A solution contain in 1.9 g …

According to the question, a solution containing 1.9 g per 100 mL of KCl (M = 74.5 g mol^–1 ) is isotonic with a solution containing 3 g per 100 mL of urea (M = 60 g mol^–1 ).
As solution is isotonic, 
{tex}\pi_1{/tex} (urea) = {tex}\pi _2 {/tex} (KCI)
{tex}\Rightarrow {/tex}C₁RT = iC₂RT
{tex}\Rightarrow {/tex}{tex}\frac { n _ { 1 } } { V 1 } = \mathrm { i } \frac { n _ { 2 } } { V 2 }{/tex} ({tex}\because {/tex}V₁ = V₂)
{tex}\Rightarrow {/tex}{tex}\frac { 3 } { 60 } = \mathrm { i } \times \frac { 1.9 } { 74.5 }{/tex}
{tex}\Rightarrow {/tex}i = 1.96
We know that, {tex}\alpha = \frac { i - 1 } { n - 1 }{/tex}
{tex}= \frac { 1.96 - 1 } { 2 - 1 } \\= 0.96{/tex}
= 96%