Osmotic pressure-

source : hindi.theindianwire.com

### Solution )

molarity of solution = decimolar means 0.1 mole/litre = 1/10 mole /litre = no. of moles / volume of solution (litre)

π   =  ? atm  , S = 8.314  J K-1mole-1 , T  =300 K ,n / v = 0.1 mole / litre = [0.1 / 10 -3 ]mole /m3 = 100 mole/m3

π N = n ST / V

π N =  100 x 8.314 x 300

π N=  2.49 x 105 N/m2

K4 [Fe (CN)6]                 ——–>   4 K +        +     [Fe (CN)6] 4-

mole before dissociation                     1                                                  0                       0

moles after  dissociation                    1 – ∝                                          4 ∝                       ∝

Total moles after  dissociation  = 1 – ∝ + 4∝ + ∝ = 1 + 4∝

% of ∝ = 50 % , ∝ = 50 / 100 = 0.50

π  exp / π N = (1 + 4∝ ) / 1

π exp /  2.49 x 105 = ( 1 + 4 x 0.50 )

π exp =   2.49 x 105 x 3

π exp =  7.47 x 105 N/m2   Ans.

### Solution )

π exp = 1.804 atm.

w = 7.6 gm , V = 1250 ml = 1250 / 1000 = 1.25 litre

π   =  ? atm , m of  KBr =  39  + 80  = 119   , S = 0.0821 Litre atm K-1mole-1 , T = 273 + 27 =300 K

π N = w ST /m V

π N = 7.6  x 0.0821 x 300 / 119 x 1.25

π N= 1.258 atm. Ans.

KBr    ——–>    K+        +      Br

mole before dissociation                     1                           0                  0

moles after dissociation                     1 – ∝                    ∝                    ∝

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

π  expπ N = (1 + ∝ ) / 1

1.804 / 1.258 = ( 1 +∝ )

1.804 =1.258 (1 + ∝)

1.804 =  1.258 + 1.258 ∝   Ans.

1.258 ∝ = 1.804 – 1.258

1.258 ∝ = 0.546

∝ = 0.546/ 1.258

∝ = 0.434

% of  ∝ = 0.434 x 100 = 43.4 %  Ans.

i = no. of particles  after  dissociation / no. of particles  before  dissociation

i = ( 1 +∝ ) / 1

i = 1 + 0.434

i = 1.434 Ans.

Alternative method for Vant Hoff factor –

i= π exp N

i = 1.804 / 1.258

i = 1.434 Ans.

### Solution )

w = 7.0 gm , V = 100 ml = 100 / 1000 = 0.1 litre

π   =  20 mm = 20 /760 atm  , m  of protein = ?  , S = 0.0821 Litre atm K-1mole-1 , T = 273 + 37 =310 K

π = w ST /m V

m = wST /π V

m  = 7.0  x 0.0821 x 310 x 760 / 20 x 0.1

m = 67699.66 atm. Ans.

### Solution )

w / V = 3.42 gm / dm3 = [3.42 x 10 -3 /10 -3 ]Kg / m3 ,

π   =  ? N / m2   , m  of  sucrose [C12H22O11] = 342  , S = 8.314 J K-1mole-1 , T = 273 + 27 =300 K

π = w ST /m V

π  = 3.42 x 10 -3 x  8.314 x 300  /  342 x 10 -3

π = 24.95  N/m2 Ans.