Test: MCQs (One or More Correct Option): Solutions | JEE Advanced - JEE MCQ

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]

[Mathematical statement of Raoult's law]

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 ΔT_f increases.
[NOTE : : Depression in freezing point is a colligative property]

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 ΔT_f or maximum F. pt.

TIPS/Formulae : The salt that ionises to least extent will have highest freezing point. [ i.e., minimum Δ T_f ]
Glucose, being non electrolyte, gives minimum no. of particles and hence minimum DTf or maximum F. pt

Depression in freezing point, 

Van’t Hoff factor,   where n = no. of ions

[ For 20 % ionisation, α = 20/100= 0.2]
∴ΔT_f = 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]

Benzoic acid exists as dimer in benzene.

[Normal molecular mass = 122 amu
observed molecular mass = 244 amu, in case of complete association] 

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
increase.

TIPS/Formulae :

Assuming 100% ionization So, i = 1 + 2 = 3

Molecular weight of naphthoic acid
C₁₁H₈O₂ = 172 gmol^–1.
The theoretical value of depression in freezing point

Van't Hoff factor,

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

ΔT_f = i × K_f × m
Where m = Molality of the solution
(i.e. number of moles of solute per 1000 g of the solvent)

From Raoult law relation,

When the concentration of solute is much lower than the concentration of solvent,

           .... (i)

From elevation in boiling point, ΔT_b = K_b × m
2 = 0.76 × m
m = 2/0.76              ...(ii)
From (i) and (ii), p = 724 mm

(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.

(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.