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प्रश्न
Find the percentage of phosphorus in the fertilizer superphosphate Ca(H2PO4)2.
उत्तर
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%
संबंधित प्रश्न
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\[\ce{NaCl + AgNO3 -> AgCl + NaNO3}\]
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\[\ce{Na2SO4 + BaCl2 -> BaSO4 + 2NaCl}\]
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