0.10g of an organic compound containing phosphorus gave 0.222g of $M g_{2} P_{2} O_{7}$ . Then, the percentage of phosphorus in it is:

62%

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6.2%

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31%

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13%

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Solution

The correct option is A 62%
$% o f P = \frac{A t o m i c m a s s o f p}{m o l . m a s s o f M g_{2} P_{2} O_{7}} \times \frac{m a s s o f M g_{2} P_{2} O_{7} \times 100}{M a s s o f c o m p o u n d}$

$% P = \frac{62}{222} \times \frac{M a s s o f M g_{2} P_{2} O_{7} \times 100}{M a s s o f C o m p o u n d}$

$= \frac{62}{222} \times \frac{0.222 \times 100}{0.10}$ $= 62 %$

Option A is correct.

Suggest Corrections

Similar questions

M g_{2} P_{2} O_{7}

0.1 g

0.222 g of M g_{2} P_{2} O_{7}

M g_{2} P_{2} O_{7}

0.124 g m

0.222 g m

M g_{2} P_{2} O_{7}

(M g = 24, P = 31)

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