As per hardy-schulze formulation, the flocculation values of the following for haemoglobin sol are in the order

Options

A. `AlCl_(3) gt K_(3)[Fe(CN)_(6)] gt K_(2)CrO_(4) gt KBr = KNO_(3)`
B. `K_(3)[Fe(CN)_(6)] lt K_(2)CrO_(4) lt KBr = KNO_(3)=AlCl_(3)`
C. `K_(3)[Fe(CN_(6))]gtAlCl_(3) gt K_(2)CrO_(4) gt KBr gt KNO_(3)`
D. `K_(3)[Fe(CN)_(6)] lt K_(2)CrO_(4) lt AlCl_(3) lt KBr lt KNO_(3)`

Answer

B

Explanation

Haemoglobin is a positive charged sol.
So, flocculation power will be in the order
`[Fe(CN)_(6)]^(-3)gt CrO_(4)^(-2)gt Br^(Theta)=NO_(3)^(Theta)=Cl^(Theta)`