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An azeotropic solution of two liquids has boiling point lowerthan either of them when it (1981 - 1 Mark)
Lower the B. pt., higher will be the V.P. The V.P. of the mixture is greater than either of the two liquids.
[NOTE : : In case of positive deviation from Roult's law the partial vapour pressure of each liquid and total vapour pressure of solution will be greater as compared to initial solution]
For a dilute solution, Raoult’s law states that :(1985 - 1 Mark)
[Mathematical statement of Raoult's law]
When mercuric iodide is added to the aqueous solution ofpotassium iodide then (1987 - 1 Mark)
Added HgI2 forms a complex with KI in the solution as follows
2KI + HgI2 → K2[HgI4]
As a result, number of particles decreases and so ΔTf increases.
[NOTE : : Depression in freezing point is a colligative property]
Which of the following 0.1 M aqueous solutions will havethe lowest freezing point? (1989 - 1 Mark)
NOTE : The salt producing highest number of ions will have lowest freezing point.
gives highest number of particles (2 + 1 = 3).
Glucose, being non-electrolyte gives minimum no. of particles and hence minimum ΔTf or maximum F. pt.
The freezing point of equimolal aqueous solutions will behighest for : (1990 - 1 Mark)
TIPS/Formulae : The salt that ionises to least extent will have highest freezing point. [ i.e., minimum Δ Tf ]
Glucose, being non electrolyte, gives minimum no. of particles and hence minimum DTf or maximum F. pt
0.2 molal acid HX is 20% ionised in solution. Kf = 1.86 Kmolality–1. The freezing point of the solution is : (1995S)
Depression in freezing point,
Van’t Hoff factor, where n = no. of ions
[ For 20 % ionisation, α = 20/100= 0.2]
∴ΔTf = 1.2 × 1.86 × 0.2 = 0.45 [ ∵ m = 0.2]
Hence freezing point of solution is 0 – 0.45 = –0.45
[∵ F.P of water = 0.0 C]
The molecular weight of benzoic acid in benzene asdetermined by depression in freezing point methodcorresponds to : (1996 - 1 Mark)
Benzoic acid exists as dimer in benzene.
[Normal molecular mass = 122 amu
observed molecular mass = 244 amu, in case of complete association]
During depression of freezing point in a solution thefollowing are in equililbrium (2003S)
NOTE : At the freezing point liquid and solid remain in equilibrium. If a solution of a non-volatile solute is cooled to a temperature below the freezing point of solution, some of liquid solvent will separate as a solid solvent and thus the concentration of solution will
The elevation in boiling point of a solution of 13.44 g ofCuCl2 in 1 kg of water using the following information will be(Molecular weight of CuCl2 = 134.4 and Kb = 0.52 K molal-1) (2005S)
Assuming 100% ionization So, i = 1 + 2 = 3
When 20 g of naphthoic acid (C11H8O2) is dissolved in 50g of benzene (Kf = 1.72 K kg mol–1), a freezing pointdepression of 2K is observed. The Van't Hoff factor (i) is (2007)
Molecular weight of naphthoic acid
C11H8O2 = 172 gmol–1.
The theoretical value of depression in freezing point
Van't Hoff factor,
The Henry’s law constant for the solubility of N2 gas inwater at 298 K is 1.0 × 105 atm. The mole fraction of N2 in airis 0.8. The number of moles of N2 from air dissolved in10 moles of water at 298 K and 5 atm pressure is (2009)
Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gavea solution of density 1.15 g/mL. The molarity of the solutionis (2011)
Number of moles of urea = 120/60 = 2
Total mass of solution = 1000 + 120 = 1120 g
Total volume of solution (in L) = Mass/density
Molarity of the solution
The freezing point (in °C) of a solution containing 0.1 gof K3[Fe(CN)6] (Mol. wt. 329) in 100 g of water (Kf = 1.86 Kkg mol–1) is (2011)
ΔTf = i × Kf × m
Where m = Molality of the solution
(i.e. number of moles of solute per 1000 g of the solvent)
For a dilute solution containing 2.5 g of a non-volatile nonelectrolytesolute in 100 g of water, the elevationin boiling point at 1 atm pressure is 2°C. Assumingconcentration of solute is much lower than the concentrationof solvent, the vapour pressure (mm of Hg) of the solutionis (take Kb = 0.76 K kg mol–1) (2012)
From Raoult law relation,
When the concentration of solute is much lower than the concentration of solvent,
From elevation in boiling point, ΔTb = Kb × m
2 = 0.76 × m
m = 2/0.76 ...(ii)
From (i) and (ii), p = 724 mm
In the depression of freezing point experiment, it is foundthat the (1999 - 3 Marks)
(a, d) The freezing point of a solvent depresses as a nonvolatile solute is added to a solvent. According to Raoult's law, when a non-volatile solute is added to a solvent the vapour pressure of the solvent decreases.
At the freezing point it will be only the solvent
molecules which will solidify.
Benzene and naphthalene form an ideal solution at roomtemperature. For this process, the true statement(s) is(are) (JEE Adv. 2013)
Mixture(s) showing positive deviation from Raoult’s law at35 °C is (are) (JEE Adv. 2016)
(A) H-bonding of methanol breaks when CCl4 is added so bonds become weaker, resulting positive deviation.
(B) Mixing of polar and non-polar liquids will produce a solution of weaker interaction, resulting positive deviation
(C) Ideal solution
(D) –ve deviation because stronger H-bond is formed.