Question

For coagulating 200 ml of arsenious sulphide sol, 10 ml of 1 m NaCl solution is required find out the flocculation value of NaCl.

पर्याय

• 55 mill moles

• 40 mill moles

• 22 mill moles

• 50 mill moles

Answer 1

50 mill moles

Explanation:

10 ml of 1 m NaCl solution contain 10 mill-moles

New, 200 ml of arsenious sulphide (As₂S₃) required for coagulation NaCl = 10 mill-moles 1000 ML of solution required for coagulation

NaCl = ${\displaystyle \frac{105}{200}\times 1000}$ = 50 mill moles