Question

what is the percentage dissociation of srcl2 if van't hoff factor for srcl2 at 0.01M is 1.6.

Answer 1

Using the equation given below:

$i=1+ \alpha (n-1)$  

Where,  

i=van't hoff factor

n=number of particles that dissociate

$\alpha$=degree of dissociation

Dissociation of SrCl₂ , can be seen in the equation given below:

SrCl₂----->Sr²⁺ + 2Cl⁻  

So number of paticles dissociate=3

i=1.6

putting all the values in the equation:

$1.6=1+ \alpha (3-1)$  

$\alpha$ =0.3

As $\alpha$ is the degree of dissociation

then percent dissociation=0.3×100

                                          =30 percent

Answer 2

hope it helpsss~~~~~