Question

van't hoff factor for SrCl2 at 0.01M is 1.6.Percent dissociation of SrCl2 is?

Answer 1

Answer is 30 %

Explanation. The vant hoff factor is defined as number of particles after the dissociation by the number particles before dissociation.

now according to question ,

given

               $SrCl_{2}\rightleftharpoons Sr^{2+} +2Cl^{1-}$

initial               c                  0               0

at eq'm      c-c$\alpha$              c$\alpha$         2c$\alpha$

where, $\alpha$ = Degree of Dissociation = $\frac{\text{Number of moles dissociated}}{\text{total number of moles}}$

c = Total Number of moles before dissociation

Now,

Total moles after dissocaition=  $c-c\alpha+c\alpha +2c\alpha$    ..........(1)

Total moles before dissociation= c    ......(2)

According to definition of vanthoff factor:    

i= $\frac{c-c\alpha +c\alpha +2c\alpha}{c}$

i= 1 + 2$\alpha$

Now putting the value of 'i' in above eqaution:

1.6= 1+ 2$\alpha$

$\alpha$= 0.3

Converting 0.3 into percentage by multiply it by 100

$\alpha=0.3\times 100$

    $\alpha$= 30%

So percentage of dissociation is 30%

Answer 2

Answer:

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