Question

Calculating solubility!



The molar solubility of Ag2CO3 is 1.2 x 10^-4 M at 25 degrees Celsius (room temperature).

Express this value in grams per 100.0 ml.

Answer 1

the solubility is given in \(M\), or \(\frac{mol}{L}\), you need to convert that into a \(different\) unit, \(\frac{g}{100.mL}\)

Answer 2

At the end when I get grams, can I just divide it by 100 mL, would that work?

Answer 3

you actually divide it by 10, because in 1000mL, there are 10 100mL increments