Van't Hoff factor of 60% dissociated K₄[Fe(CN)₆] solution is

For electrolytes like….A₂B₃(tri-bivalent), A₃B₂(bi-trivalent), A₄B(uni-tetravalent)…..n=5

                     Van't Hoff factor, i = $1-\mathrm{\alpha }+\mathrm{n\alpha }$ ;

    when n= 5

                    Van't Hoff factor, i = $1+4\mathrm{\alpha }$ ; 

Ex….Al₂(SO₄)₃, Ca₃(PO₄)₂, K₄[Fe(CN)₆], [Co(NH₃)₆]₂(SO₄)₃

$\mathrm{\alpha }$= Degree of ionization  (or) dissociation ;

Ex ; For 5% ionized electrolyte  $\mathrm{\alpha }=\frac{5}{100}= 0.05 ;$

0.1, 0.15, 0.2, 0.25, 0.3…..for 10%, 15%, 20%, 25%, 30% ionized electrolyte

$\mathrm{\alpha }$ for x% electrolyte = $\frac{\mathrm{x}}{100}$; 

a) i= 1.2….5% ionized  b) i= 1.4…..10% ionized c) i=1.6….15% ionized d) i= 1.8…..20% ionized…..

Hint: For every 5% increase in ionization i.e for every 0.05 units increase in $\text{'}\mathrm{\alpha }\text{'}$ value ‘i’ increases by 0.2 units ;