QJEEM23Ch000d: JEE Main 2023 Chemistry - Class 12 JEEM23Ch000d Solution: 0.004 M K SO solution is isotonic with 0.01 M glucose solution. …

Given that isotonic solutions have the same osmotic pressure, we equate the osmotic pressures of the K₂SO₄ and glucose solutions :

$i \times 0.004 \, \text{M} = 0.01 \, \text{M}$

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 $i = 2.5$ .

For the compound K₂SO₄, which dissociates into 3 ions (2K⁺ and 1 SO₄^2-), the formula for $i$ is given by

$i = 1 + (n-1)\alpha$

where $n$ is the number of ions the solute dissociates into and $\alpha$ is the degree of dissociation.

Substituting $n=3$ and $i=2.5$ into this formula and solving for $\alpha$ gives

$\alpha = \frac{i-1}{n-1} = \frac{2.5-1}{3-1} = 0.75$

To convert this into a percentage, we multiply by 100, yielding a percentage dissociation of 75%.