Question

0.1M solution of K2SO4 is dissolved to the extent of 90%.What would be its osmotic pressure at 27°C?

Answer 1

π(osmotic pressure) = i x CRT

R(ideal gas constant) = 0.08206 L atm mol^-1 K^-1
T = 27°C = 27 + 273 = 300 K

degree of dissociation = 90/100 = 0.9 
for dissociation 
i = 1-α + nα
for K2SO4 , n = 3 
i = 1-0.9 + 3 x 0.9 = 0.1 + 2.7 = 2.8 
 
π = 2.8 x 0.1 x 0.08206 x 300
π = 6.89304 pascal

i hope it will help you 
regards

Answer 2

Explanation:

hope it helps............