QJEEM23Ch8b24: JEE Main 2023 Chemistry - Class 12 JEEM23Ch8b24 Solution: Consider the following pairs of solution which will be isotonic …

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:

$\boxed{4}.$