$2.79 \mathrm{~g}$ of an organic compound when heated in Carius tube with conc. $\mathrm{HNO}_{3}$ and…
percentage of $\mathrm{P}$ in the compound is
- $23.33 \%$
- $13.33 \%$
- $33.33 \%$
- $26.66 \%$
Solution
$=\frac{62}{222} \times \frac{\text { wt.of } \mathrm{Mg}_{2} \mathrm{P}_{2} \mathrm{O}_{7}}{\text { wt.of compound }} \times 100$
$=\frac{62}{222} \times \frac{1.332}{2.79} \times 100=13.33 \%$
Asked in: JEE-TOPICTESTS-CHEMISTRY
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