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Q.

A compound X undergoes 100% pentamerisation in a given solvent Y. Correct statements are

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a

Experimental elevation in boiling point $= \frac{Calculated elevation in boiling point}{5}$

b

Observed molar mass of solute $= \frac{Normal molar mass}{5}$

c

van’t Hoff factor of compound is 0.20

d

Observed freezing point × 5 = Normal freezing point

answer is A, B.

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Detailed Solution

For pentamerization $5 A ⇌ A_{5}$
$\begin{array}{l} i = 1 - α + \frac{α}{5} = 1 + α (\frac{1}{5} - 1) = \frac{5 - 4 α}{5} \\ 0 .2 = i = \frac{1}{5}; (α = 1) \end{array}$
$∴$ van’t-Hoff factor is equal to 0.2 and 5 × (Experimental elevation in Boiling point) = Calculated elevation in B.P.

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