Consider the following pairs of solution which will be isotonic at the same temperature. The number of pairs of solutions is / are ___________.

A. $1 ~\mathrm{M}$ aq. $\mathrm{NaCl}$ and $2 ~\mathrm{M}$ aq. urea

B. $1 ~\mathrm{M}$ aq. $\mathrm{CaCl}_{2}$ and $1.5 ~\mathrm{M}$ aq. $\mathrm{KCl}$

C. $1.5 ~\mathrm{M}$ aq. $\mathrm{AlCl}_{3}$ and $2 ~\mathrm{M}$ aq. $\mathrm{Na}_{2} \mathrm{SO}_{4}$

D. $2.5 ~\mathrm{M}$ aq. $\mathrm{KCl}$ and $1 ~\mathrm{M}$ aq. $\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}$

<p>We say that two solutions are isotonic (at the same temperature) if they have the <strong>same osmotic pressure</strong>, $\pi$. For dilute solutions at the same temperature $T$,</p> <p>$ \pi = i\, M\, R\, T, $</p> <p>where </p> <p><p>$M$ = molarity of the solution, </p></p> <p><p>$i$ = van 't Hoff factor (number of particles the solute dissociates into), </p></p> <p><p>$R$ = universal gas constant, </p></p> <p><p>$T$ = absolute temperature.</p></p> <p>Since $R$ and $T$ are common for both solutions (same temperature), two solutions will be isotonic if</p> <p>$ i_1 \, M_1 \;=\; i_2 \, M_2. $</p> <p>Let us check each pair:</p> <hr /> <h3>1. Pair A: $1\,\mathrm{M}$ NaCl and $2\,\mathrm{M}$ urea</h3> <p><p><strong>NaCl</strong> dissociates into $ \mathrm{Na}^+ $ and $ \mathrm{Cl}^- $. </p> <p>Hence $i(\mathrm{NaCl}) = 2$. </p> <p>So $i \, M = 2 \times 1 = 2.$</p></p> <p><p><strong>Urea</strong> ($\mathrm{(NH_2)_2CO}$) does <strong>not</strong> dissociate in water. </p> <p>Hence $i(\mathrm{urea}) = 1$. </p> <p>So $i \, M = 1 \times 2 = 2.$</p></p> <p>Both solutions have $i M = 2.$ </p> <p>$\therefore$ <strong>They are isotonic</strong>.</p> <hr /> <h3>2. Pair B: $1\,\mathrm{M}$ $\mathrm{CaCl_2}$ and $1.5\,\mathrm{M}$ $\mathrm{KCl}$</h3> <p><p><strong>Calcium chloride</strong> ($\mathrm{CaCl_2}$) dissociates into $ \mathrm{Ca}^{2+} $ and $ 2\,\mathrm{Cl}^- $. </p> <p>Hence $i(\mathrm{CaCl_2}) = 3.$ </p> <p>So $i \, M = 3 \times 1 = 3.$</p></p> <p><p><strong>Potassium chloride</strong> ($\mathrm{KCl}$) dissociates into $ \mathrm{K}^+ $ and $ \mathrm{Cl}^- $. </p> <p>Hence $i(\mathrm{KCl}) = 2.$ </p> <p>So $i \, M = 2 \times 1.5 = 3.$</p></p> <p>Both solutions have $i M = 3.$ </p> <p>$\therefore$ <strong>They are isotonic</strong>.</p> <hr /> <h3>3. Pair C: $1.5\,\mathrm{M}$ $\mathrm{AlCl_3}$ and $2\,\mathrm{M}$ $\mathrm{Na_2SO_4}$</h3> <p><p><strong>Aluminum chloride</strong> ($\mathrm{AlCl_3}$) dissociates into $ \mathrm{Al}^{3+} $ and $ 3\,\mathrm{Cl}^- $. </p> <p>Hence $i(\mathrm{AlCl_3}) = 4.$ </p> <p>So $i \, M = 4 \times 1.5 = 6.$</p></p> <p><p><strong>Sodium sulfate</strong> ($\mathrm{Na_2SO_4}$) dissociates into $ 2\,\mathrm{Na}^+ $ and $ \mathrm{SO_4}^{2-} $. </p> <p>Hence $i(\mathrm{Na_2SO_4}) = 3.$ </p> <p>So $i \, M = 3 \times 2 = 6.$</p></p> <p>Both solutions have $i M = 6.$ </p> <p>$\therefore$ <strong>They are isotonic</strong>.</p> <hr /> <h3>4. Pair D: $2.5\,\mathrm{M}$ $\mathrm{KCl}$ and $1\,\mathrm{M}$ $\mathrm{Al_2(SO_4)_3}$</h3> <p><p><strong>Potassium chloride</strong> ($\mathrm{KCl}$) has $i(\mathrm{KCl}) = 2.$ </p> <p>So $i \, M = 2 \times 2.5 = 5.$</p></p> <p><p><strong>Aluminum sulfate</strong> $\mathrm{Al_2(SO_4)_3}$ dissociates into </p> <p>$ 2\,\mathrm{Al}^{3+} \;+\; 3\,\mathrm{SO_4}^{2-} \quad \Longrightarrow i(\mathrm{Al_2(SO_4)_3}) = 2 + 3 = 5. $</p> <p>So $i \, M = 5 \times 1 = 5.$</p></p> <p>Both solutions have $i M = 5.$ </p> <p>$\therefore$ <strong>They are isotonic</strong>.</p> <hr /> <h2>Conclusion</h2> <p>All <strong>four</strong> given pairs $(A, B, C, D)$ satisfy $i_1 M_1 = i_2 M_2$. Therefore, <strong>all of them are isotonic pairs</strong>. </p> <p>Hence, the <strong>number of isotonic pairs</strong> is:</p> <p>$ \boxed{4}. $</p>