Azeotropes — Maximum and minimum boiling
Solutions
10
JEE Qs
8%
Hard
40
min
Master the direct correlation between positive/negative deviations from Raoult's Law, vapor pressure, and whether an azeotrope will be minimum or maximum boiling, often tested with examples and graphical interpretations.
🧮 Key Formulas
✅ Key Points for JEE
- 1Azeotropes are binary mixtures that boil at a constant temperature and distil without change in composition, behaving like pure liquids.
- 2Minimum boiling azeotropes exhibit large positive deviations from Raoult's Law, meaning they have a higher vapor pressure and thus a lower boiling point than either of their pure components.
- 3Maximum boiling azeotropes exhibit large negative deviations from Raoult's Law, meaning they have a lower vapor pressure and thus a higher boiling point than either of their pure components.
- 4The composition of an azeotrope is fixed at a specific temperature and pressure, making it impossible to separate its components by fractional distillation.
- 5Examples include ethanol-water (minimum boiling) and nitric acid-water (maximum boiling).
⚠️ Common Mistakes
- ✕Confusing the relationship between deviation type (positive/negative), vapor pressure, and boiling point (e.g., assuming positive deviation means higher boiling point).
- ✕Incorrectly assuming that azeotropes can be separated into their pure components by fractional distillation.
- ✕Not understanding that the azeotropic composition is specific and constant at its boiling point, unlike other non-ideal mixtures.
📝 Practice Questions
See allQ73.Consider the following cases of standard enthalpy of reaction (ΔH∘r in kJmol−1) 7 C2H6( g) + O2( g) →2CO2( g) + 3H2O(l)ΔH∘1 = −1550 2 C (graphite) + O2( g) →CO2( g) ΔH∘2 = −393.5 The magnitude of ΔH∘fC2H6( g) is _______ 1 H2( g) + O2( g) →H2O(l) ΔH∘3 = −286 2 kJmol−1 (Nearest integer).
Q70.When a non-volatile solute is added to the solvent, the vapour pressure of the solvent decreases by 10 mm of Hg . The mole fraction of the solute in the solution is 0.2 . What would be the mole fraction of the solvent if decrease in vapour pressure is 20 mm of Hg ? (1) 0.8 (2) 0.4 (3) 0.2 (4) 0.6 Q71.0.01 mole of an organic compound (X) containing 10% hydrogen, on complete combustion produced 0.9 gH2O. Molar mass of (X) is _____ gmol−1 .
Q69.Arrange the following solutions in order of their increasing boiling points. (i) 10−4M NaCl (ii) 10−4M Urea (iii) 10−3M NaCl (iv) 10−2M NaCl (1) (i) < (ii) < (iii) < (iv) (2) (iv) < (iii) < (i) < (ii) (3) (ii) < (i ) ≡(iii) < (iv) (4) (ii) < (i) < (iii) < (iv)
Q52.Density of 3 M NaCl solution is 1.25 g/mL. The molality of the solution is : (1) 1.79 m (2) 2.79 m (3) 2 m (4) 3 m
Q64.Consider a binary solution of two volatile liquid components 1 and 2. x1 and y1 are the mole fractions of component 1 in liquid and vapour phase, respectively. The slope and intercept of the linear plot of 1 vs 1 x1 y1 are given respectively as: P02−P01 (1) P0 P02−P01 (2) P0 2 , 1 , P01 P02 P02 P02 P01−P02 (3) P0 P01−P02 (4) P0 1 , 2 , P02 P02 P01 P02
Q69.Consider the given plots of vapour pressure (VP) vs temperature(T/K). Which amongst the following options is correct graphical representation showing ΔTf , depression in the freezing point of a solvent in a solution? (1) (2) (3) (4)
NCERT Chapters
- Class 12 Chemistry Ch 2: Solutions