50 g of saturated aqueous solution of potassium chloride at 30°C is evaporated to dryness, when 13.2 gm of dry KCl was obtained. The solubility of KCl in water at 30°C is
35.87 g
25.62 g
28.97 g
27.81 g
A 0.001 molal solution of [Pt(NH3)4Cl4] in water had a freezing point depression of 0.0054°C. If Kf for water is 1.80, the correct formulation of the above molecules is
[Pt(NH3)4Cl3]Cl
[Pt(NH3)4Cl2]Cl2
[Pt(NH3)4Cl]Cl2
[Pt(NH3)4Cl4]
If M is molecular weight of solvent, Kb is molal elevation constant, Tb is its boling point p° is its vapour pressure at temperature T and ps is vapour pressure of its solution having a non- volatile solute at T K , then
For a non-volatile solute,
vapour pressure of solution is more than vapour pressure of solvent
Vapour pressure of solvent is zero
Vapour pressure of solvent is zero
All of the above
What will happen if a cell is placed into 0.4% (mass/volume) NaCl solution?
There will be no change in cell volume
Cell will dissolve
cell will swell
cell will shrink
Calculate the osmotic pressure of 0.01 M solution of cane sugar at 300 K (R = 0.08212 atm degree-1 mol-1)
0.3568 atm
0.2463 atm
0.1562 atm
0.5623 atm
An aqueous dilute solution containing non-volatile solute boils at 100.52° C. What is the molality of solution?(Kb = 0.52 kg mol-1K, boiloing temperature of water = 100° C)
0.1 m
0.01 m
0.001 m
1.0 m
The experimental depression in freezing point of a dilute solution is 0.025 K. If the van't Hoff factor (i) is 2.0, the calculated depression in freezing point (in K) is
0.00125
0.025
0.0125
0.05
The molality of an aqueous dilute solution containing non-volatile solute is 0.1 m. What is the boiling temperature (in °C) of solution? (Boiling point elevation constant, Kb = 0.52 kg mol-1K; boiling temperature of water = 100°C).
100.0052
100.052
100.0
100.52
The vapour pressure of a non-ideal two component solution is given below
A.
option (a) is correct answer.