NaCl (aq) ⇌ Na^+ + Cl^ Initially: C 0 0 Eqb: C Cα Cα Cα nbsp; i=No. of moles after dissociationNo. of m ...

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Van't Hoff Factor

Assuming 25

Question

Assuming 25% ionisation in aqueous solution, what will be the van't Hoff factor for $N a C l$ ?

0.25

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1.25

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0.75

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Solution

The correct option is C 1.25
$N a C l (a q) ⇌ N a^{+} + C l^{-} I n i t i a l l y : C 00 E q b : C - C α C α C α$

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

$i = \frac{C - C α + C α + C α}{C} i = \frac{C + C α}{C} = 1 + α$
Given that :
$α = 0.25$
$i = 1 + (0.25) i = 1.25$

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