JEE Main 2025 Jan 29 Shift 1, Chemistry Q12: Colligative Properties

JEE Main2025Jan 29, Shift 1Chemistry

1.24 g of $A X_{2}$ (molar mass 124 g mol $^{- 1}$ ) is dissolved in 1 kg of water to form a solution with boiling point of 100.0156°C, while 25.4 g of $A Y_{2}$ (molar mass 250 g mol $^{- 1}$ ) in 2 kg of water constitutes a solution with a boiling point of 100.0260°C.

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

Which of the following is correct?

A
$A X_{2}$ and $A Y_{2}$ (both) are completely unionised.
B
$A X_{2}$ and $A Y_{2}$ (both) are fully ionised.
C
$A X_{2}$ is completely unionised while $A Y_{2}$ is fully ionised.
D
$A X_{2}$ is fully ionised while $A Y_{2}$ is completely unionised.

Solution

Use $Δ T_{b} = i K_{b} m$ for each solute to extract the van't Hoff factor $i$ .

$A X_{2}$ : molality $m = \frac{1.24/124}{1} = 0.01 mol k g^{- 1}$ , and $Δ T_{b} = 0.0156 K$ . Hence $i_{A X_{2}} = \frac{0.0156}{0.52 \times 0.01} = 3$ . An $A X_{2}$ salt giving 3 particles per formula unit means complete ionisation ( $A^{2 +} + 2 X^{-}$ ).

$A Y_{2}$ : molality $m = \frac{25.4/250}{2} = 0.0508 mol k g^{- 1}$ , and $Δ T_{b} = 0.0260 K$ . Hence $i_{A Y_{2}} = \frac{0.0260}{0.52 \times 0.0508} \approx 1$ . This is the signature of a non-ionising (molecular) solute.

$∴$ $A X_{2}$ is fully ionised; $A Y_{2}$ is completely unionised.

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JEE Main 2025 Jan 29 Shift 1, Chemistry Q12: Colligative Properties

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