$2.79 \mathrm{~g}$ of an organic compound when heated in Carius tube with conc. $\mathrm{HNO}_{3}$ and…

$2.79 \mathrm{~g}$ of an organic compound when heated in Carius tube with conc. $\mathrm{HNO}_{3}$ and $\mathrm{H}_{3} \mathrm{PO}_{4}$ formed is converted into $\mathrm{MgNH}_{4} \cdot \mathrm{PO}_{4}$ ppt. The ppt. on heating gave $1.332 \mathrm{~g}$ of $\mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7} .$ The
percentage of $\mathrm{P}$ in the compound is
  1. $23.33 \%$
  2. $13.33 \%$
  3. $33.33 \%$
  4. $26.66 \%$

Solution

Percentage of $\vec{F}$
$=\frac{62}{222} \times \frac{\text { wt.of } \mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}}{\text { wt.of compound }} \times 100$
$=\frac{62}{222} \times \frac{1.332}{2.79} \times 100=13.33 \%$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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