0.004 M K$_2$SO$_4$ solution is isotonic with 0.01 M glucose solution. Percentage dissociation of K$_2$SO$_4$ is ___________ (Nearest integer)

Given that isotonic solutions have the same osmotic pressure, we equate the osmotic pressures of the K₂SO₄ and glucose solutions : <br/><br/>$i \times 0.004 \, \text{M} = 0.01 \, \text{M}$ <br/><br/>Here, $i$ is the van't Hoff factor, which accounts for the number of ions the compound dissociates into in the solution. Solving this equation for $i$, we get $i = 2.5$. <br/><br/>For the compound K₂SO₄, which dissociates into 3 ions (2K⁺ and 1 SO₄^2-), the formula for $i$ is given by <br/><br/>$i = 1 + (n-1)\alpha$ <br/><br/>where $n$ is the number of ions the solute dissociates into and $\alpha$ is the degree of dissociation. <br/><br/>Substituting $n=3$ and $i=2.5$ into this formula and solving for $\alpha$ gives <br/><br/>$\alpha = \frac{i-1}{n-1} = \frac{2.5-1}{3-1} = 0.75$ <br/><br/>To convert this into a percentage, we multiply by 100, yielding a percentage dissociation of 75%.