K2 HD I4 is 40% ionized in aqueous solution the value of its vont off factor
Answer (1)
23rd Dec, 2019
Sujal, there is no compound as K2 HD I4 but in case you were asking about K2HgI4 or Potassium mercuric iodide, I have provided the solution below:
Van't Hoff factor is given by the formula: 1+(n-1).(alpha), where alpha is the degree of disassociation and n is the number of molecules on the product side of the reaction.
In the given question, alpha = 40% or 0.4
Also, the reaction upon ionization of K2HgI4 in aqueous solution is:
K2HgI4 -> 2K(+) + HgI4(2-)
So the number of molecules on product side is 2+1=3. Hence n=3.
So van't Hoff factor is 1+(n-1).(alpha)
= 1+(3-1).(0.4)
= 1+2.(0.4)
= 1+0.8
= 1.8
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