Question

In order to cause coagulation of 100 mL arsenious sulphide sol. 111.7 mg of 2 m $NaCl$ solution is used what will be the flocculation value of $NaCl$? (density of the $NaCl$ solution is 1.117 g/ml)

Answer 1

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 molal $NaCl$ $1000\times 1.117$ = 1117 g.
Moles of NaCl = 2
Mass of $NaCl$ present in 1117 g of the solution $=2\times 58.5=117g$
Mass of NaCl present 111.7 mg of the solution $\frac{117\times 0.1117}{1117}=0.0117g=11.7mg$
mmoles of $NaCl$ = $\frac{11.7}{58.5}=0.2$

0.2 mmole of $NaCl$ required for 100 mL of the solution
Millimole of $NaCl$ for 1 litre of the aresenious sulphide solution
$=\frac{0.2}{100}\times 1000=2$