The coagulation of $200ml$ of a positive colloid took place when $0.73 gHCl$ was added to it without changing the volume much. The flocculation value of $HCl$ for the colloid is

From the given data,

200ml of the sol requires 0.73 g of $HCl$. Molecular weight of $HCl$ is 36.5 g/mol. Moles present in 0.73 g of HCl.

$$\begin{gathered} \mathrm{Moles}=\frac{0.76 \mathrm{g}}{36.5 \mathrm{g}/\mathrm{mol}} \\ =0.02 \mathrm{mol} \mathrm{or} 20 \mathrm{mmol} \end{gathered}$$

If 200 ml of sol requires 20 mmol of HCl, then mmol (HCl) requires by 1 L or 1000 mL of sol:

$\mathrm{mmol}=\frac{1000 \mathrm{mL}}{200 \mathrm{mL}}\times 20 \mathrm{mmol}=100 \mathrm{mmol}$

Therefore 1000ml will require 100mmol of HCl.

Hence, the correct option is (C).