Is Solutions part of the EAPCET Chemistry syllabus?

Yes. Solutions is taught in Intermediate second year Chemistry, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Chemistry questions are there in EAPCET?

40 of the 160 questions in the Engineering paper are Chemistry, spread across 24 chapters.

What mistakes do students make in Solutions?

Forgetting the van ’t Hoff factor for electrolytes — NaCl doubles and CaCl₂ roughly triples every colligative effect. Using molarity in ΔT_b = K_b m — boiling and freezing point formulas need molality (mol per kg of solvent). Reversing the deviations — weaker solute–solvent attraction gives positive deviation and a minimum-boiling azeotrope; stronger attraction gives negative deviation and a maximum-boiling azeotrope. Writing osmotic pressure with T in °C or V in mL — use π = CRT with C in mol L⁻¹, R = 0.0821 L atm K⁻¹ mol⁻¹ and T in kelvin.

EAPCET Chemistry · Intermediate second year

Solutions

Solutions covers concentration terms, Raoult’s and Henry’s laws, ideal and non-ideal solutions, and colligative properties.

Subject: Chemistry
Taught in: Intermediate second year
Chemistry in the paper: 40 of 160 questions

Key formulas and results

Raoult’s law: $p = p^\circ x$
Relative lowering of vapour pressure = mole fraction of solute
$\Delta T_b = K_b m$ and $\Delta T_f = K_f m$
Osmotic pressure $\pi = CRT$
The van ’t Hoff factor $i$ multiplies every colligative property
Henry’s law: $p = K_H x$
Raoult’s law for two volatile liquids: $P_{\text{total}} = p^\circ_A x_A + p^\circ_B x_B = p^\circ_B + x_A(p^\circ_A - p^\circ_B)$
Mole fraction of A in the vapour: $y_A = \frac{p^\circ_A x_A}{P_{\text{total}}}$
Relative lowering of vapour pressure for a dilute solution: $\frac{p^\circ - p}{p^\circ} \approx \frac{w_B \times M_A}{M_B \times w_A}$
Molal elevation constant: $K_b = \frac{M_A R T_b^2}{1000\,\Delta H_{\text{vap}}}$ ; molal depression constant: $K_f = \frac{M_A R T_f^2}{1000\,\Delta H_{\text{fus}}}$ , with $M_A$ the molar mass of the solvent in g/mol
van ’t Hoff factor for dissociation into $n$ ions: $i = 1 + (n - 1)\alpha$
van ’t Hoff factor from experiment: $i = \frac{\text{observed colligative property}}{\text{calculated value}} = \frac{\text{normal molar mass}}{\text{observed molar mass}}$
Osmotic pressure of mixed solutions: $\pi = \frac{(n_1 + n_2)RT}{V_{\text{total}}}$

What this chapter covers

Concentration terms
Henry’s law
Raoult’s law
Ideal and non-ideal solutions
Colligative properties
van ’t Hoff factor

How it is usually asked

Colligative property calculations
Molar mass from colligative data
Positive and negative deviations
Using i for dissociation or association

Understanding Solutions

A solution is a homogeneous mixture, and its composition is stated as mass percent, mole fraction, molarity or molality. Henry’s law says the solubility of a gas in a liquid is proportional to its partial pressure above the liquid, which is why soda bottles are sealed under high CO₂ pressure. Gas solubility falls as temperature rises. Raoult’s law says each volatile component contributes a vapour pressure equal to its mole fraction times its pure vapour pressure.

Ideal solutions obey Raoult’s law exactly, with ΔH(mix) = 0 and ΔV(mix) = 0, as in benzene–toluene. If the new A–B attractions are weaker than A–A and B–B, vapour pressure is higher than expected (positive deviation, such as ethanol–water or acetone–CS₂). If they are stronger, as in acetone–chloroform with hydrogen bonding, the deviation is negative. Large deviations produce azeotropes, which boil at constant composition and cannot be separated by simple distillation.

Colligative properties depend only on the number of solute particles: relative lowering of vapour pressure, boiling point elevation, freezing point depression and osmotic pressure. The van ’t Hoff factor i corrects for dissociation (i > 1, as for NaCl with i ≈ 2) or association (i < 1, as for acetic acid dimerising in benzene). EAPCET questions usually ask you to compute ΔT or π, or to find molar mass from colligative data.

Worked example

EAPCET-style question

The freezing point of a 0.1 m aqueous NaCl solution, assuming complete dissociation, is ( $K_f$ of water $= 1.86\ \text{K kg mol}^{-1}$ ):

(1)−0.186 °C
(2)−0.372 °C
(3)−0.558 °C
(4)+0.372 °C

Solution

$\ce{NaCl -> Na+ + Cl-}$ gives two particles per formula unit, so $i = 2$ .
$\Delta T_f = i K_f m = 2 \times 1.86 \times 0.1 = 0.372\ \text{K}$ .
A depression of 0.372 K equals 0.372 °C.
Freezing point $= 0 - 0.372 = -0.372\ {}^\circ\text{C}$ .
−0.186 °C ignores dissociation, and −0.558 °C uses $i = 3$ .

Answer: −0.372 °C

Shortcuts and results to remember

Positive deviation (A–B attraction weaker than A–A and B–B): $\Delta H_{\text{mix}} > 0$ and $\Delta V_{\text{mix}} > 0$ , e.g. ethanol + water, acetone + ethanol, CCl₄ + CHCl₃.
Negative deviation (A–B attraction stronger): $\Delta H_{\text{mix}} < 0$ and $\Delta V_{\text{mix}} < 0$ , e.g. chloroform + acetone, HNO₃ + water, HCl + water.
Ideal solutions have $\Delta H_{\text{mix}} = 0$ and $\Delta V_{\text{mix}} = 0$ , e.g. benzene + toluene, n-hexane + n-heptane, ethyl bromide + ethyl chloride.
An azeotrope boils at constant composition (vapour has the same composition as the liquid), so it cannot be separated by fractional distillation.
Isotonic solutions have equal osmotic pressure; a hypertonic solution has the higher and a hypotonic one the lower osmotic pressure; for non-electrolytes such as glucose or urea, $i = 1$ .

Common mistakes to avoid

Forgetting the van ’t Hoff factor for electrolytes — NaCl doubles and CaCl₂ roughly triples every colligative effect.
Using molarity in ΔT_b = K_b m — boiling and freezing point formulas need molality (mol per kg of solvent).
Reversing the deviations — weaker solute–solvent attraction gives positive deviation and a minimum-boiling azeotrope; stronger attraction gives negative deviation and a maximum-boiling azeotrope.
Writing osmotic pressure with T in °C or V in mL — use π = CRT with C in mol L⁻¹, R = 0.0821 L atm K⁻¹ mol⁻¹ and T in kelvin.

How to prepare it

Decide first whether the solute dissociates or associates — that fixes i before any calculation.

Once the formulas feel familiar, test them the way the exam does: practise Solutions questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Chemistry chapters

See every chapter in the EAPCET Chemistry syllabus.

Practise Solutions chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Solutions part of the EAPCET Chemistry syllabus?: Yes. Solutions is taught in Intermediate second year Chemistry, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Solutions for EAPCET?: Decide first whether the solute dissociates or associates — that fixes i before any calculation.
How many Chemistry questions are there in EAPCET?: 40 of the 160 questions in the Engineering paper are Chemistry, spread across 24 chapters.
What mistakes do students make in Solutions?: Forgetting the van ’t Hoff factor for electrolytes — NaCl doubles and CaCl₂ roughly triples every colligative effect. Using molarity in ΔT_b = K_b m — boiling and freezing point formulas need molality (mol per kg of solvent). Reversing the deviations — weaker solute–solvent attraction gives positive deviation and a minimum-boiling azeotrope; stronger attraction gives negative deviation and a maximum-boiling azeotrope. Writing osmotic pressure with T in °C or V in mL — use π = CRT with C in mol L⁻¹, R = 0.0821 L atm K⁻¹ mol⁻¹ and T in kelvin.