Consider the following pairs of solution which will be isotonic at the same temperature. The number of pairs of solutions is/are ______.
A. 1 M aq. NaCl and 2 M aq. urea
B. 1 M aq. CaCl$$_2$$ and 1.5 M aq. KCl
C. 1.5 M aq. AlCl$$_3$$ and 2 M aq. Na$$_2$$SO$$_4$$
D. 2.5 M aq. KCl and 1 M aq. Al$$_2$$(SO$$_4$$)$$_3$$

Two solutions are isotonic (have the same osmotic pressure) at the same temperature when they have the same effective concentration, i.e., $$i \times M$$ (van't Hoff factor times molarity) is equal.

For Pair A, 1 M $$\text{NaCl} \to \text{Na}^+ + \text{Cl}^-$$ with $$i = 2$$ so $$i \times M = 2 \times 1 = 2$$; 2 M urea is a non-electrolyte with $$i = 1$$ so $$i \times M = 1 \times 2 = 2$$. Since $$2 = 2$$, they are isotonic.

For Pair B, 1 M $$\text{CaCl}_2 \to \text{Ca}^{2+} + 2\text{Cl}^-$$ with $$i = 3$$ so $$i \times M = 3 \times 1 = 3$$; 1.5 M $$\text{KCl} \to \text{K}^+ + \text{Cl}^-$$ with $$i = 2$$ so $$i \times M = 2 \times 1.5 = 3$$. Since $$3 = 3$$, they are isotonic.

For Pair C, 1.5 M $$\text{AlCl}_3 \to \text{Al}^{3+} + 3\text{Cl}^-$$ with $$i = 4$$ so $$i \times M = 4 \times 1.5 = 6$$; 2 M $$\text{Na}_2\text{SO}_4 \to 2\text{Na}^+ + \text{SO}_4^{2-}$$ with $$i = 3$$ so $$i \times M = 3 \times 2 = 6$$. Since $$6 = 6$$, they are isotonic.

For Pair D, 2.5 M $$\text{KCl}$$ has $$i = 2$$ so $$i \times M = 2 \times 2.5 = 5$$; 1 M $$\text{Al}_2(\text{SO}_4)_3 \to 2\text{Al}^{3+} + 3\text{SO}_4^{2-}$$ with $$i = 5$$ so $$i \times M = 5 \times 1 = 5$$. Since $$5 = 5$$, they are isotonic.

All four pairs are isotonic; therefore the number of isotonic pairs is 4.