In order to cause the coagulation of $100 \mathrm{ml}$ Arsenious sulphide solution, $111.7 \mathrm{mg}$ of $2 \mathrm{M}$ $\mathrm{NaCl}$ solution is used. What will be the coagulation value of $\mathrm{NaCl}$?

Flocculation value is the number of millimoles of the electrolyte needed for 1 litre of the colloidal solution.
Assuming 1 litre of the solution,
mass of 1litre of 2 m NaCl = 1000×1.117 = 1117 g.
Moles of NaCl = 2
Mass of NaCl present in 1117 g of the solution = 2 × 58.5 = 117 g
Mass of NaCl present in 111.7 mg of the solution is $\frac{117 \times 0.1117}{1117}$= 0.0117 g =11.7 mg
milli-moles of NaCl = $\frac{11.7}{58.5}$=0.2

0.2 milli-moles of NaCl required for 100 mL of the solution
Millimoles of NaCl for 1 L of the arsenious sulphide solution
= $\frac{0.2}{100}\times 1000$ = 2