Question

$0.124gm$ of an organic compound containing phosphorus gave $0.222gm$ of $M{g}_2{P}_2{O}_7$ the usual analysis. Calculate the percentage of phosphorous in the compound. $(Mg=24,P=31)$

• A. $25$
• B. $15$
• C. $62$
• D. $50$

Answer 1

Answer: (D)

$50$

Mass of the compound=0.12g

Mass of $M{g}_2{P}_2{O}_7$=0.22g

Since 1 mole of $M{g}_2{P}_2{O}_7$ has 2g atoms of P, or

222g of $M{g}_2{P}_2{O}_7$=62g of P

So,0.22 g of $M{g}_2{P}_2{O}_7$ contains P=$\frac{62}{222}X0.22g$

So % of P present in the compound=$\frac{62}{222}X0.22X\frac{100}{0.12}$

=51.2$\approx$50 is the answer.