JEE Main 2023 — Solutions Question with Solution

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 $T$,

$\pi =iMRT,$

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 $iM=2\times 1=2.$

Urea ($(\mathrm{N}{\mathrm{H}}_2{)}_2\mathrm{CO}$) does not dissociate in water.

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

So $iM=1\times 2=2.$

Both solutions have $iM=2.$

$\therefore$ They are isotonic.

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2. Pair B: $1\mathrm{M}$ $\mathrm{CaC}{\mathrm{l}}_2$ and $1.5\mathrm{M}$ $\mathrm{KCl}$

Calcium chloride ($\mathrm{CaC}{\mathrm{l}}_2$) dissociates into ${\mathrm{Ca}}^{2+}$ and $2{\mathrm{Cl}}^{-}$.

Hence $i(\mathrm{CaC}{\mathrm{l}}_2)=3.$

So $iM=3\times 1=3.$

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

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

So $iM=2\times 1.5=3.$

Both solutions have $iM=3.$

$\therefore$ They are isotonic.

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3. Pair C: $1.5\mathrm{M}$ $\mathrm{AlC}{\mathrm{l}}_3$ and $2\mathrm{M}$ $\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4$

Aluminum chloride ($\mathrm{AlC}{\mathrm{l}}_3$) dissociates into ${\mathrm{Al}}^{3+}$ and $3{\mathrm{Cl}}^{-}$.

Hence $i(\mathrm{AlC}{\mathrm{l}}_3)=4.$

So $iM=4\times 1.5=6.$

Sodium sulfate ($\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4$) dissociates into $2{\mathrm{Na}}^{+}$ and ${\mathrm{S}{\mathrm{O}}_4}^{2-}$.

Hence $i(\mathrm{N}{\mathrm{a}}_2\mathrm{S}{\mathrm{O}}_4)=3.$

So $iM=3\times 2=6.$

Both solutions have $iM=6.$

$\therefore$ They are isotonic.

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4. Pair D: $2.5\mathrm{M}$ $\mathrm{KCl}$ and $1\mathrm{M}$ $\mathrm{A}{\mathrm{l}}_2(\mathrm{S}{\mathrm{O}}_4{)}_3$

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

So $iM=2\times 2.5=5.$

Aluminum sulfate $\mathrm{A}{\mathrm{l}}_2(\mathrm{S}{\mathrm{O}}_4{)}_3$ dissociates into

$2{\mathrm{Al}}^{3+}+3{\mathrm{S}{\mathrm{O}}_4}^{2-}⟹i(\mathrm{A}{\mathrm{l}}_2(\mathrm{S}{\mathrm{O}}_4{)}_3)=2+3=5.$

So $iM=5\times 1=5.$

Both solutions have $iM=5.$

$\therefore$ 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:

$\boxed{4}.$