osmotic pressure source : SlidePlayer

## osmotic pressure-

### 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.

### 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. source : Biology Stack Exchange

### 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.

### 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)

Kf = 1.86

molality = 0.3