A decimolar solution of K₄[Fe(CN)₆] at 300K is 50% dissociated, then, osmotic pressure of the solution is

For  K₄[Fe(CN)₆] , no. of particles after dissociation(n)=5

Given, $\alpha =50\%=\frac{50}{100}=0.5$  ,  C=0.1M

$$\begin{gathered} \mathrm{\alpha }=\frac{\mathrm{i}-1}{\mathrm{n}-1} \\ 0.5=\frac{\mathrm{i}-1}{5-1} \\ \mathrm{i}=3 \end{gathered}$$

$\mathrm{\pi }=\mathrm{iCRT}=3\times 0.1\times 0.0821\times 300=7.38\mathrm{atm}.$