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

User Dierk
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1 Answer

5 votes

Answer:

= 331.81 g

Step-by-step 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

User Iago
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