0.29g of an organic compound containing phosphorous gave 0.887 g of Mg2P2O7 by the usual analysis. Calculate the percentage of phosphorous in the compound.
A
45.8%
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B
85.48%
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C
83.7%
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D
72.2%
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Solution
The correct option is A 83.7%
The mass of the compd = 0.29 g
Mass of Mg2P2O7 = 0.88 g
Since 1 mole of Mg2P2O7 has 2 g atoms of P, or
222 g of Mg2P2O7 = 62 g of P
So, 0.88 g of Mg2P2O7 contains P = 62/222 x 0.88 g
So % P present in the compound = 62/222×0.88×100/0.29