NEET ChemistryUpdated 19 Oct 2026

Solutions: NEET PYQs and weightage

Solutions carries 5.7 of 100 Chemistry marks in NEET by past-paper share: 40 of the 707 tagged NEET Chemistry PYQs in the ViJEEta bank (NEET 2013–2026). It ranks 4 of 26 Chemistry chapters.

Take the 25-question Solutions test Formula sheet ↓All NEET Chemistry chapters

5.7/100 Weight in Chemistry

Exam importance (4/5, by past-paper share)

40 NEET PYQs

461 Practice questions

437 With measured difficulty

How the weight is computed: this chapter's tagged NEET PYQs divided by all tagged NEET Chemistry PYQs in the bank (40 of 707), times 100, with repeated questions counted once. The bank does not store the year of each NEET question, so none is printed. Computed from the bank at build time. Updated 19 Oct 2026.

Solutions: formula sheet for NEET

14 formulas, each with when to use it. Copy it onto one page of your notebook: that page is this chapter's handbook.

Depression of freezing point
$Δ T_{f} = i K_{f} m (Δ T_{f} = T_{f}^{0} - T_{f})$
Use it when: A solute lowers the freezing point and you need $Δ T_{f}$ , the new freezing point, or $m$ . Water: $K_{f} = 1.86 K kg mol^{- 1}$
Elevation of boiling point
$Δ T_{b} = i K_{b} m (Δ T_{b} = T_{b} - T_{b}^{0})$
Use it when: A non-volatile solute raises the boiling point. Water: $K_{b} = 0.52 K kg mol^{- 1}$
Osmotic pressure
$π = i C R T = \frac{i n _{2} R T}{V}$
Use it when: The question gives a membrane, isotonic solutions or a protein/polymer. $C$ is molarity and $V$ is in litres. With $R = 0.083 L bar mol^{- 1} K^{- 1}$ , $π$ comes out in bar
Relative lowering of vapour pressure
$\frac{p _{1}^{0} - p _{1}}{p _{1}^{0}} = x_{2} = \frac{n _{2}}{n _{1} + n _{2}}$
Use it when: A non-volatile solute lowers the solvent's vapour pressure. With an electrolyte, $\frac{p _{1}^{0} - p _{1}}{p _{1}^{0}} = \frac{i n _{2}}{n _{1}}$
Dilute-solution form of item 4
$\frac{p _{1}^{0} - p _{1}}{p _{1}^{0}} \approx \frac{n _{2}}{n _{1}} = \frac{w _{2} \times M _{1}}{M _{2} \times w _{1}}$
Use it when: You need the solute's molar mass $M_{2}$ from a vapour pressure drop and $n_{2} ≪ n_{1}$
Molality
$m = \frac{moles of solute}{mass of solvent in kg} = \frac{w _{2} \times 1000}{M _{2} \times w _{1}}$
Use it when: You feed $m$ into $Δ T_{b}$ or $Δ T_{f}$ . Here $w_{1}$ is the solvent mass in grams
Molar mass from ΔT
$M_{2} = \frac{K _{f} \times w _{2} \times 1000}{Δ T _{f} \times w _{1}} (same form with K_{b} and Δ T_{b})$
Use it when: You know $Δ T$ and the masses, and need the solute's molar mass
Molar mass from osmotic pressure
$M_{2} = \frac{w _{2} R T}{π V}$
Use it when: The question is about the molar mass of a protein or polymer (macromolecules)
Raoult's law, two volatile liquids
$p_{total} = p_{1}^{0} x_{1} + p_{2}^{0} x_{2} = p_{1}^{0} + (p_{2}^{0} - p_{1}^{0}) x_{2}$
Use it when: Both liquids evaporate and you need the total vapour pressure of an ideal mixture
Vapour-phase mole fraction
$y_{i} = \frac{p _{i}}{p _{total}} (p_{i} = x_{i} p_{i}^{0})$
Use it when: You need the composition of the vapour. The vapour is always richer in the more volatile liquid
Henry's law
$p = K_{H} x$
Use it when: A gas dissolves in a liquid. A higher $K_{H}$ means lower solubility, and $K_{H}$ rises as $T$ rises
van't Hoff factor
$i = \frac{normal molar mass}{abnormal molar mass} = \frac{observed colligative property}{calculated colligative property}$
Use it when: The experimental value doesn't match the expected one. $i > 1$ for dissociation and $i < 1$ for association
Degree of dissociation
$i = 1 + α (weak electrolyte splitting into 2 ions, e.g. C H_{3} COOH \to H^{+} + C H_{3} CO O^{-})$
Use it when: You get $i$ from $Δ T_{f}$ and need $α$ (and then $K_{a} = \frac{C α ^{2}}{1 - α}$ )
Degree of association (dimerisation)
$i = 1 - α + \frac{α}{2} = 1 - \frac{α}{2} (2 A ⇌ A_{2})$
Use it when: An acid forms a dimer in benzene. Full dimerisation gives $i = 0.5$

Trap

forgetting i for electrolytes. For a salt like KCl or NaCl, i ≈ 2 and for K₂SO₄, i ≈ 3, so every colligative value comes out about 2× or 3× the non-electrolyte value. Also note that m is per kg of solvent, not per kg of solution.

How does NEET ask Solutions?

8 real previous-year questions from this chapter, as printed in the paper, each with its worked solution. Answer the first one here with no sign-in.

Q1ChemistryNEET UG PYQTypes of Solutions

Try right here · No sign-in

Match List-I with List-II List-I (Example) List-II (Type of Solution) A. Humidity I. Solid in solid B. AlloysII. Liquid in gas C. Amalgams III. Solid in gas D. Smoke IV. Liquid in solid Choose the correct answer from the options given below :

Answer key (D) & worked solution

Correct answer: Option D

A-II, B-I, C-IV, D-III

Q2ChemistryNEET UG PYQColligative Properties of Dilute Solutions

The boiling point of 0.2 mol kg

^{- 1}

solution of X in water is greater than equimolal solution of Y in water. Which one of the following statements is true in this case ?

AY is undergoing dissociation in water while X undergoes no change.

BX is undergoing dissociation in water.

CMolecular mass of X is greater than the molecular mass of Y.

DMolecular mass of X is less than the molecular mass of Y.

Answer key (B) & worked solution

Correct answer: Option B

Molality of solution of x = molality of solution of y = 0.2 mol/kg By elevation in boiling point relation ΔT

_{b}

= i K

_{b}

m or ΔT

_{b}

∝ i ∵ ΔT

_{b}

of solution of ‘x’ > ΔT

_{b}

of solution of ‘y’. ∴ ′ i ′of solution of x > i of solution of ‘y’ ∴ Solute of ‘x’ undergoing dissociation.

Q3ChemistryNEET UG PYQIdeal and Non-Ideal Solution

Which one is not equal to zero for an ideal solution?

AΔP = Pobserved − P

_{R a o u l t}

BΔH

_{mi x}

CΔS

_{mi x}

DΔV

_{mi x}

Answer key (C) & worked solution

Correct answer: Option C

For ideal solution ΔH

_{mi x}

= 0 and ΔG

_{mi x}

_{mi x}

> 0

Q4ChemistryNEET UG PYQConcentration Terms

Identify the correct statements : (A) The molality of 2.5 g of ethanoic acid (Molar mass :

60 g mol^{- 1}

) in 75 g of benzene solution is 0.556 m . (B) The molarity of a solution containing 5 g of NaOH (molar mass :

40 g mol^{- 1}

) in 450 mL of solution is 0.278 M at 298 K . (C) Aquatic species are more comfortable in cold water. (D) The solubility of gas increases with decrease in pressure. (E) For a binary mixture of

A

and

B

, the number of moles of

A

and

B

are

n_{A}

and

n_{B}

respectively. The mole fraction of

B

will be

x_{B} = \frac{n _{A}}{n _{A} + n _{B}}

. Choose the correct answer from the options given below :

AA and C only

BA, B and C only

CA, D and E only

DA and B only

Answer key (B) & worked solution

Correct answer: Option B

(A) Molality

= \frac{2.5}{60} \times \frac{1000}{75} = 0.556 molal

(B) Molarity

= \frac{5}{40} \times \frac{1000}{450} = 0.278 M

K_{H} \propto

Temp

\propto \frac{1}{solubility}

(D) According to Henry's Law,

P = K_{H} X

P \propto x

\propto

Solubility So as

P ↑

, solubility ↑ (E)

x_{B} = \frac{n _{B}}{n _{A} + n _{B}}

Q5ChemistryNEET UG PYQRaoult's Law

The correct option for the value of vapour pressure of a solution at 45°C with benzene to octane in molar ratio 3 : 2 is: [At 45°C vapour pressure of benzene is 280 mm Hg and that of octane is 420 mm Hg. Assume Ideal gas]

A336 mm of Hg

B350 mm of Hg

C160 mm of Hg

D168 mm of Hg

Answer key (A) & worked solution

Correct answer: Option A

Let benzene is B and octane is O X

_{B}

= 3/5 = 0.6 X

_{O}

= 2/5 = 0.4 Now, PT = P

_{B}

^{0}

_{B}

+ P

_{O}

^{0}

_{O}

= 0.6(280)+0.4(420) = 336 mm Hg

Q6ChemistryNEET UG PYQColligative Properties of Dilute Solutions

On which of the following properties does the coagulating power of an ion depend?

ABoth magnitude and sign of the charge on the ion

BSize of the ion alone

CThe magnitude of the charge on the ion alone

DThe sign of charge on the ion alone

Answer key (A) & worked solution

Correct answer: Option A

Coagulation of colloidal solution by using an electrolyte depends on the charge present (positive or negative) on colloidal particles as well as on its size. Coagulating power of an electrolyte depends on the magnitude of charge present on effective ion of electrolyte.

Q7ChemistryNEET UG PYQIntroduction to Solutions

Assertion (A): In a particular point defect, an ionic solid is electrically neutral, even if few of its cations are missing from its unit cells. Reason(R): In an ionic solid, Frenkel defect arises due to dislocation of cation from its lattice site to interstitial site, maintaining overall electrical neutrality. Choose the most appropriate answer from the options given below:

ABoth (A) and (R) are correct and (R) is the correct explanation of (A)

BBoth (A) and (R) are correct but (R) is not the correct explanation of (A)

C(A) is correct but (R) is not correct

D(A) is not correct but (R) is correct

Answer key (B) & worked solution

Correct answer: Option B

Both (A) and (R) are correct, but (R) is not correct expression of (A)

Q8ChemistryNEET UG PYQIdeal and Non-Ideal Solution

For an ideal solution, the correct option is:

Δ_{mi x} V \neq = 0

at constant T and P

Δ_{mi x} H = 0

at constant T and P

Δ_{mi x} G = 0

at constant T and P

Δ_{mi x} S = 0

at constant T and P

Answer key (B) & worked solution

Correct answer: Option B

For ideal solution,

Δ_{mi x} H = 0 Δ_{mi x} S > 0 Δ_{mi x} G < 0 Δ_{mi x} V = 0

Which topics in Solutions carry the most questions?

The 12 busiest of 13 topics, by tagged PYQs, then practice questions. Counted from the bank. Updated 19 Oct 2026.

Topic	PYQs	Practice questions
Introduction to Solutions	12	14
Colligative Properties of Dilute Solutions	8	181
Ideal and Non-Ideal Solution	7	41
Concentration Terms	5	71
Raoult's Law	3	48
Van't Hoff Factor and Its Significance	2	38
Solubility of Gases and Solids	1	31
Concept of Boiling and Azeotropes	1	12
Types of Solutions	1	4
Vapour Pressure	0	15
Abnormal Value of Molar Mass	0	4
Miscellaneous Examples	0	1

Why does Solutions show "–" until you answer 3 questions?

Unknown is not zero. Until you answer at least 3 graded questions in Solutions, ViJEEta shows – instead of a zero or a 50% default. Book exercises can raise chapter mastery up to 0.60, adaptive practice up to 0.75, and only a timed chapter test or mock can certify 0.80 and above.

How ViJEEta scores you

What happens when you test Solutions?

Every chapter test is 25 questions. It starts at moderate difficulty and adapts after each saved answer. If you open a solution before answering, that attempt earns no credit, so your mastery stays honest.

Start the Solutions test

Chapters either side of Solutions by weight

Coordination Compounds Weight 6.5/100 Hydrocarbons Weight 5.9/100 Alcohols, Phenols, and Ethers Weight 5.0/100 Chemical Kinetics Weight 4.5/100

Solutions: NEET PYQs and weightage

Solutions: formula sheet for NEET

Depression of freezing point

Elevation of boiling point

Osmotic pressure

Relative lowering of vapour pressure

Dilute-solution form of item 4

Molality

Molar mass from ΔT

Molar mass from osmotic pressure

Raoult's law, two volatile liquids

Vapour-phase mole fraction

Henry's law

van't Hoff factor

Degree of dissociation

Degree of association (dimerisation)

How does NEET ask Solutions?

Which topics in Solutions carry the most questions?

Why does Solutions show "–" until you answer 3 questions?

What happens when you test Solutions?

Chapters either side of Solutions by weight