Chemistry · Solutions and Colligative Properties

JEE Main 2025 — 29 January, Morning Shift — Question 21

If $A_{2} B$ is $30 \%$ ionised in an aqueous solution, then the value of van't Hoff factor (i) is $\qquad$ $\times 10^{-1}$ .

Answer: 16

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{A}_{2} \mathrm{~B} \rightarrow 2 \mathrm{~A}^{+}+\mathrm{B}^{-2} ; \mathrm{y}=3$ $\alpha=0.3$ $i=1+(y-1) \alpha$ $=1+(3-1)(0.3)=1.6=16 \times 10^{-1}$ 0.1 mole of compound ' $S$ ' will weigh $\qquad$ g. (Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16$ )

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Exam: JEE Main 2025
Paper: 29 January, Morning Shift
Subject: Chemistry
Chapter: Solutions and Colligative Properties
Topic: Abnormal Colligative Properties - van't Hoff Factor

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 100.0156^circ C , while…

Answer: 16

Step-by-step solution

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