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एक यौगिक की प्रतिशत रचना निम्नलिखित है।
`Ca=38.72%`, `P=20.0%` एवं `O=41.28%` , लवण का सूत्र निर्धारित करे और उसका नाम बताएं। [परमाणु द्रव्यमान `Ca=40`, `P=31`, `O=16`]

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अतः लवण का मूलानुपाती सूत्र `=Ca_(3)P_(2)O_(8)=Ca_(3)(PO_(4))_(2)`
अतः लवण कैल्शियम फ़ास्फ़ोट है।

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