What mass of Ba3(PO4)2 is contained in 1575 mL of a 0.35M solution of Ba3(PO4)2
1 answer:
Answer:
= 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
<u>=331.81 g</u>
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