Question

For 0.1 M AICI solution, van't hoff factor is (considering 100% dissociation)

1) i = 0 2) i = 2 3) i = 3 4) i = 4 ​

Answer 1

Answer:

The correct answer is option 4) i = 4

Explanation:

considering 100% dissociation that means it is a strong electrolyte

i.e it's degree of dissociation (alpha) = 1

for van't Hoff factor (i) in case of dissociation

apply formula

i = 1 + (n-1) alpha

as we know already alpha = 1

n (number of ions produced on the dissociatio of one formula unit of electrolyte) = 4

AlCl3 = Al3+ & 3Cl-

i = 1 + ( 4-1) 1

finally i = 4