Question

Osmotic pressure of a 4.44% solution of anhydrous CaCl2 was found to be 16.42 atm at 27°C. What is the degree of dissociation of CaCl2 ?

Answer 1

4.44% (w/V) of CaCl2

Answer 2

\[\large \bf \pi=cRT\]
where,
\[\bf \pi=osmotic~pressure,R=Universal~gas~constant,T=temp.(K)\]

Answer 3

\[\large \bf c=\frac{no.of~moles~of CaCl_2}{Vol.of~Solution}\]

Answer 4

but ,you have to determine degree of dissociation,so
\[\large \bf \pi=i \times cRT\]
where,
\[\large \bf i=van't~hoff~factor\]

Answer 5

solve for `i`

Answer 6

and we know that
\[\large \bf i=1+(n-1)\alpha~~~~~~~\rightarrow For~Dissociation\]

Answer 7

where, n=3

Answer 8

and solve for `alpha`

you will get your answer !

Answer 9

hope you understand. @Tushi