Question

In the quantitative estimation of phosphorous by using magnesia mixture i.e. $\left(M{g}_2{P}_2{O}_7\right)$, the formula used is:

(where W is the mass of $M{g}_2{P}_2{O}_7$ and w is the mass of the compound)

• A. Percentage of $P=\frac{62}{222}\times \frac{W\times 100}{w}$
• B. Percentage of $P=\frac{31}{222}\times \frac{W\times 100}{w}$
• C. Percentage of $P=\frac{62}{222}\times \frac{w\times 100}{W}$
• D. Percentage of $P=\frac{31}{222}\times \frac{w\times 100}{W}$

Answer 1

Answer: (A)

Percentage of $P=\frac{62}{222}\times \frac{W\times 100}{w}$

Molecular weight of $M{g}_2{P}_2{O}_7=222gmo{l}^{-1}$

Atomic weight of $P=31gmo{l}^{-1}$

Since,

222g of $M{g}_2{P}_2{O}_7$ contain 62g ($2\times 31$) of $P$

$\therefore$ 1 g of $M{g}_2{P}_2{O}_7$ contain $\frac{62}{222}g$ of $P$

Thus,

Mass of $P$ in W g of $M{g}_2{P}_2{O}_7=W\times \frac{62}{222}g$

Percentage of P in the compound = $W\times \frac{62}{222}\times \frac{100}{w}$

Hence, option $\left(A\right)$ is the correct answer.