Consider the following pairs of solutions which will be isotonic… — Solution | JEE Chemistry

NVAEasyJEE 2023Abnormal Molar Mass & van't Hoff Factor

Consider the following pairs of solutions which will be isotonic at the same temperature. The number of isotonic pairs is:

$1 \, \text{M}$ NaCl and $2 \, \text{M}$ urea
$1 \, \text{M}$ CaCl $_2$ and $1.5 \, \text{M}$ KCl
$1.5 \, \text{M}$ AlCl $_3$ and $2 \, \text{M}$ Na $_2$ SO $_4$
$2.5 \, \text{M}$ KCl and $1 \, \text{M}$ Al $_2$ (SO $_4$ ) $_3$

Answer

Correct answer:4

Step-by-step solution

Standard Method

Given: Pairs of solutions are to be checked for isotonicity at the same temperature.

Find: The number of isotonic pairs.

Isotonic solutions have the same osmotic pressure. Since osmotic pressure depends on the total concentration of solute particles, we compare the effective particle concentration after dissociation.

For pair 1:

\text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^-

So, $1 \, \text{M}$ NaCl gives

1 \times 2 = 2 \, \text{M particles}

Urea does not dissociate, so $2 \, \text{M}$ urea gives

2 \, \text{M particles}

Hence, pair 1 is isotonic.

For pair 2:

\text{CaCl}_2 \rightarrow \text{Ca}^{2+} + 2\text{Cl}^-

So, $1 \, \text{M}$ CaCl $_2$ gives

1 \times 3 = 3 \, \text{M particles}

Also,

\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-

Thus $1.5 \, \text{M}$ KCl gives

1.5 \times 2 = 3 \, \text{M particles}

Hence, pair 2 is isotonic.

For pair 3:

\text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^-

So, $1.5 \, \text{M}$ AlCl $_3$ gives

1.5 \times 4 = 6 \, \text{M particles}

Also,

\text{Na}_2\text{SO}_4 \rightarrow 2\text{Na}^+ + \text{SO}_4^{2-}

Thus $2 \, \text{M}$ Na $_2$ SO $_4$ gives

2 \times 3 = 6 \, \text{M particles}

Hence, pair 3 is isotonic.

For pair 4:

\text{KCl} \rightarrow \text{K}^+ + \text{Cl}^-

So, $2.5 \, \text{M}$ KCl gives

2.5 \times 2 = 5 \, \text{M particles}

Also,

\text{Al}_2(\text{SO}_4)_3 \rightarrow 2\text{Al}^{3+} + 3\text{SO}_4^{2-}

Thus $1 \, \text{M}$ Al $_2$ (SO $_4$ ) $_3$ gives

1 \times 5 = 5 \, \text{M particles}

Hence, pair 4 is isotonic.

All four pairs are isotonic.

Therefore, the number of isotonic pairs is $4$ .

Ion Count Comparison

Given: We need to compare isotonic pairs.

Find: How many pairs have the same total ion concentration.

Use the shortcut that isotonic solutions at the same temperature must have equal value of concentration $\times$ van't Hoff factor.

Check each pair directly:

$1 \, \text{M}$ NaCl $\Rightarrow 2$ particles, $2 \, \text{M}$ urea $\Rightarrow 2$ particles
$1 \, \text{M}$ CaCl $_2$ $\Rightarrow 3$ particles, $1.5 \, \text{M}$ KCl $\Rightarrow 3$ particles
$1.5 \, \text{M}$ AlCl $_3$ $\Rightarrow 6$ particles, $2 \, \text{M}$ Na $_2$ SO $_4$ $\Rightarrow 6$ particles
$2.5 \, \text{M}$ KCl $\Rightarrow 5$ particles, $1 \, \text{M}$ Al $_2$ (SO $_4$ ) $_3$ $\Rightarrow 5$ particles

Why All Four Pairs Match

Given: Isotonic solutions have the same osmotic pressure.

Find: Number of isotonic pairs.

At the same temperature,

\pi = iCRT

Since $R$ and $T$ are the same, two solutions are isotonic when

i_1 C_1 = i_2 C_2

Here $i$ is the number of particles formed after dissociation.

Applying $iC$ to each pair:

\begin{aligned} \text{NaCl:} &\quad i = 2, \; iC = 2 \times 1 = 2 \\ \text{urea:} &\quad i = 1, \; iC = 1 \times 2 = 2 \\ \text{CaCl}_2: &\quad i = 3, \; iC = 3 \times 1 = 3 \\ \text{KCl:} &\quad i = 2, \; iC = 2 \times 1.5 = 3 \\ \text{AlCl}_3: &\quad i = 4, \; iC = 4 \times 1.5 = 6 \\ \text{Na}_2\text{SO}_4: &\quad i = 3, \; iC = 3 \times 2 = 6 \\ \text{KCl:} &\quad i = 2, \; iC = 2 \times 2.5 = 5 \\ \text{Al}_2(\text{SO}_4)_3: &\quad i = 5, \; iC = 5 \times 1 = 5 \end{aligned}

Conclusion

Each pair has equal effective particle concentration, so every listed pair is isotonic. Hence, the required number is $4$ .

Common mistakes

Ignoring dissociation of ionic solutes is incorrect because osmotic pressure depends on the total number of particles present. Always multiply molarity by the number of ions formed for strong electrolytes.
Treating urea like an ionic solute is wrong because urea does not dissociate in solution. Use van't Hoff factor $i = 1$ for non-electrolytes like urea.
Counting the ions in salts incorrectly can change the result. For example, CaCl $_2$ gives $3$ ions, AlCl $_3$ gives $4$ ions, and Al $_2$ (SO $_4$ ) $_3$ gives $5$ ions, so write the dissociation first before comparing.

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