Question

1)At 25.0∘C, the molar solubility of silver chromate in water is 1.10×10⁻¹²M . Calculate the solubility in grams per liter.

Answer 1

given

molar solubility of silver chromate = 1.1 x 10-12 M

the units of molar sulibility is M = mol / L

so

molar solubility of silver chromate = 1.1 x 10-12 mol/L

we know that

mass in grams = moles x molar mass

so

to get the solubility in grams

the formula is

solubility in grams = molar solubility x molar mass

we know that

molar mass silver chromate = 331.73 g/mol

so

solubility in grams = 1.1 x 10-12 x 331.73

solubility in grams = 3.65 x 10-10 g/L

so

the solubility in grams per liter of silver chromate is 3.65 x 10-10