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0.5 gm of an organic substance containing phosphorous was heated with conc.. $H N O_{3}$ in the
Carius tube. The phosphoric aid thus formed was precipitated with magnesia mixture $(M g N H_{4} P O_{4})$ which on irnition gave a residue of 1.0 gm of magnesium pyrophosphate $(M g_{2} P_{2} O_{7})$ . The percentage of phosphorous in the organic compound is :

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29.72%

55.85%

19.18%

20.5%

answer is A.

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Detailed Solution

$% o f P = \frac{A t o m i c m a s s o f P}{M o l e c u l a r m a s s o f M g_{2} P_{2} O_{7}} \times \frac{M a s s o f M g_{2} P_{2} O_{7}}{M a s s o f c o m p o u n d} \times 100$

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

$= \frac{62}{222} \times \frac{1}{0.5} \times 100$

= 55.85%

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