Maharashtra BoardEnglish MediumSTD 12 ScienceChemistrySolutions4 Marks
Question
How is van’t Hoff factor related to degree of ionization?
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Answer
Consider $1\ dm^3$ of a solution containing m moles of an electrolyte AxBy.
The electrolyte on dissociation gives $x$ number of $A^{y+}$ ions and y number of $B^{x-}$ ions. Let α be the degree of dissociation.At equilibrium,
$AxBy ⇌ xA^{y+} + yB^{x-}$
For 1 mole of electrolyte : $1 – α, xα, yα$
and For ‘m’ moles of an electrolyte : $m(1 – α), mxα, myα$ are the number of particles.
Total number of moles at equilibrium, will be,
Total moles $= m(1 – α) + mxα + myα$
$= m[(1 – α) + xα + yα]$
$= m[1 + xα + yα – α]$
$= m[1 + α(x + y – 1)]$
The van’t Hoff factor i will be,
$i =\frac{\text { Observed colligative property }}{\text { Theoretical colligative property }} $
$ =\frac{m[1+\alpha(x+y-1)]}{m} \text { } $
$ i =1+\alpha(x+y-1)$
If total number of ions from one mole of electrolyte is denoted by $n$, then $(x + y) = n$
$\therefore i =1+\alpha( n -1)$
$\therefore \alpha( n -1)= i -1$
$\therefore \alpha=\frac{i-1}{n-1} \ldots \ldots(1)$
This is a relation between van’t Hoff factor i and degree of dissociation of an electrolyte.
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