Question

 A solution 0.1 M of Na₂SO₄ is dissolved to the extent of 95%. What would be its osmotic pressure at 027 ? C (R = 0.0821 L atm K^-1 mol^-1) .

Answer 1

 Conc. of  Na₂SO₄ = 0.1M 
Extent of dissociation = 95% 
        α = 0.95
\[\ce Na_2\underset{1}SO4 = 0.1M \]

α = 0.95
\[\begin{array}{C}
\ce Na_2 \underset{1}SO4   → 2\underset{0}Na^+ + SO_4^{2-}\\
{(1 -α)\phantom{---}  2\alpha \phantom{---}   \alpha}\end{array}
\]
After dissociation osmotic pressure  (π) = i × CRT 
Total number of particles (i) = 1 - α + 3α 
                                             =1+2α
i =  1+2 × 0.95 = 1 + 1.90 = 2.90

π = 2.90  × 0.0821 × 300 atm

π = 71.427 atm