Question

The coagulation of 200 ML of position sol took place when 0.73 HCL was added to its without changing the volume much. The flocculation value of HCL for the colloid is.

Options

• 0.365

• 36.5

• 100

• 200

Answer 1

100

Explanation:

No. of moles of HCL = `(0.73  g)/(36.5  g  "mol"^-1)` = 0.00 mol^–1

No. of mill mole of HCL = 0.020

1000 = 20 milli moles

No. of mill moles of HCl required for `20/200 xx 1000` = 100.