1.24 g of AX₂ (molar mass 124 g mol⁻¹) is dissolved in 1 kg of water to form a solution with boiling point of 100.015°C, while 25.4 g of AY₂ (molar mass 250 g mol⁻¹) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C.

K_b(H₂O) = 0.52 kg mol⁻¹

Which of the following is correct?

<p>Mass of $A x_2=1.24 \mathrm{~g}$ (solute)</p> <p>Molarmass of $A X_2=124 \mathrm{~g} \mathrm{~mol}^{-1}$</p> <p>Mass of water $=1 \mathrm{~kg}$ (solvent.)</p> <p>Boiling point of water $=100^{\circ} \mathrm{C}$</p> <p>Boiling point of water after adding solute $A X_2=100.0156^{\circ} \mathrm{C}$</p> <p>Mass of $A Y_2=25.4 \mathrm{~g}$ (solute)</p> <p>Molarmass of $A Y_2=250 \mathrm{~g~mol}^{-1}$</p> <p>Mass of water $=2 \mathrm{~kg}$ (Solvent)</p> <p>Boiling point of water $\mp 100^{\circ} \mathrm{C}$</p> <p>Boiling point of water after adding solute $A y_2=100.0260^{\circ} \mathrm{C}$</p> <p>$\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_2 \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}{ }^{-1}=0.520^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}$</p> <p>The ionisation of $A x_2$ and $A Y_2$ can be determined by calculating Van't Hoff factor.</p> <p>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.</p> <p>Relation: $\Delta T_b=K_b \cdot i \cdot m$</p> <p>$\Delta T_b=T_b-T_b^0$</p> <p>$\Delta T_b \rightarrow$ Boiling point elevation</p> <p>$T_b \rightarrow$ Boiling point of solution (solvent + solute)</p> <p>$T_b^0 \rightarrow$ Boiling point of solvent</p> <p>$K_b \rightarrow$ Molal elevation constant</p> <p>$i \rightarrow$ Van't Hoff factor</p> <p>$$m \rightarrow \text { Molality }=\frac{\text { Number of moles }}{\mathrm{kg} \text { of solvent }}, \text { Moles }=\frac{\text { Mass }}{\text { Molarmass }}$$</p> <p>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).</p> <p>For non-electrolytes, $i=1$</p> <p>$$\begin{aligned} &\begin{aligned} & A X_2: \\ & \text { Moles }=\frac{\text { mass }}{\text { molarmass }}=\frac{1.249}{124 \mathrm{~g~mol^{-1}}}=0.01 \mathrm{~mol} \\ & \text { Molality }=\frac{\text { Moles }}{k_g \text { of solvent }}=\frac{0.01 \mathrm{~mol}}{1 \mathrm{~kg}}=0.01 \mathrm{~mol} \mathrm{~kg}^{-1} \\ & \Delta T_b=K_b \times i \times m \\ & i=\frac{\Delta T_b}{k_b \times m} \\ &=\frac{T_b-T_b^0}{K_b \times m} \end{aligned}\\ &\text { Substitute values as., } \end{aligned}$$</p> <p>$$\begin{aligned} i & =\frac{\left(100.0156^{\circ} \mathrm{C}-100^{\circ} \mathrm{C}\right)}{0.52^{\circ} \mathrm{C} \mathrm{~kg} \mathrm{~mol}^{-1} \times 0.01 \mathrm{~mol} \mathrm{~kg}^{-1}} \\ & =\frac{0.0156^\circ \mathrm{C}}{0.52 \times 0.01{ }^{\circ} \mathrm{C}}=3 \end{aligned}$$</p> <p>$i=3$ means, there are 3 particles in solution after $A X_2$ dissolved in water.</p> <p>$A x_2 \rightarrow 1 A^{+}+2 X^{-} \quad(1+2=3)$</p> <p>$$\begin{aligned} &A x_2 \text { is completely ionised. }\\ &A Y_2: \end{aligned}$$</p> <p>$$\begin{aligned} &\begin{aligned} \text { Moles } & =\frac{\text { Mass }}{\text { Molarmass }}=\frac{25.4 \mathrm{~g}}{250 \mathrm{g~mol}}=0.1016 \mathrm{~mol} \\ \text { Molality } & =\frac{\text { Moles }}{\mathrm{kgof}^{-1} \text { solvent }}=\frac{0.1016 \mathrm{~mol}}{2 \mathrm{~kg}}=0.050 .8 \mathrm{~mol} \mathrm{~kg}^{-1} \\ \Delta T_b & =k_b \times \mathrm{l}^2 \times \mathrm{m} \\ i & =\frac{\Delta T_b}{k_b \times m} \\ & =\frac{T_b-T_b^0}{k_b \times m} \end{aligned}\\ &\text { Substitute values as, } \end{aligned}$$</p> <p>$$i = {{(100.0260^\circ C - 100^\circ C)} \over {0.52^\circ C\,kg\,mo{l^{ - 1}} \times 0.0508\,mol\,k{g^{ - 1}}}}$$</p> <p>$=\frac{0.0260 ^\circ \mathrm{C}}{0.52 \times 0.0508 ^\circ \mathrm{C}}$</p> <p>$= 0.98$</p> <p>$\approx 1$</p> <p>$i=1$ means $A Y_2$ is completely unionised.</p> <p>$A y_2$ not give ionised particles when dissolved in water.</p> <p>So, $A X_2$ is completely ionised and $A Y_2$ is completely unionised.</p> <p>Answer: Option 4) $A x_2$ is fully ionised, $A Y_2$ is completely unionised.</p>