JEE Main 2025 — Solutions Question with Solution

Mass of $A{x}_2=1.24\mathrm{g}$ (solute)

Molarmass of $A{X}_2=124\mathrm{g}{\mathrm{mol}}^{-1}$

Mass of water $=1\mathrm{kg}$ (solvent.)

Boiling point of water $=100^{\circ }\mathrm{C}$

Boiling point of water after adding solute $A{X}_2=100.0156^{\circ }\mathrm{C}$

Mass of $A{Y}_2=25.4\mathrm{g}$ (solute)

Molarmass of $A{Y}_2=250{\mathrm{g}\mathrm{mol}}^{-1}$

Mass of water $=2\mathrm{kg}$ (Solvent)

Boiling point of water $\mp 100^{\circ }\mathrm{C}$

Boiling point of water after adding solute $A{y}_2=100.0260^{\circ }\mathrm{C}$

${\mathrm{K}}_{\mathrm{b}}\left({\mathrm{H}}_2\mathrm{O}\right)=0.52\mathrm{K}\mathrm{kg}\mathrm{mol}{}^{-1}=0.520^{\circ }\mathrm{C}\mathrm{kg}{\mathrm{mol}}^{-1}$

The ionisation of $A{x}_2$ and $A{Y}_2$ can be determined by calculating Van't Hoff factor.

The phenomenon given in the question is elevation in boiling point. The boiling point of solvent increases when another compound (solute) is added to it.

Relation: $\Delta T_b=K_b\cdot i\cdot m$

$$\Delta T_b=T_b-T_b^0$$

$\Delta T_b\to$ Boiling point elevation

$T_b\to$ Boiling point of solution (solvent + solute)

$T_b^0\to$ Boiling point of solvent

$K_b\to$ Molal elevation constant

$i\to$ Van't Hoff factor

$$m\to \text{ Molality }=\frac{\text{ Number of moles }}{\mathrm{kg}\text{ of solvent }},\text{ Moles }=\frac{\text{ Mass }}{\text{ Molarmass }}$$

For $A{X}_2$ and $A{Y}_2$ (solute), Van't Hoff factor represents how many particles a solute dissociates into when dissolved in a solvent (water).

For non-electrolytes, $i=1$

$$\begin{array}{rl}& \begin{array}{rl}& A{X}_2:\\ & \text{ Moles }=\frac{\text{ mass }}{\text{ molarmass }}=\frac{1.249}{124\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}}=0.01\mathrm{mol}\\ & \text{ Molality }=\frac{\text{ Moles }}{k_g\text{ of solvent }}=\frac{0.01\mathrm{mol}}{1\mathrm{kg}}=0.01\mathrm{mol}{\mathrm{kg}}^{-1}\\ & \Delta T_b=K_b\times i\times m\\ & i=\frac{\Delta T_b}{k_b\times m}\\ & =\frac{T_b-T_b^0}{K_b\times m}\end{array}\\ & \text{ Substitute values as., }\end{array}$$

$$\begin{array}{rl}i& =\frac{\left(100.0156^{\circ }\mathrm{C}-100^{\circ }\mathrm{C}\right)}{0.52^{\circ }\mathrm{C}\mathrm{kg}{\mathrm{mol}}^{-1}\times 0.01\mathrm{mol}{\mathrm{kg}}^{-1}}\\ & =\frac{0.0156^{\circ }\mathrm{C}}{0.52\times 0.01{}^{\circ }\mathrm{C}}=3\end{array}$$

$i=3$ means, there are 3 particles in solution after $A{X}_2$ dissolved in water.

$$A{x}_2\to 1A^{+}+2X^{-}(1+2=3)$$

$$\begin{array}{rl}& A{x}_2\text{ is completely ionised. }\\ & A{Y}_2:\end{array}$$

$$\begin{array}{rl}& \begin{array}{rl}\text{ Moles }& =\frac{\text{ Mass }}{\text{ Molarmass }}=\frac{25.4\mathrm{g}}{250\mathrm{g}\mathrm{mol}}=0.1016\mathrm{mol}\\ \text{ Molality }& =\frac{\text{ Moles }}{{\mathrm{kgof}}^{-1}\text{ solvent }}=\frac{0.1016\mathrm{mol}}{2\mathrm{kg}}=\mathrm{0.050.8}\mathrm{mol}{\mathrm{kg}}^{-1}\\ \Delta T_b& =k_b\times {\mathrm{l}}^2\times \mathrm{m}\\ i& =\frac{\Delta T_b}{k_b\times m}\\ & =\frac{T_b-T_b^0}{k_b\times m}\end{array}\\ & \text{ Substitute values as, }\end{array}$$

$$i=\frac{(100.0260^{\circ }C-100^{\circ }C)}{0.52^{\circ }Ckgmo{l}^{-1}\times 0.0508molk{g}^{-1}}$$

$=\frac{0.0260^{\circ }\mathrm{C}}{0.52\times 0.0508^{\circ }\mathrm{C}}$

$=0.98$

$\approx 1$

$i=1$ means $A{Y}_2$ is completely unionised.

$A{y}_2$ not give ionised particles when dissolved in water.

So, $A{X}_2$ is completely ionised and $A{Y}_2$ is completely unionised.

Answer: Option 4) $A{x}_2$ is fully ionised, $A{Y}_2$ is completely unionised.