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Question
Find the percentage of phosphorus in the fertilizer superphosphate Ca(H2PO4)2.
Solution
The molecular mass of Ca(H2PO4)2 = 40 + 2(1 × 2 + 31 + 16 × 64)
= 234 g
Mass of phosphorus in 234 g superphosphate = 62 g
∴ Percentage of phosphorus in superphosphate = `62/234 xx 100`
= 26.5%
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