Which of the following pairs of solutions are isotonic?

0.1 M $Ca(NO_3)_2$ and 0.1 M $Na_2SO_4$

The correct answer is Option (3) → 0.1 M $Ca(NO_3)_2$ and 0.1 M $Na_2SO_4$

Isotonic solutions have equal osmotic pressure.

$π = iCRT$

• $\pi$ (Osmotic Pressure): Often measured in atmospheres (atm) or Pascals (Pa).
• $i$ (van't Hoff factor): Dimensionless factor for the number of particles produced in solution (e.g., $i = 1$ for glucose, $i = 2$ for $NaCl$).
• $C$ (Molar Concentration): Molarity of the solute in mol/L.
• $R$ (Gas Constant): $0.0821 \text{ L} \cdot \text{atm}/(\text{mol} \cdot \text{K})$ or $8.314 \text{ J}/(\text{mol} \cdot \text{K})$.
• $T$ (Absolute Temperature): Temperature in Kelvin (K).

For same concentration and temperature, isotonic condition depends on van’t Hoff factor (i).

Explanation of Each Option

1. In the first pair, $NH_2CONH_2$ is a non-electrolyte and does not dissociate in water, so its van't Hoff factor is $i = 1$. $NaCl$ dissociates into $Na^+$ and $Cl^-$ giving $i = 2$. Since the number of particles in solution is different, osmotic pressures will not be equal and the solutions are not isotonic.
2. In the second pair, $NH_2CONH_2$ again has $i = 1$, while $MgCl_2$ dissociates into $Mg^{2+}$ and $2Cl^-$ giving $i = 3$. The large difference in particle number means osmotic pressures are not equal, so they are not isotonic.
3. In the third pair, $Ca(NO_3)_2$ dissociates into $Ca^{2+}$ and $2NO_3^-$ giving $i = 3$. $Na_2SO_4$ dissociates into $2Na^+$ and $SO_4^{2-}$ also giving $i = 3$. Since both solutions produce the same number of particles at the same molarity, they will have equal osmotic pressure and hence are isotonic.
4. In the fourth pair, $NaCl$ dissociates into two ions ($i = 2$), whereas glucose does not dissociate ($i = 1$). Therefore, the osmotic pressures are different and the solutions are not isotonic.