0.57g of an organic compound containing Phosphorus gave 2.47g of ammonium phosphomolybdate by usual analysis. Calculate the percentage of Phosphorus in the given compound.

78.5%

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30%

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5.41%

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None of these

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Solution

The correct option is D 5.41%
The mass of the compd = 0.57 g
Mass of Mg2P2O7 = 2.47 g
Since 1 mole of $(N H 4)_{3} P M o_{12} O_{40}$ has 1 g atoms of P, or
1876 g of $(N H_{4})_{3} P M o_{12} O 40$ = 31 g of P
So, 2.47 g of $(N H 4)_{3} P M o_{12} O_{40}$ contains P = 31/1876 x 2.47 g
So % P present in the compound = $\frac{31}{1876} \times 2.47 \times \frac{100}{0.57}$
= 5.41 %
C is correct answer

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