R.C. Mukherjee Test: Solutions - JEE MCQ

(A) NaCl → Na⁺ + Cl^–
ΔT_b = K_b × ms
ΔTb = 2K_b × m
elavation of b.p. will be double in case of NaCl not b.p.
(B) Will be correct because b.p. elavation will be double here in comprasion to glucose.
(C) Elavation of b.p. is colligative property not b.p

As, we know that

= 25 + 45 mm Hg
= 70 mm Hg
To mm Hg < 75 mm Hg thus there is positive deviation from Raoults law,
so (A) is correct
(B) If v.p. is increasing boiling point will be lowered in that case
(C) is correct that force of attraction between A & B is smaller than that between A and A or between B and B.
So All statement are corrrect

As van't Holf factor increases RLVP increases i.e.,V,P. decreases y > x > z

π = iCRT
K₃[Fe(CN)₆] = 3k⁺ + [Fe(CN)₆]^3–

i = (1 + 3α)

α = 0.2
so  20%

The correct answer is Option B.

Density is the ratio of mass to volume.
Density is 1.052 g/ml.
1500 ml of solution corresponds to       
= 1.052 g/ml×1500ml
= 1578 g     
Molar mass of urea is 60 g/mol. Number of moles of urea is the ratio of mass to molar mass. It is 18g / 60g/mol = 0.3mol
 
Molality of solution is the ratio of the number of moles of urea to the mass of solvent in kg.
Mass of water =1578 g − 18 g
                          =1560g
                          =1.560kg (as 1 kg = 1000g)
Molality = 0.3 / 1.560
              = 0.192m
 

M = 256
M = 8 × 32
Thus atomicity = 8

(A) All solutions are not isotonic because all solution are not at equal concentration so (A) is incorrect
(B) As benzoic acid dimerises so III is hypotonic of I, II, IV
(C) Is also correct as I, II & IV are hypertonic of III
(D) As ammoniun sulphate has maximum number of particale so it will be hypertonic of I, II, III

As number of moles is maximum in case urea > glucose > sucrose
π = CRT
It depends on number of moles so osmotic pressure
P2 > P1 > P3

As benzoic acid dimerises so number of moles decreases so osmotic pressure of benzoic acid is less than benzene.

Higher the number of particles higher will be osmoatic pressure so (A) will be answer in case of (D) precipitaiton will take place so number of particles will decrease.

Isotonic means equal osmotic pressure so, π₁ = π₂
i₁C₁RT = i₂C₂RT
so i₁C₁ = i₂C₂
In case of (D) both the volume of (i) & (C) are equal.

As we know relative lowering of v.p. is equal to the mole fraction of the solute so (I) is correct it doesn’t depend upon the number of moles so (II) is wrong mole fraction of solvent will be 0.8 so number of moles of solvent will be 4 so (III) will be correct, (IV) will be also wrong so I & III will be correct

In option (B) oxidaton state of platinum is (iv)
x – 6 = –2
x = + 4

i = 1 + α
10^–4 = Cα²

i = 1 + 2 = 1 + 0.1 = 1.1

HA = H⁺ + A^–
(1 – α) α       α
PH = 2 = [H⁺]  : 10^–2  = C²
a = 0.1
i = 1 + α = 1 + 0.1 = 1.01

(a) As we know
 i = 1 + α

i = 1 + α =  50 % dissociation
i = 1.5

1 + Zα ,  α = .4
i = 1.8

1 + 3α
i = 1.9

1 + 4α
i = 1.8

18 gm of glucose means 0.1mole of glucose as it present in 1000 gm of solvent so it is 0.1 mole

Molality of BaCl₂ = 0.1 × 0.25 = 0.025
by calculation we get the values of (D)

Mole of HCl = 25 × 3 +75 × 0.05 = 75 + 3.75 = 78.75

so concentration of solution = 5 ppm

= 40 + 120 = 160

π 40 = y_A × 160
⇒ y_A = ¼

The exact mathematical expression of Raoult's law is 
Here, P^o represents the vapour pressure of the pure solvent, P represents the vapour pressure of the solution, n represents the number of moles of solute and N represents the number of moles of the solvent.

= 85.88 mm Hg

less no of particle of solute means maximum vapour pressure.

Glucose does not dissociate
⇒ i = 1

AlCl₃ = no. of particle = 4
CaCl2 = no. of particle = 3
(vapour pressure of AlCl₃) < (vapour pressure of CaCl₂ solution)
T_B.P. (AlCl₃) > T_B.P. (CaCl₂)
T₁ > T₂