Calculate the mass of anhydrous $C a_{3} (P O_{4})_{2}$ present in 250 ml of 0.25 M solution.

15.42 gm

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21.89 gm

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16.65 gm

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19.37 gm

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Solution

The correct option is D 19.37 gm
Moles of $C a_{3} (P O_{4})_{2}, n = M o l a r i t y \times V o l u m e = 0.25 \times 0.25 = 0.0625$ mol

Mass of $C a_{3} (P O_{4})_{2} = n \times M = 0.0625 \times 310 = 19.37 g m$

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