31 Years NEET Previous Year Questions: Solutions - 2 - NEET MCQ

For ideal solution,

C₂H₅I and C₂H₅OH form non-ideal solution.

According to Raoult's law, the relative lowering in vapour pressure of a dilute solution is equal to the mole fraction of the solute present in the solution.

 mole fraction of solute

ΔT_f = i × K_f × m Van't Hoff factor, i = 2 for NaCl, hence ΔT_f = 0.02 K_f which is maximum in the present case.
Hence ΔT_f is maximum or freezing point is minimum.

Blood cells neither swell n or shrink in isotonic solution. As isotonic solutions have equal concentration therefore there is no flow of solvent is found and hence solvent neither enters nor flow out of the blood cells.

The molality involves weights of the solute and the solvent. Since weight does not change with the temperature, therefore molality does not depend upon the temperature.

Osmotic pressure is a colligative property.

As both the solutions are isotonic hence there is no net movement of the solvent through the semipermeable membrane between two solutions.

Conc. of particles in CaCl₂ sol. will be max. as i = 3 is max.
Note : Glucose and Urea do not dissociate into ions, as they are nonelectrolytes.

K₄ [Fe(CN)₆] and Al₂(SO₄)₃ both dissociates to give 5 ions or i = 5

and

Percentage volume of oxygen = 21%. (Given)
∵ 100 ml of air contains = 21ml of O₂
∴ Volume of oxygen in one litre of air

Therefore no. of moles 

(∵ volume of 1 litre of gas at S.T.P. is 22400 ml)

Give vapour pressure of  pur e solute (P⁰) = 121.8 mm;
Weight of solute (w) = 15 g
Weight of solvent (W) = 250 g;
Vapour pressure of pure solvent (P) = 120.2 mm and Molecular weight of solvent (M) = 78 From Raoult’s law

Relative lowering of vapour pressure depends upon the mole fraction of solute.

mole fraction of solute

According to Raoult's law

Depression in F.P. ∝ No. of particles.
Al₂(SO₄)₃ provides five ions on ionisation

while KCl provides two ions
KCl     K    Cl
C₆H₁₂O₆ and C₁₂H₂₂O₁₁ are not ionised so they have single particle.
Hence, Al₂(SO₄)₃ have maximum value of depression in F.P or lowest F.P

Osmotic pressure of 5% cane sugar solution 5% cane sugar solution means 100ml of solution contain cane sugar = 5g

∴    1000 of solution contain cane sugar

Similarily 1% is solution  

Osmotic pressure of 1% solution of susbtance  (π₂)

As both are isotonic, So
π₁ = π₂

∴  M (mol. wt. of X) 

According to Raoult's law

 (mole fraction of solute)

For other solution of same solvent

(Mole fraction of solute)

mole fraction of solute

⇒   Mole fraction of solute = 0.4 As mole praction of solute + mole traction of solvent = 1

Hence, mole fraction of solvent = 1 – 0.4    = 0.6

Mole fraction of any component A

As total no. of moles > no. of moles of A thus x can never be equal to one on zero.

Molecular masses of polymers are best determined by osmotic pressure method .
Firstly because other colligative properties give so low values that they cannot be measured accurately and secondly, osmotic pressure measurements can be made at room temperature and do not require heating which may change the nature of the polymer.

The beans are cooked earlier in pressure cooker because boiling point increases with increasing pressure.

From molarity equation M₁V₁ + M₂V₂ = MV
1× 2.5 + 0.5 × 3 = M × 5.5

These two components A and B follows the condition of Raoult’s law if the force of attraction between A and B is equal to the force of attraction between A and A or B and B.

For an ideal solution, ΔH_mixing = 0
  (Accroding to Hess's law) i.e., for ideal solutions there is no change in magnitude of the attractive forces in the two components present.

Solvent having high cryoscopic constant can be used in determination of molecular mass by cryoscopic method.

Given V.P_P = 80 torr  V.PQ = 60 torr

P_total = 48 + 24 = 72 torr

As ΔT_f = K_f. m
ΔT_b = K_b. m

 Hence, we have  

= [ΔT_b = 100.18 - 100 = 0.18°C]

As the Freezing Point of pure water is 0°C,
ΔT_f = 0 –T_f
0.654 = 0 – T_f
∴  T_f = – 0.654 thus the freezing point of solution will be – 0.654°C.

One molal solution means one mole solute present in 1 kg (1000 g) solvent i.e., mole of solute = 1

Mole of solvent 

Mole fraction of solute  =

A solution of acetone in ethanol shows positive deviation from Raoult's law. It is because ethanol molecules are strongly hydrogen bonded. When acetone is added, these molecules break the hydrogen bonds and ethanol becomes more volatile. Therefore its vapour pressure is increased.