Question

0.248 g of an organic compound was heated in carius tube with conc. $HN{O}_3$. Produced $H_{3}P{O}_4$ was precipitated as $Mg\left(N{H}_4\right)P{O}_4$, which formed 0.444 g $M{g}_2{P}_2{O}_7$ on ignition What percentage of 'P' is present in compound? (Mg = 24, P = 31)

• A. 50%
• B. 25%
• C. 82%
• D. 32%

Answer 1

Answer: (A)

50%

Organic compound + $HN{O}_3\to H_{3}P{O}_4+$ other product

$H_{3}P{O}_4\to 2Mg\left(N{H}_4\right)\underset{PPt}{P{O}_4}\downarrow \underset{PPt}{M{g}_2{P}_2{O}_7}$

Amount of $M{g}_2{P}_2{O}_7=12\times 2+$

Molecular weight $M{g}_2{P}_2{O}_7=24\times 2+2\times 31+7\times 16$

$=222g$

No. of mole of PPt $M{g}_2{P}_2{O}_7=\frac{0.444}{222}$

$=2\times {10}^{-3}$ mole

$2$ Mole of $Mg\left(N{H}_4\right)P{O}_4$ produce one mole of $M{g}_2{P}_2{O}_{7}PPt$

$2\times {10}^{-3}$ mole $M{g}_2{P}_2{O}_7$ require $2\times 2\times {10}^{-3}$

$Mg\left(N{H}_4\right)P{O}_4$

$=4\times {10}^{-3}$ mole

Amount of phosphorous = $4\times {10}^{-3}\times 31g$

$=224gm\times {10}^{-3}$

$=0.224gm$

% of phosphorous in organic compound $=\frac{0.224}{0.448}\times 100\%$

$=50\%$