The osmotic pressure of a lithium chloride $0.118\;M$ aqueous solution at $27^oC$ is $5.49\;atm. $ Calculate the percentage of dissociation of the salt!
Lithium chloride: $LiCl$ Van't Hoff factor: $\Pi=i[LiCl]RT$ $i=\frac{\Pi}{[LiCl]RT}=\frac{5.49}{0.118\cdot 0.082\cdot 300}=1.89$ Degree of dissociation: $i=1+\alpha(n-1)$ $1.89=1+\alpha(2-1)$ $\alpha=0.89$ Percentage of dissociation: $\%=\alpha\cdot 100= 89$
