0.5 gm of an organic substance containing phosphorous was heated with conc. HNO 3 in the Carius tube. The phosphoric acid thus formed was precipitated with magnesia mixture( .MgNH 4 PO 4) which on ignition gave a residue of 1.0 gm of magnesium pyrophosphate ( Mg 2 P 2 O 7). The percentage of phosphorous in the organic compound is:

Question

0.5 gm of an organic substance containing phosphorous was heated with conc. HNO 3 in the Carius tube. The phosphoric acid thus formed was precipitated with magnesia mixture( .MgNH 4 PO 4) which on ignition gave a residue of 1.0 gm of magnesium pyrophosphate ( Mg 2 P 2 O 7). The percentage of phosphorous in the organic compound is: (A) 55.85 % (B) 29.72 % (C) 19.18 % (D) 20.5 %

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/edb26d731f20ca0db89ed3005ce7b58e-.png" /> Mw of Mg 2 P 2 O 7=24 × 2+31 × 2+16 × 7=222 Percentage of P =(62/222) × ( text Weight of Mg 2 P 2 O 7/ text Weight of compound ) × 100 =(62/222)=(1.0/0.5) × 100=55.85 %

$55.85%$

$29.72%$

$19.18%$

$20.5%$

Mw of $M g_{2} P_{2} O_{7} = 24 \times 2 + 31 \times 2 + 16 \times 7 = 222$
Percentage of $P = \frac{62}{222} \times \frac{Weight of M g _{2} P _{2} O _{7}}{Weight of compound} \times 100$
$= \frac{62}{222} = \frac{1.0}{0.5} \times 100 = 55.85%$

55.85%

29.72%

19.18%

20.5%

Mw of Mg2​P2​O7​=24×2+31×2+16×7=222 Percentage of P=22262​× Weight of compound Weight of Mg2​P2​O7​​×100 =22262​=0.51.0​×100=55.85%

$55.85%$

$29.72%$

$19.18%$

$20.5%$

Mw of $M g_{2} P_{2} O_{7} = 24 \times 2 + 31 \times 2 + 16 \times 7 = 222$
Percentage of $P = \frac{62}{222} \times \frac{Weight of M g _{2} P _{2} O _{7}}{Weight of compound} \times 100$
$= \frac{62}{222} = \frac{1.0}{0.5} \times 100 = 55.85%$