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A 0.002 molar solution of NaCl having degree of dissociation of 90% at 270C, has osmotic pressure equal to:-
= i-1/n-1
0.9 = i-1/2-1
i=1.9
=iCRT = 1.9x0.002x0.082x300
=0.094 bar
The density of NH4OH solution is found to be 0.6 g/ml. It contains 35% by mass of NH4OH. The normality of the solution is
dsol= mass
volume
vsol= 100 ml
0.6
N= W X 1
E V(L)
N= 35 x 1000 x 0.6
35 100
N = 6
The relative lowering of vapour caused by dissolving 71.3g of a substance in 1000g of water is 7.13x10-3. The molecular mass of substance is:-
P/P0 = nB/nA
7.13x10-3=71.3x18/mBx1000
msolute = 180
Total vapour pressure of mixture of 1 mol A(p0A = 150 torr) and 2 mol B(P0B = 240 torr) is 200 torr. In this case:-
P = P0AXA + P0BXB
P = 150(1/3)+240(2/3)
P = 210 torr
Pexp. < Pca; Negative derivation from Raoult's law
Which of the following aqueous solution should have the highest osmotic pressure?
π = iCRT T ↑ , T ↑ so, π ↑
50 g of antifreeze (ethylene glycol) is added to 200g water. What amount of ice will separate out at
-9.3°C. (Kf = 1.86 K Kg mol-1) :-
ΔTf = 1000 Kf w ⇒ 9.3 = 1000 x 1.86 x 50
mW 62 x W
⇒ W= 161.29g
Separated ice= 200-161.29= 38.71g
40 g NaOH (i=2) is dissolved in 1L volatile solvent vapour pressure of this solution becomes equal to its solid phase at 300K. What is the freezing point of pure solvent.
(Density = 1g/mL) (kf=1.8 k kgmol-1)
Tf = iKfm
Tf - 300 = 2x1.8x(40/40)/1
Tf = 303.6K
100 g solute is dissolved in 1400 g of solvent. Density of resultant solution is 1.5 g/mL. The ratio of its molarity and molality will be :-
The molarity of the solution is given by the number of moles of solute divided by the volume of the solution in liters. The molality is given by the number of moles of solute divided by the mass of the solvent in kilograms. Since the density of the solution is 1.5 g/mL, the ratio of molarity to molality will be (1.5 g/mL) / (1000 g/kg) = 1.4.
As scuba divers come towards water surface from underwater, solubility of gases in their blood :-
According to Henry‘s law, solubility Partial directly proportional to the pressure.
Saturated solution of Ag2SO4 shows 0.003K rise in boiling point, Ksp will be :(Kb = 5 K kg mol-1)(1m=1M)
Tb = iKbm
0.003=3x5xS
S=2x10-4
Ksp = 4S3=3.2x10-11
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