KCl (aq) ⇌ K^+ + Cl^ Initially: C 0 0 Eqb: C Cα Cα Cα nbsp; i=No. of moles after dissociationNo. of mol ...

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Van't Hoff Factor

If KCl is 50

Question

If $K C l$ is 50% ionised in the aqueous solution what will be the van't Hoff factor of $K C l$ in the solution?

2

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1.5

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1.25

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1

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Solution

The correct option is B $1.5$
$K C l (a q) ⇌ K^{+} + C l^{-} I n i t i a l l y : C 00 E q b : C - C α C α C α$

$i = \frac{No. of moles after dissociation}{No. of moles before dissociation}$

$i = \frac{C - C α + C α + C α}{C} i = \frac{C + C α}{C} = 1 + α$
Given that :
$α = 0.5$
$i = 1 + (0.5) i = 1.5$

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