Two liquids containing non-volatile substance a(ionizable) and b(non-ionizable) respectively are isotonic. The molarity of solution containing b is 10-3 then what will be the molarity of solution containing a?

Let A dissociates into two parts P and Q

So A -> P +Q

If c be the number of moles of A dissolved in 1L solution initially and a is the degree of dissociation then

Dissociation constant 

K = (c²a²)/(c(1-a))

=>c = (K(1-a))/a²........(1)

Now if i be the vant hoff's factor of the solution of A . So  effective molarity of the solution will be = ic. As the solution of A is isotonic with solution of B having molarity 10^-3. We can say 

ic= 10^-3............(2)

 Again i= 1+ a.........(3)

Given K = 10^-5........(4)

Combining (1),(2),(3) and (4) we get

10^-5×(1-a²)/a² =10^-3

=> 1/a²=101

So  c = (K(1-a))/a²

c=10^-5(1/a²-1/a)

=10^-5(101-✓101)

=5.9×10^-4