Question

The solubility product of AgCl is 10–10. The minimum volume (in L) of water required to dissolve 1.722 mg of AgCl is (molecular weight of AgCl = 143.5).

• A. 10 lt.
• B. 2.2 lt.
• C. 1.2 lt.
• D. 20 lt.

Answer 1

Answer: (C)

1.2 lt.

Answer 2

Solubilility of AgCl in water = 10–5 mol/lt.

so volume of water required = $\frac{1.722 \times 10^{- 3}}{143.5 \times 10^{- 5}}$= 1.2 lt.