`K_(2)HgI_(4)` is 40% ionised in aqueous solution. The value of its van't Hoff factor (i) is :

Options

A. 1.8
B. 2.2
C. 2
D. 1.6

Answer

A

Explanation

`K_(2)[HgI_(4)] = 2K^(+) + [HgI_(4)]^(-2)`
Moles of solute left undissoicated =`1 - (40)/(100) = 0.6`
Moles of solute formed `= 0.4 xx 3 = 1.2`
Total moles of solute ` = 1.2 +0.6=1.8`
`I = (1.8)/(1) = 1.8`