JEE Chemistry 2023 Question with Solution

Given: $0.004 \, \text{M}$ K₂SO₄ solution is isotonic with $0.01 \, \text{M}$ glucose solution.

Find: Percentage dissociation of K₂SO₄.

For isotonic solutions:

$$i(\text{glucose}) = i(\text{K}_2\text{SO}_4)$$

Since glucose is a non-electrolyte, its van't Hoff factor is $1$. Therefore,

$$0.01 = i(\text{K}_2\text{SO}_4) \times 0.004$$

So,

$$i(\text{K}_2\text{SO}_4) = \frac{0.01}{0.004} = 2.5$$

For $\text{K}_2\text{SO}_4$, the total number of ions on complete dissociation is $3$. Using

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

we get

$$2.5 = 1 + (3-1)\alpha$$ $$2.5 = 1 + 2\alpha$$ $$2\alpha = 1.5$$ $$\alpha = 0.75$$

Hence, percentage dissociation is

$$0.75 \times 100 = 75\%$$

Therefore, the percentage dissociation of K₂SO₄ is $75\%$, so the correct option is B.

Given: Two solutions are isotonic.

Find: Degree of dissociation of $\text{K}_2\text{SO}_4$.

Isotonic solutions have equal osmotic pressure, so at the same temperature:

$$i_1 M_1 = i_2 M_2$$

For glucose:

$$i_1 = 1, \quad M_1 = 0.01$$

For $\text{K}_2\text{SO}_4$:

$$M_2 = 0.004$$

Thus,

$$1 \times 0.01 = i \times 0.004$$ $$i = 2.5$$

Now $\text{K}_2\text{SO}_4 \rightarrow 2\text{K}^+ + \text{SO}_4^{2-}$, so $n = 3$. Using

$$i = 1 + (n-1)\alpha$$ $$2.5 = 1 + 2\alpha$$ $$\alpha = \frac{1.5}{2} = 0.75$$

Therefore,

$$\%\text{ dissociation} = 75\%$$

So the correct option is B.