Question #142544
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.
1
Expert's answer
2020-11-09T13:51:03-0500

Solubility in mol/mL =1.3×10−5molL×143.5gAgClmol×1L10×100mL\frac{1.3×10^-5mol }{L} × \frac{143.5g AgCl}{mol} ×\frac{1L}{10×100mL}

= 1.87×10-4 mol/ 100mL

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