Question

Van't Hoff factor for K₃[Fe(CN)₆] is 3.333. What is it's percentage dissociation in water?

Options

• 77.7 %

• 70 %

• 58 %

• 83 %

Answer 1

77.7 %

Explanation:

K₃[Fe(CN)₆],  i = 3.33

\[\ce{K3[Fe(CN)6] -> 3K+ + [Fe(CN)6]^{3-}}\]

1        0     0

1 - α  3α  α

Total no. of Particles = I - α + 3α + α

= 1 + 3α

1 + 3α = 3.33

3α = 2.33

α = `2.33/3`

α % = `2.33/3 xx 100` = 77.7 %