Test: Solutions (October 22) - NEET MCQ

For 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

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.

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

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)

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

mole fraction of solute

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) 

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.

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

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

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.