h · p, and partial pressures from mole fractions

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Concentration Terms: Which Ones Change With Temperature
Concentration terms
$M = \frac{n}{V(\text{L})} \ (\text{T-dependent}),\qquad m = \frac{n}{W_{\text{solvent}}(\text{kg})},\qquad x = \frac{n}{n_{\text{total}}}$
Solubility of Solids: Like Dissolves Like, ΔH of Solution, and the Salt That Dissolves Less on Heating
Enthalpy of solution
$\Delta H_{\text{sol}} = \Delta H_{\text{lattice}} + \Delta H_{\text{hyd}}$
Henry's Law: S = K_H · P, and Partial Pressures From Mole Fractions
Henry and Dalton
$S = K_H\,P \qquad K_H = \frac{S}{P} \qquad P_i = x_i\,P_{\text{total}}$

Reference tables (1)

Types of Solutions by the States of Solute and Solvent9 rows

Solute	Solvent	Example
Gas	Liquid	Carbonated water ( $\text{CO}_2$ in water), oxygen in water The gas is the solute even though it is what the drink is named for.
Liquid	Liquid	Ethanol in water; gasoline (a liquid-in-liquid mixture of hydrocarbons)
Solid	Liquid	Sea water (salt in water), sugar in water
Solid	Gas	Iodine vapour in air; camphor in nitrogen Iodine in air is solid-in-GAS — air is the solvent, whatever the amount of iodine.
Liquid	Gas	Chloroform mixed with nitrogen; water vapour in air (humidity)
Gas	Gas	Air (oxygen in nitrogen)
Solid	Solid	Alloys — brass (zinc in copper), bronze (tin in copper) An alloy is a solid solution; bronze is NOT solid-in-liquid.
Gas	Solid	Hydrogen adsorbed in palladium
Liquid	Solid	Amalgam — mercury in sodium or in silver

Name the solute first, then the solvent: 'solid in gas' means a solid solute in a gaseous solvent.

Watch out for (4)

Naming the solvent first
→ Types of Solutions by the States of Solute and Solvent
Ignoring the water of crystallisation
→ Concentration Terms: Which Ones Change With Temperature
Subtracting the hydration enthalpy
→ Solubility of Solids: Like Dissolves Like, ΔH of Solution, and the Salt That Dissolves Less on Heating
Dividing by the pressure
→ Henry's Law: S = K_H · P, and Partial Pressures From Mole Fractions

Test yourself on Solutions and Colligative Properties

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