Question

$0.5gm$ of an organic substance containing phosphorous was heated with conc.$H{NO}_3$ in the Carius tube. The phosphoric acid thus formed was precipitated with magnesia mixture $\left(Mg{NH}_4{PO}_4\right)$ which on ignition gave a residue of $1.0gm$ of magnesium pyrophosphate $\left({Mg}_2{P}_2{O}_7\right)$. The percentage of phosphorous in the organic compound is:

• A. $55.85$%
• B. $29.72$%
• C. $19.18$%
• D. $20.5$%

Answer 1

The correct option is A $55.85$%

$P+HN{O}_3⟶H_{3}P{O}_{4}N{H}_{4}Cl+MgC{l}_2---⟩MgN{H}_4\left(P{O}_4\right)$

${Mg}_2{P}_2{O}_7$

Magnesium Pyrrophosphate

Molecular weight of ${Mg}_2{P}_2{O}_7=(24\times 21+31\times 2+16\times 7)$

$=222.$

$\therefore \%$ of $P=\frac{62}{222}\times \frac{\text{Weight of}{Mg}_2{P}_2{O}_7}{\text{Weight of compound}}\times 100$

$=\frac{62}{222}\times \frac{1.0}{0.5}\times 100$

$=55.85\%$

$\therefore$ Option $\left(A\right)$ is correct.