To find the number of pairs of solutions that are isotonic with each other, we need to consider the van 't Hoff factor ($i$) which represents the number of particles a substance dissociates into in solution. We then use this to find the osmotic pressure of each solution in the pairs, using the formula:

$ \Pi = i \cdot C \cdot R \cdot T $

where:

$ \Pi $ is the osmotic pressure
$ i $ is the van 't Hoff factor
$ C $ is the concentration of the solution
$ R $ is the ideal gas constant
$ T $ is the temperature in Kelvin

Two solutions are isotonic if they have the same osmotic pressure. Therefore, the $i \cdot C$ products for each solution in the pair should be equal.

The van 't Hoff factors for the substances given are:

$ \text{NaCl} $: 2 (NaCl dissociates into Na⁺ and Cl⁻)

$ \text{Urea (NH}_2\text{CONH}_2\text{)} $: 1 (Urea does not dissociate)

$ \text{CaCl}_2 $: 3 (CaCl₂ dissociates into Ca²⁺ and 2 Cl⁻)

$ \text{KCl} $: 2 (KCl dissociates into K⁺ and Cl⁻)

$ \text{AlCl}_3 $: 4 (AlCl₃ dissociates into Al³⁺ and 3 Cl⁻)

$ \text{Na}_2\text{SO}_4 $: 3 (Na₂SO₄ dissociates into 2 Na⁺ and SO₄²⁻)

$ \text{Al}_2(\text{SO}_4)_3 $: 5 (Al₂(SO₄)₃ dissociates into 2 Al³⁺ and 3 SO₄²⁻)

Now, we will compare the $i \cdot C$ products for each pair:

Pair A: $2 \cdot 1 = 2$ and $1 \cdot 2 = 2$, so these are isotonic.

Pair B: $3 \cdot 1 = 3$ and $2 \cdot 1.5 = 3$, so these are isotonic.

Pair C: $4 \cdot 1.5 = 6$ and $3 \cdot 2 = 6$, so these are isotonic.

Pair D: $2 \cdot 2.5 = 5$ and $5 \cdot 1 = 5$, so these are isotonic.

Therefore, all four pairs of solutions are isotonic at the same temperature. The number of pairs of solutions is 4.