Answer to Question #142544 in General Chemistry for Sehra Abdi

Question

The solubility of AgCl at 25 C is 1.3 x 10 -5 moles per liter. Express this as a solubility in grams per 100mL. Show the calculations here.

Answer 1

Solubility in mol/mL =$\frac{1.3×10^-5mol }{L} × \frac{143.5g AgCl}{mol} ×\frac{1L}{10×100mL}$

= 1.87×10^-4 mol/ 100mL