A solution 0.1 M of Na_2SO_4 is dissolved to the extent of 95%. What would be its osmotic pressure at 27^oC?

Given:-Concentration of Na_2SO_4( C ) = 0.1 M
Temperature ( T ) = 27 ℃ = ( 27 + 273 ) K = 300 K
Degree of dissociation ( α) = 0.95
Now,
i = 1 + α( n - 1 )
For Na_2SO_4, n = 3.
Therefore,
i = 1 + 0.95 ( 3 - 1 ) = 2.9
As we know that,
π = i CRT
⇒π = 2.9 × 0.1 × 0.0821 × 300 = 7.14 atm
Hence the osmotic pressure is 7.14 atm.