Question

Van't hoff factor for srcl2 at 0.01 m is 1.6. Percent dissociation of srcl2 is

Answer 1

Answer:

Using the equation given below:

 

Where,  

i=van't hoff factor

n=number of particles that dissociate

=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:

 

=0.3

As  is the degree of dissociation

then percent dissociation=0.3×100

                                          =30 percent

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Answer 2

Answer:

Given,

Degree of dissociation of binary electrolyte =  = 40% = 0.40

Explanation: