What mass of Ba3 (PO4) 2 is contained in 1575 mL of a 0.35 M solution of Ba3 (PO4) 2?

= 331.81 g

Explanation:

Molarity is calculated by the formula;

Molarity = Moles/volume in liters

Therefore;

Moles = Molarity * Volume in liters

= 0.35 M * 1.575 L

= 0.55125 Moles

But; Molar mass of Ba3 (PO4) 2 is 601.93 g/mol

Thus;

Mass = 0.55125 moles * 601.93 g/mol

= 331.81 g