You visited us 1 times! Enjoying our articles? Unlock Full Access!

Dipole Moment

0.5 g of an o...

Question

$0.5$ g of an organic substance containing phosphorous was heated with conc. $H N O_{3}$ in the Carius tube. The phosphoric acid thus formed was precipitated with magnesia mixture, $(M g N H_{4} P O_{4})$ , which on ignition gave a residue of $1.0$ g of magnesium pyrrophosphate $(M g_{2} P_{2} O_{7})$ . The percentage of phosphorous in the organic compound is :

55.85 %

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

29.72 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

19.18 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

20.5 %

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $55.85 %$
$P + H N O_{3} ⟶ H_{3} P O_{4} ⟶ (N H_{4} C l + M g C l_{2}) M g N H_{4} (P O_{4})$
$\begin{matrix} ↓ △ M g_{2} P_{2} O_{7} M g n e s i u m p y r r o p h o s p h a t e \end{matrix}$
Molecular weight of $M g_{2} P_{2} O_{7} = 24 \times 2 + 31 \times 2 + 16 \times 7 = 222$
Percentage of $P = \frac{62}{222} \times \frac{W e i g h t o f M g P_{2} O_{7}}{W e i g h t o f c o m p o u n d} \times 100$ $= \frac{62}{222} = \frac{1.0}{0.5} \times 100 = 55.85$ %

Suggest Corrections

Similar questions

0.5

H N O_{3}

(M g N H_{4} P O_{4})

1.0

(M g_{2} P_{2} O_{7})

0.5 g m

H {N O}_{3}

(M g {N H}_{4} {P O}_{4})

1.0 g m

({M g}_{2} P_{2} O_{7})

H N O_{3}

H_{3} P O_{4}

M g (N H_{4}) P O_{4}

M g_{2} P_{2} O_{7}

M g N H_{4} P O_{4}

M g_{2} P_{2} O_{7}

M g_{2} P_{2} O_{7}

M g N H_{4} P O_{4}

M g_{2} P_{2} O_{7}

M g_{2} P_{2} O_{7}

Join BYJU'S Learning Program