1.24 g of $\mathrm{AX}_{2}$ (molar mass $124 \mathrm{~g} \mathrm{~mol}^{-1}$) is dissolved in 1 kg of water to form a solution with boiling point of $100.0156^{\circ} \mathrm{C}$, while 25.4 g of $\mathrm{AY}_{2}$ (molar mass $250 \mathrm{~g} \mathrm{~mol}^{-1}$) in 2 kg of water constitutes a solution with a boiling point of $100.0260^{\circ} \mathrm{C}$. $\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_{2} \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ Which of the following is correct?

(D) $\mathrm{AX}_{2}$ is fully ionised while $\mathrm{AY}_{2}$ is completely unionised.

(A) $\mathrm{AX}_{2}$ and $\mathrm{AY}_{2}$ (both) are completely unionised.

(B) $\mathrm{AX}_{2}$ and $\mathrm{AY}_{2}$ (both) are fully ionised.

(C) $\mathrm{AX}_{2}$ is completely unionised while $\mathrm{AY}_{2}$ is fully ionised.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Solutions

1.24 g of

AX_{2}

(molar mass

124 g mol^{- 1}

) is dissolved in 1 kg of water to form a solution with boiling point of

100.015 6^{\circ} C

, while 25.4 g of

AY_{2}

(molar mass

250 g mol^{- 1}

) in 2 kg of water constitutes a solution with a boiling point of

100.026 0^{\circ} C

K_{b} (H_{2} O) = 0.52 K kg mol^{- 1}

Which of the following is correct?

Answer: (D)

Both solutions are analysed through

Δ T_{b} = i K_{b} m

.

For

AX_{2}

: moles

= 1.24/124 = 0.01

, molality

= 0.01 m

in 1 kg water, and

Δ T_{b} = 100.0156 - 100 = 0.0156 K

i = \frac{0.0156}{0.52 \times 0.01} = 3

An i value of 3 matches complete dissociation

AX_{2} \to A^{2 +} + 2 X^{-}

AX_{2}

is fully ionised.

For

AY_{2}

: moles

= 25.4/250 = 0.1016

, molality

= 0.1016/2 = 0.0508 m

, and

Δ T_{b} = 100.0260 - 100 = 0.0260 K

i = \frac{0.0260}{0.52 \times 0.0508} \approx 1

An i value of 1 means no dissociation occurs at all:

AY_{2}

stays completely unionised.

AX_{2}

is fully ionised while

AY_{2}

is completely unionised.

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option