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Solutions & Colligative Properties revision notes

Every idea in the Solutions & Colligative Properties chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 21 and download the PDF.

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Concentration and Henry's law

Concentration terms
12× asked, last 2026
m=nBWA (kg),M=nBV (L),m=1000 M1000 d−M MBm = \frac{n_B}{W_A\,(\text{kg})},\qquad M = \frac{n_B}{V\,(\text{L})},\qquad m = \frac{1000\,M}{1000\,d - M\,M_B}
dd in g mL⁻¹, MBM_B in g mol⁻¹. Mass %, ppm, xx, mm use masses: fixed with TT. MM, % w/v fall on heating; % v/v also varies.
water 1074.5 gNaCl 175.5 g1 L of solution =1000×1.25=1250= 1000\times1.25 = 1250 gm=3/1.0745=2.79m = 3/1.0745 = 2.79, not 3
3 M NaCl, d=1.25d=1.25: molality divides by the water alone
  • Not kg of solution for molality. From 1 L of solution the solvent is 1000d−MMB1000d-MM_B grams. 3 M NaCl with d=1.25d=1.25: 1250−175.5=1074.51250-175.5=1074.5 g water, so m=2.79m=2.79, not 3 or 2.4.
Henry's law
9× asked, last 2026
pgas=KH xgas,ngas≈pgasKH nsolventp_{\text{gas}} = K_H\,x_{\text{gas}},\qquad n_{\text{gas}} \approx \frac{p_{\text{gas}}}{K_H}\,n_{\text{solvent}}
Dilute, fixed TT. KHK_H has pressure units; larger KHK_H = less soluble. Uses: soda bottles, divers' bends, anoxia at altitude.
log⁡x\log xlog⁡p\log p (mm Hg)−5-5−4-4−3-32346080 mm0.835 barNX2\ce{N2}COX2\ce{CO2}
Each gas is a line of slope 1, raised by log⁡KH\log K_H: at the same pp, NX2\ce{N2} (larger KHK_H) dissolves less
  • KHK_H belongs to one gas–solvent pair at one TT: constant over the dilute range, different in another solvent, larger when warmer (less gas dissolves). log⁡p=log⁡KH+log⁡x\log p=\log K_H+\log x is a line of slope 1.
  • A constant in mol L⁻¹ bar⁻¹ (c=kHpc=k_Hp) is a solubility constant: there a larger kHk_H means MORE soluble. Check the unit before ranking gases.
  • Not the total pressure. Use the gas's own partial pressure, in the unit of KHK_H. NX2\ce{N2} at 80% of 10 atm: p=8p=8 atm =6080=6080 mm Hg, so x=6080/(6.5×107)=9.35×10−5x=6080/(6.5\times10^7)=9.35\times10^{-5}.
  • Quick method. p/KHp/K_H is a mole fraction, not a molarity. For moles of gas multiply by moles of water (1 L ≈ 55.5 mol): COX2\ce{CO2} at 0.835 bar with KH=1.67K_H=1.67 kbar gives x=5×10−4x=5\times10^{-4}, 27.8 mmol per litre.

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Practise Solutions & Colligative Properties

19 more concepts in 5 topics

Vapour pressure and Raoult's law; Non-ideal solutions and azeotropes; Colligative properties: vapour pressure, boiling and freezing points; Osmosis and osmotic pressure; and van 't Hoff factor and abnormal molar mass. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Vapour pressure and Raoult's law

  • Raoult's law for two volatile liquids6×
  • Composition of the vapour8×
  • Ideal solutions1×

Non-ideal solutions and azeotropes

  • Positive deviation4×
  • Negative deviation4×
  • Azeotropes5×

Colligative properties: vapour pressure, boiling and freezing points

  • Relative lowering of vapour pressure14×
  • Elevation of boiling point19×
  • Depression of freezing point22×
  • Molar mass from ΔTb or ΔTf10×

Osmosis and osmotic pressure

  • Osmotic pressure11×
  • Molar mass of macromolecules8×
  • Isotonic, hypertonic and hypotonic12×
  • Reverse osmosis2×

van 't Hoff factor and abnormal molar mass

  • van 't Hoff factor i7×
  • Dissociation14×
  • Weak electrolytes: i and Ka9×
  • Association8×
  • Comparing colligative properties14×

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