JEE PYQ: Surface Chemistry Question 21
Question 21 - 2020 (08 Jan Shift 1)
Type: MCQ
As per Hardy-Schulze formulation, the flocculation values of the following for ferric hydroxide sol are in the order:
(a) $\text{K}_3[\text{Fe(CN)}_6] < \text{K}_2\text{CrO}_4 < \text{KBr} = \text{KNO}_3 = \text{AlCl}_3$
(b) $\text{K}_3[\text{Fe(CN)}_6] < \text{K}_2\text{CrO}_4 < \text{AlCl}_3 < \text{KBr} < \text{KNO}_3$
(c) $\text{AlCl}_3 > \text{K}_3[\text{Fe(CN)}_6] > \text{K}_2\text{CrO}_4 > \text{KBr} = \text{KNO}_3$
(d) $\text{K}_3[\text{Fe(CN)}_6] > \text{AlCl}_3 > \text{K}_2\text{CrO}_4 > \text{KBr} > \text{KNO}_3$
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Answer: 0
Solution
According to Hardy-Schulte, coagulation value is inversely proportional to flocculation power.
Order of coagulation power: $\text{K}_3[\text{Fe(CN)}_6] > \text{K}_2\text{CrO}_4 > \text{KBr} = \text{KNO}_3 = \text{AlCl}_3$
So order of flocculation value: $\text{K}_3[\text{Fe(CN)}_6] < \text{K}_2\text{CrO}_4 < \text{KBr} = \text{KNO}_3 = \text{AlCl}_3$
BETA
