100 ml Fe(OH)3 is coagulated by 10 ml 1N Na2SO4. Then what will be coagulation value of Na2SO4?

(1)	5
(2)	10
(3)	50
(4)	100

Dear student,

Coagulation value is the minimum millimoles of the electrolyte that brings about the coagulation in 1 litre of the colloidal solution.
$$\begin{gathered} Molarity of N{a}_{2}S{O}_4 solution=\frac{Normality}{Valency} \\ =\frac{1}{2}=0.5M \\ 0.5M solution of N{a}_{2}S{O}_4 contains 0.5 moles of N{a}_{2}S{O}_4 in 1000 ml of solution \\ 0.5 moles is equal to 500 millimoles of N{a}_{2}S{O}_4. \\ 500 millimoles of N{a}_{2}S{O}_4 is present in 1000 ml of solution. \\ Thus, 5 millimoles of N{a}_{2}S{O}_4 is present in 10 ml of solution. \\ 10 ml of solution coagulates 100 ml of Fe{\left(OH\right)}_3 solution. \\ Therefore, 100 ml of N{a}_{2}S{O}_4 solution coagulates 1000 ml of Fe{\left(OH\right)}_3 solution. \\ Therefore, 5\times 10=50 millimoles of N{a}_{2}S{O}_4 coagulates 1000 ml of Fe{\left(OH\right)}_3 solution. \\ The correct option is \left(3\right). \\ Regards \end{gathered}$$