(a) Shweta mixed two liquids A and B of 10 mL each. After mixing, the volume of the solution was found to be 2

Question

Q. (a) Shweta mixed two liquids A and B of 10 mL each. After mixing, the volume of the solution was found to be 20\cdot2 mL.

(i) Why was there a volume change after mixing the liquids ?

(ii) Will there be an increase or decrease of temperature after mixing ?

(iii) Give one example for this type of solution.

(OR)

(b)

(i) How does sprinkling of salt help in clearing the snow covered roads in hilly areas ?

(ii) What happens when red blood cells are kept in 0·5% (mass/vol) NaCl solution ? Justify your answer.

(iii) Write an application of reverse osmosis.

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Part (a)Concept understanding — Ideal and Non-Ideal Solutions

Ideal and Non-Ideal Solutions

Imagine you have two friends, A and B. When they work together, sometimes they get along perfectly — each does exactly their share, no extra effort, no friction. Other times, they either pull apart (making the job harder) or cling together (making it easier than expected). A solution of two liquids behaves the same way.

The Intuition: What "Ideal" Really Means

An ideal solution is the "perfect teamwork" case. The molecules of A and B are so similar that they don't care whether they are next to an A or a B. They interact with each other exactly as they would with their own kind. Think of mixing two grades of petrol — the molecules are nearly identical, so the mixture behaves predictably.

In an ideal solution, two things happen:

No heat is absorbed or released when you mix them (zero enthalpy change, $\Delta H_{\text{mix}} = 0$ ).
No volume change occurs — the total volume is exactly the sum of the individual volumes ( $\Delta V_{\text{mix}} = 0$ ).

Why? Because the forces between A-A, B-B, and A-B are all the same. No energy is needed to break old contacts or form new ones; no space is saved or wasted.

The Precise Statement: Raoult's Law

Raoult's law is the mathematical definition of an ideal solution. For a mixture of two volatile liquids A and B, the partial vapour pressure of each component above the solution is proportional to its mole fraction in the liquid:

p_A = x_A \, p_A^0 \quad \text{and} \quad p_B = x_B \, p_B^0

where $p_A^0$ and $p_B^0$ are the vapour pressures of pure A and pure B at that temperature. The total vapour pressure is simply the sum:

P_{\text{total}} = p_A + p_B = x_A p_A^0 + x_B p_B^0

P_{\text{total}} = x_A p_A^0 + x_B p_B^0

This is a straight line when plotted against mole fraction. Every ideal solution obeys this law at all compositions and temperatures.

Non-Ideal Solutions: When Molecules Disagree

Real solutions are rarely ideal. The molecules of A and B are different — different sizes, polarities, or hydrogen-bonding abilities. Now the A-B interaction is not the same as A-A or B-B. This causes deviations from Raoult's law.

Positive deviation occurs when A-B interactions are weaker than A-A and B-B. The molecules "prefer their own company" and escape into the vapour more easily than expected. The actual vapour pressure is higher than Raoult's law predicts. Examples: ethanol + water (initially), acetone + carbon disulphide. The mixture absorbs heat ( $\Delta H_{\text{mix}} > 0$ ) and expands slightly ( $\Delta V_{\text{mix}} > 0$ ).

Negative deviation occurs when A-B interactions are stronger than A-A and B-B. The molecules "hold on to each other" and escape less easily. The actual vapour pressure is lower than Raoult's law predicts. Examples: chloroform + acetone, nitric acid + water. The mixture releases heat ( $\Delta H_{\text{mix}} < 0$ ) and contracts ( $\Delta V_{\text{mix}} < 0$ ).

Watch out

A common mistake: thinking "positive deviation means the solution is better" or "negative means worse." The words refer only to the direction of the pressure deviation from Raoult's law — nothing about quality.

The Key Table

Property	Ideal Solution	Positive Deviation	Negative Deviation
A-B interaction	= A-A, B-B	< A-A, B-B	> A-A, B-B
Vapour pressure	Follows Raoult's law	Higher than predicted	Lower than predicted
$\Delta H_{\text{mix}}$	0	> 0 (endothermic)	< 0 (exothermic)

Part (b)Concept understanding — Osmosis

Osmotic Pressure and Molar Mass: From Intuition to Formula

Imagine you have a glass of pure water, and you carefully place a tea bag into it. After a while, the water turns brown. The tea molecules have moved from the bag into the water. That's simple diffusion. But now imagine a different setup: you have a U-shaped tube with a special membrane at the bottom that only lets water molecules pass through — not larger molecules like sugar. On one side you put pure water, on the other side you put a sugar solution. What happens?

Water will spontaneously move from the pure water side into the sugar solution side, pushing the liquid level higher on the sugar side. That rising column of liquid is a direct physical effect — it's osmotic pressure trying to equalise concentrations. The taller the column gets, the more hydrostatic pressure it exerts back. Eventually, that back-pressure exactly balances the "pull" of the sugar, and the system stops.

That balancing pressure — the pressure you would need to apply to the solution side to prevent the water from moving — is the osmotic pressure ( $\Pi$ ).

The Intuition Behind Molar Mass from Osmotic Pressure

Here's the key insight: the osmotic pressure depends only on the number of solute particles in a given volume of solution, not on what those particles are. A big protein molecule and a tiny sugar molecule, if present in the same number per litre, produce the same osmotic pressure.

This is incredibly useful. If you dissolve an unknown substance (say, a polymer or a protein) in water and measure the osmotic pressure, you can work backwards to find how many moles of it are present. And if you know the mass you dissolved, you can calculate the molar mass:

\text{Molar mass} = \frac{\text{mass of solute (g)}}{\text{number of moles}}

So osmotic pressure becomes a direct window into the molecular weight of substances that are too large or too fragile to vaporise (like proteins, polymers, or enzymes).

The Precise Statement

For dilute solutions, osmotic pressure follows a law that looks exactly like the ideal gas law:

\Pi V = nRT

This is the van't Hoff equation for osmotic pressure. It tells you that osmotic pressure is directly proportional to the molar concentration of the solute:

\Pi = \frac{n}{V}RT = cRT

where $c$ is the molar concentration (mol/L).

From Osmotic Pressure to Molar Mass

If you dissolve a known mass $w$ (in grams) of an unknown substance in a volume $V$ of solvent, and measure the osmotic pressure $\Pi$ at temperature $T$ , you can find the molar mass $M$ as follows:

From $\Pi V = nRT$ , we get $n = \frac{\Pi V}{RT}$
But $n = \frac{w}{M}$ (mass divided by molar mass)
Equating: $\frac{w}{M} = \frac{\Pi V}{RT}$
Rearranging:

M = \frac{wRT}{\Pi V}

This is the working formula. Every quantity on the right is measurable in the lab.

Why This Method is Special

Osmotic pressure measurements are extraordinarily sensitive. For a substance with a very large molar mass (say, 100,000 g/mol), the freezing point depression or boiling point elevation would be too tiny to measure accurately. But osmotic pressure can still give a measurable reading because it's a colligative property that depends only on particle count, and the effect is large even at low concentrations.

Important

Osmotic pressure is the most sensitive colligative property for determining molar masses of macromolecules. It can detect concentrations as low as $10^{-4}$ M, which is 100–1000 times more sensitive than freezing point depression.

A Worked Example

Problem: 0.50 g of a protein is dissolved in enough water to make 100 mL of solution at 25°C. The osmotic pressure is measured as 0.012 atm. Find the molar mass of the protein.

Solution:

Given: …

Why this formula?

Osmosis: Why the Key Formulas Hold

Osmosis is the net movement of solvent molecules (usually water) across a semipermeable membrane from a region of lower solute concentration to higher solute concentration. The membrane allows solvent to pass but blocks solute.

Let’s build the reasoning step-by-step — from the physical picture to the formulas.

1. The Physical Picture: Why Does Water Move?

Imagine a U-shaped tube divided by a semipermeable membrane. Left side: pure water. Right side: water + dissolved sugar.

Water molecules on both sides are in constant random motion.
On the pure water side, every molecule hitting the membrane can pass through (if it fits the pore).
On the sugar side, sugar molecules block some water molecules from reaching the membrane — effectively reducing the number of water molecules that can cross per second.

Result: More water molecules cross from pure side to sugar side than the reverse. This net flow continues until equilibrium is reached.

Key insight: The driving force is the difference in chemical potential of water across the membrane — not a "desire" to dilute the sugar.

2. Chemical Potential: The Real Driver

For an ideal dilute solution, the chemical potential of water ( $\mu_w$ ) is:

\mu_w = \mu_w^0 + RT \ln x_w

Where:

$\mu_w^0$ = chemical potential of pure water
$x_w$ = mole fraction of water
$R$ = gas constant
$T$ = absolute temperature

Since $x_w < 1$ in a solution, $\ln x_w < 0$ , so $\mu_w$ is lower in the solution than in pure water.

Water flows spontaneously from higher $\mu_w$ (pure side) to lower $\mu_w$ (solution side) — this is the fundamental thermodynamic reason.

3. Osmotic Pressure: The Formula $\Pi = iCRT$

What is osmotic pressure ( $\Pi$ )?

It is the external pressure that must be applied to the solution side to prevent net water flow into it — i.e., to make the chemical potential of water equal on both sides.

Derivation sketch:

At equilibrium under applied pressure $\Pi$ :

\mu_w^{\text{pure}}(P) = \mu_w^{\text{solution}}(P + \Pi)

For the solution side, we have two contributions:

Dilution effect (lower mole fraction): $RT \ln x_w$
Pressure effect: $V_w \cdot \Pi$ (where $V_w$ = partial molar volume of water)

So:

\mu_w^0(P) = \mu_w^0(P) + RT \ln x_w + V_w \Pi

Cancel $\mu_w^0(P)$ from both sides:

0 = RT \ln x_w + V_w \Pi

\Pi = -\frac{RT}{V_w} \ln x_w

For dilute solutions:

$x_w \approx 1 - x_{\text{solute}}$ (where $x_{\text{solute}}$ is small)
$\ln(1 - x_{\text{solute}}) \approx -x_{\text{solute}}$ (using Taylor expansion)

Thus:

\Pi \approx \frac{RT}{V_w} \cdot x_{\text{solute}}

Now, $x_{\text{solute}} = \frac{n_{\text{solute}}}{n_{\text{solute}} + n_{\text{water}}} \approx \frac{n_{\text{solute}}}{n_{\text{water}}}$ for dilute solutions.

And $n_{\text{water}} \cdot V_w \approx V$ (total volume of solution).

So:

\Pi \approx \frac{RT}{V} \cdot n_{\text{solute}} = CRT

Where $C = \frac{n_{\text{solute}}}{V}$ = molar concentration. …

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Part (a): Volume rises to 20.2 mL — positive deviation; weaker A–B forces expand volume, mixing is endothermic so temperature falls; e.g. ethanol + water.

Part (b): Salt depresses water's freezing point (melts snow); RBCs in 0.5% NaCl (hypotonic) swell/burst; reverse osmosis is used for desalination.

Part (a)

Mixing 10 mL + 10 mL gives 20.2 mL, so $\Delta V_{mix}>0$ . This marks a non-ideal solution with positive deviation from Raoult's law.

(i) In such solutions the new A–B intermolecular attractions are weaker than the original A–A and B–B attractions. The loosely held molecules occupy slightly more space, so the volume increases.

(ii) Breaking the stronger A–A / B–B interactions costs more energy than is released by forming the weaker A–B interactions, so mixing is endothermic ( $\Delta H_{mix}>0$ ). Heat is absorbed from the surroundings, so the temperature decreases. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 25 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. Example of an ideal solution is$
(a) Mixture of n-hexane and n-heptane
(b) Mixture of acetone and chloroform
(c) Mixture of acetone and ethanol
(d) Mixture of phenol and aniline
›Reveal solutionSolution
An ideal solution obeys Raoult's law over the entire range of concentration, with ΔHmix = 0 and ΔVmix = 0; this happens when the two components are chemically very similar.

For a solution to be ideal, the A-A, B-B and A-B intermolecular interactions must be nearly identical, so mixing causes no enthalpy or volume change.

n-Hexane and n-heptane are both non-polar straight-chain alkanes of similar size, shape and polarity, so their mutual interactions closely match their pure-component interactions - this pair behaves ideally.

Acetone + chloroform show strong negative deviation (H-bonding between the two). …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Assertion [A]: Nitric acid and water form maximum boiling Azeotrope. Reason [R]: Azeotropes are binary mixture showing the same composition in liquid and vapour phase.$
(a) Both [A] and [R] are true and [R] is the correct explanation of [A].
(b) Both [A] and [R] are true, but [R] is not the correct explanation of [A].
(c) [A] is true, but [R] is false.
(d) [A] is false, but [R] is true.
›Reveal solutionSolution

Nitric acid–water is a genuine example of a maximum boiling azeotrope, and the reason given (the defining property of azeotropes) correctly explains why such a mixture cannot be separated by simple fractional distillation.

Assertion: $HNO_3$ – $H_2O$ shows large negative deviation from Raoult's law (due to strong H-bonding/attraction between unlike molecules), and such systems form a maximum boiling azeotrope at a specific composition (≈68% $HNO_3$ by mass, boiling at ~120.5°C, higher than either pure component's boiling point). This is TRUE.

…

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CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank- The solutions which obey ......... law over the entire range of concentration are known as ideal solutions.$
›Reveal solutionSolution

The solutions that obey Raoult's law over the entire concentration range are called ideal solutions.

Concept. For an ideal solution, the partial vapour pressure of each component equals its mole fraction times its pure-component vapour pressure, at all compositions:

$p_A = x_A\,p_A^{0}, \qquad p_B = x_B\,p_B^{0}$

…

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CBSE 2026Set ANNUAL1 mark
$Q. When water and ethanol are mixed together the volume of the solution is not equal to the sum of the volumes of the two liquid components. Give reason.$
›Reveal solutionSolution

Ethanol + water is a non-ideal solution; the A–B (ethanol–water) interactions differ in strength from the A–A and B–B interactions of the pure liquids, so mixing changes the packing and the volume is not additive.

In an ideal solution the interactions between the two kinds of molecules are identical to those in the pure components, so volumes add up ( $V_{mix}=V_1+V_2$ ) and Raoult's law is obeyed exactly.

…

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CBSE 2025Set ANNUAL1 mark
$Q. Explain the following — Ideal solution$
›Reveal solutionSolution

Ideal solutions have identical A–A, B–B and A–B interactions.

An ideal solution is one that obeys Raoult's law over the entire range of concentration at all temperatures, i.e. $p_A = x_A p_A^0$ and $p_B = x_B p_B^0$ for every composition. In such a solution, the intermolecular forces between solute-solute (A-A), solvent-solvent (B-B), and solute-solvent (A-B) molecules are of similar magnitude, so there is no enthalpy change ( $\Delta H_{mix} = 0$ ) and no volume cha …

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CBSE 2025Set D1 markMCQ
$Q. The solution which shows positive or negative deviation from Raoult's law is called$
(a) Ideal solution
(b) True solution
(c) Non-ideal solution
(d) Colloidal solution
›Reveal solutionSolution

Deviation from Raoult's law (positive or negative) defines a non-ideal solution.

An ideal solution obeys Raoult's law over the entire range of concentration; its enthalpy of mixing and volume of mixing are zero, and A-B interactions equal A-A and B-B interactions.

When solute-solvent (A-B) interactions differ from A-A and B-B interactions, the solution deviates from Raoult's law: …

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CBSE 2025Set A1 mark
$Q. Write True or False: The solutions which obey Raoult's law over the entire range of concentrations are known as ideal solutions.$
›Reveal solutionSolution

This is the exact definition of an ideal solution, so the statement is True.

Raoult's law states that the partial vapour pressure of each volatile component in a solution is proportional to its mole fraction: $p_i = x_i p_i^{\circ}$ . A solution is called an ideal solution precisely when it obeys Raoult's law at every composition, from pure solvent to pure solute, over the entire concentration range, at all temperatures. Ideal solutions also have zero enthalpy of mixing ( $\Delta H_{mix}=0$ ) …

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CBSE 2025Set ANNUAL1 mark
$Q. Under what condition do non-ideal solutions show positive deviation from Raoult's law?$
›Reveal solutionSolution
Positive deviation occurs when mixing weakens the average intermolecular forces compared to the pure components, letting molecules escape into the vapour phase more easily than Raoult's law predicts.

Condition for positive deviation

In an ideal solution, A–B interactions are comparable in strength to A–A and B–B interactions. A solution shows positive deviation from Raoult's law when the solute–solvent (A–B) interactions are weaker than the solute–solute (A–A) and solvent–solvent (B–B) interactions present in the pure liquids.

When this happens, the molecules in the mixture are held less tightly than in the pure liquids, so they escape into the vapour phase more readily than expected. As a result:

the observed vapour pressure of the solution is higher than the Raoult's-law-predicted value,

mixing is typically accompanied by $\Delta_{mix}H > 0$ (endothermic) and $\Delta_{mix}V > 0$ (volume expansion). …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following pairs will not form an ideal solution?$
(a) Benzene and Toluene
(b) Ethanol and Acetone
(c) n-hexane and n-heptane
(d) Bromoethane and Chloroethane
›Reveal solutionSolution

An ideal pair needs closely similar molecular size/polarity/intermolecular forces; ethanol's hydrogen bonding is disrupted when mixed with acetone, breaking ideality.

An ideal solution requires A–A, B–B and A–B intermolecular forces to be essentially equal (Raoult's law holds exactly). Benzene–toluene, n-hexane–n-heptane, and bromoethane–chloroethane are each pairs of structurally very similar molecules (comparable size, shape and polarity), so they form essentially ideal solutions. Ethanol, however, is extensively hydrogen-bonded to itself in the pure liquid; mixing it wit …

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CBSE 2024Set D1 markMCQ
$Q. An azeotropic mixture of HCl and H2O has$
(a) 48% HCl
(b) 36% HCl
(c) 22.2% HCl
(d) 20.2% HCl
›Reveal solutionSolution

Hydrochloric acid forms a maximum-boiling azeotrope with water at about 20.2% HCl by mass (b.p. ~110 deg C), so it cannot be concentrated further by distillation.

HCl-water is a negative-deviation system that forms a maximum-boiling azeotrope. This constant-boiling mixture contains about 20.2% HCl by mass and boils around 110 deg C. Because it …

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CBSE 2024Set D1 markMCQ
$Q. Which of the following show positive deviation from Raoult's law?$
(a) C6H6 and C6H5CH3
(b) C6H6 and CCl4
(c) CHCl3 and C2H5OH
(d) CHCl3 and CH3COCH3
›Reveal solutionSolution
Chloroform + ethanol shows positive deviation from Raoult's law.

Positive deviation occurs when A–B interactions are weaker than A–A and B–B interactions, so the observed vapour pressure is higher than predicted.

In ethanol, molecules are strongly hydrogen-bonded. When chloroform is added it disrupts the ethanol H-bond network without forming equally strong new interactions, so escaping tendency increases → positive deviation. …
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CBSE 2024Set ANNUAL1 mark
$Q. What type of deviation from Raoult's law is observed when chloroform and acetone are mixed?$
›Reveal solutionSolution
Chloroform and acetone form a hydrogen bond with each other that is stronger than their own like-like interactions, so the mixture shows negative deviation from Raoult's law.

Deviation from Raoult's law depends on how the A–B (unlike-molecule) intermolecular forces compare to the A–A and B–B (like-molecule) forces:

If A–B interactions are weaker than A–A/B–B, molecules escape more easily than expected → positive deviation (higher vapour pressure than ideal).

If A–B interactions are stronger than A–A/B–B, molecules are held back more than expected → negative deviation (lower vapour pressure than ideal).

In a chloroform–acetone mixture, the hydrogen atom of chloroform ( $CHCl_3$ , made acidic by the three electronegative Cl atoms) forms a hydrogen bond with the lone pair on acetone's carbonyl oxygen: $Cl_3C-H\cdots O=C(CH_3)_2$ . This new intermolecular hydrogen bond is stronger than the (comparatively weak, dipole-dipole-only) interactions each pure liquid has with itself.

…
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Q.(a) Shweta mixed two liquids A and B of 10 mL each. After mixing, the volume of the solution was found to be 20·2 mL.

Part (a)

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