The coagulation of `200 mL` of a positive colloid took place when `0.73 g HCl` was added to it without changing the volume much. The flocculation value of `HCl` for the colloid is
a. `36.5` , b. `100` , c. `200` , d. `150`

Options

A. `0.365`
B. `36.5`
C. `100`
D. `200`

Explanation

The flocculation value is always expressed as millimoles per liter.
`n_(CHl)=(0.73g)/(36.5gmol^(-1))=0.02mol`
No. of millimoles `=(0.02mol)((10^(3)m.mol)/(1mol))`
`=20mmol`
`200mL` of the sol requires `20m` mol of `HCL` . Thus, `1000mL` of the sol will require
`(20)/(200)xx1000=100mmol`