Osmotic Pressure – Numerical Part 1

osmotic pressure

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osmotic pressure-

Question 1) At 298 K , 100 cc of a solution containing 3.02 gm. of a solute exhibits an atmospheric  pressure of 2.55 atmosphere. What is the molecular mass of the solute ?

Solution )

w = 3.02 gm , V = 100 cc = [100/1000] litre = 0.1 litre

π  = 2.55 atm , m = ? , S = 0.0821 Litre atm K^-1mole^-1 , T = 298 K

π  = w ST /m V

m = w S T /π V = 3.02 x 0.0821 x 298 /2.55 x 0.1

m = 289.75 Ans.

Question 2)  0.85 % aqueous solution of NaNO₃ is apparently 90 % dissociated at 27⁰ C . Calculate osmotic pressure of the solution  ?

(S = 0.0821 Litre atm K^-1mole^-1 )

Solution )

w = 0.85  gm , V = 100 ml = [100/1000] litre = 0.1 litre

π   =  ? atm , m of NaNO₃ =  23 + 14 +16 x 3 = 85   , S = 0.0821 Litre atm K^-1mole^-1 , T = 273 + 27 =300 K

π _N = w ST /m V

π _N = 0.85  x 0.0821 x 300 / 85 x 0.1

π _N= 2.463 atm. Ans.

                                                             NaNO₃     ——–>    Na ⁺        +      NO₃^–

mole before dissociation                     1                               0                       0

moles after  dissociation                     1 – ∝                        ∝                       ∝

Total moles after  dissociation  = 1 – ∝ + ∝ + ∝ = 1 + ∝

% of ∝ = 90 % , ∝ = 90 / 100 = 0.90

 _π  exp / π _N = (1 + ∝ ) / 1

_π exp / 2.463 = ( 1 + 0.90 )

_π exp =1.90 x 2.463

_π exp = 4.679 atm    Ans.

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Question 3 )  Calculate the osmotic pressure  of 20 % ( weight / volume ) anhy.  CaCl₂ solution at 0 ⁰ C ? . Assuming 100 % ionisation.

Solution )

w = 20  gm , V = 100 ml = [100/1000] litre = 0.1 litre

π  =  ? atm , m of CaCl₂ =  40 + 71 = 111   , S = 0.0821 Litre atm K^-1mole ^-1, T = 0 + 273 = 273 K

π N = w ST /m V

π N  = 20  x 0.0821 x 273 / 111  x 0.1

π N = 40.38  atm. Ans.

CaCl₂          ——–>    Ca⁺⁺       +     2 Cl^-1

mole before dissociation                     1                                   0                       0

moles after  dissociation                       1 – ∝                          ∝                      2 ∝

Total moles after  dissociation  = 1 – ∝ + ∝ + 2∝ = 1 + 2∝

% of ∝ = 100 % , ∝ = 100 / 100 = 1.0

π  _exp /π _N = (1 + 2∝ ) / 1

π  _exp / 40.38 = ( 1 +2 x 1)

π  _exp =  3 x  40.38

π  _exp = 121.14  atm    Ans.

Question 4) The average osmotic pressure  of human blood is 7.7 atm. at 40 ⁰ C. Calculate assuming molarity & molality be same.

a) concentration of blood in molarity

b) Freezing point of blood

Given K_f of H₂ O is 1.86

Solution )

a) π  = 7.7 atm , T = 40 + 273 = 313 K

π  = CST

C = π  /ST=7.7 /0.0821 x 313

C = 0.299 = 0.3 M Ans.

b)

ΔT_f  = K_f x molality

K_f = 1.86

molality = 0.3

ΔT_f  = 1.86 x 0.3=0.558

Depression in freezing point (ΔT_f ) = freezing point of solvent – freezing point of solution( blood)

                                                          0.558=  0 –  freezing point of solution(blood) 

  freezing point of solution(blood) =  0 – 0.558

    freezing point of solution (blood) = – 0.558 ⁰C  Ans.