A. = Normal ,molecular , mass Observed ,molecular ,mass

Van't Hoff factor i

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MCQ

Van't Hoff factor $i$

✓
$ = \frac{{{\rm{Normal \,molecular\, mass}}}}{{{\rm{Observed \,molecular \,mass}}}}$
B
$ = \frac{{{\rm{Observed \,molecular\, mass}}}}{{{\rm{Normal\, molecular\, mass}}}}$
C
Less than one in case of dissociation
D
More than one in case of association

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Answer

Correct option: A.

$ = \frac{{{\rm{Normal \,molecular\, mass}}}}{{{\rm{Observed \,molecular \,mass}}}}$

a
Van't Hoff factor ($i$) is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass.

$i=\frac{\text { normal molecular mass }}{\text { observed molecular mass }}$

For strong electrolytes which dissociate completely, it is equal to the number of ions produced in the solution.

In case of association, two or more molecules associate to form a single molecule so the van't hoff factor will be less than $1$ .

In case of dissociation one molecule is forming two or more ions so the van't Hoff factor will be more than $1$ .