JEE Advanced 2014 · Paper 1 · Chemistry Q40

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ChemistryInteger+3Question 40 of 60

$\ce{MX2}$ dissociates into $\ce{M^{2+}}$ and $\ce{X-}$ ions in an aqueous solution, with a degree of dissociation ( $\alpha$ ) of 0.5. The ratio of the observed depression of freezing point of the aqueous solution to the value of the depression of freezing point in the absence of ionic dissociation is

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Answer and worked solution

Official answer: 2

van 't Hoff factor. $\ce{MX2 -> M^{2+} + 2X-}$ gives three particles per formula unit. With degree of dissociation $\alpha$ :
$i=1-\alpha+3\alpha=1+2\alpha=1+2(0.5)=2.$

Ratio. $\Delta T_f\propto i$ , so the observed depression is twice the value without dissociation.

Answer: 2

Topics:van't Hoff factor Colligative properties

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JEE Advanced 2014 · Paper 1, Chemistry question 40