\(0.12 \mathrm{~g}\) of an organic compound ' \(X\) ' containing phosphorus on reacting with magnesia…
\(0.12 \mathrm{~g}\) of an organic compound ' \(X\) ' containing phosphorus on reacting with magnesia mixture gave \(0.22 \mathrm{~g}\) of magnesium pyrophosphate \(\left(\mathrm{Mg}_2 \mathrm{P}_2 \mathrm{O}_7\right)\). The percentage of phosphorus in the compound ' \(X\) ' is
\(45.30 \%\)
\(28.70 \%\)
\(64.25 \%\)
\(51.20 \%\)
Solution
Estimation of phosphorus in organic compound is down by using magnesia mixture. It uses the working formula.
\(\mathrm{P} \%=\frac{62}{222} \times \frac{w_1}{w} \times 100=\frac{62}{222} \times \frac{0.22}{0.12} \times 100=51.20 \%\)