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

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