We say that two solutions are isotonic (at the same temperature) if they have the same osmotic pressure, $\pi$. For dilute solutions at the same temperature $T$,

$ \pi = i\, M\, R\, T, $

where

$M$ = molarity of the solution,

$i$ = van 't Hoff factor (number of particles the solute dissociates into),

$R$ = universal gas constant,

$T$ = absolute temperature.

Since $R$ and $T$ are common for both solutions (same temperature), two solutions will be isotonic if

$ i_1 \, M_1 \;=\; i_2 \, M_2. $

Let us check each pair:

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1. Pair A: $1\,\mathrm{M}$ NaCl and $2\,\mathrm{M}$ urea

NaCl dissociates into $ \mathrm{Na}^+ $ and $ \mathrm{Cl}^- $.

Hence $i(\mathrm{NaCl}) = 2$.

So $i \, M = 2 \times 1 = 2.$

Urea ($\mathrm{(NH_2)_2CO}$) does not dissociate in water.

Hence $i(\mathrm{urea}) = 1$.

So $i \, M = 1 \times 2 = 2.$

Both solutions have $i M = 2.$

$\therefore$ They are isotonic.

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2. Pair B: $1\,\mathrm{M}$ $\mathrm{CaCl_2}$ and $1.5\,\mathrm{M}$ $\mathrm{KCl}$

Calcium chloride ($\mathrm{CaCl_2}$) dissociates into $ \mathrm{Ca}^{2+} $ and $ 2\,\mathrm{Cl}^- $.

Hence $i(\mathrm{CaCl_2}) = 3.$

So $i \, M = 3 \times 1 = 3.$

Potassium chloride ($\mathrm{KCl}$) dissociates into $ \mathrm{K}^+ $ and $ \mathrm{Cl}^- $.

Hence $i(\mathrm{KCl}) = 2.$

So $i \, M = 2 \times 1.5 = 3.$

Both solutions have $i M = 3.$

$\therefore$ They are isotonic.

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3. Pair C: $1.5\,\mathrm{M}$ $\mathrm{AlCl_3}$ and $2\,\mathrm{M}$ $\mathrm{Na_2SO_4}$

Aluminum chloride ($\mathrm{AlCl_3}$) dissociates into $ \mathrm{Al}^{3+} $ and $ 3\,\mathrm{Cl}^- $.

Hence $i(\mathrm{AlCl_3}) = 4.$

So $i \, M = 4 \times 1.5 = 6.$

Sodium sulfate ($\mathrm{Na_2SO_4}$) dissociates into $ 2\,\mathrm{Na}^+ $ and $ \mathrm{SO_4}^{2-} $.

Hence $i(\mathrm{Na_2SO_4}) = 3.$

So $i \, M = 3 \times 2 = 6.$

Both solutions have $i M = 6.$

$\therefore$ They are isotonic.

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4. Pair D: $2.5\,\mathrm{M}$ $\mathrm{KCl}$ and $1\,\mathrm{M}$ $\mathrm{Al_2(SO_4)_3}$

Potassium chloride ($\mathrm{KCl}$) has $i(\mathrm{KCl}) = 2.$

So $i \, M = 2 \times 2.5 = 5.$

Aluminum sulfate $\mathrm{Al_2(SO_4)_3}$ dissociates into

$ 2\,\mathrm{Al}^{3+} \;+\; 3\,\mathrm{SO_4}^{2-} \quad \Longrightarrow i(\mathrm{Al_2(SO_4)_3}) = 2 + 3 = 5. $

So $i \, M = 5 \times 1 = 5.$

Both solutions have $i M = 5.$

$\therefore$ They are isotonic.

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Conclusion

All four given pairs $(A, B, C, D)$ satisfy $i_1 M_1 = i_2 M_2$. Therefore, all of them are isotonic pairs.

Hence, the number of isotonic pairs is:

$ \boxed{4}. $