By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : (nearest integer) (Given, molar mass in g mol-1: O = 16, Mg = 24, P = 31)
- 40
- 30
- 20
- 50
Answer
(4) 50
Solution
Phosphorus is oxidised to H3PO4 and finally weighed as magnesium pyrophosphate, Mg2P2O7.
Molar mass of Mg2P2O7 = 2(24) + 2(31) + 7(16) = 222 g mol-1
222 g of Mg2P2O7 contain 62 g of P.
\% P = 62222 1.791.00 100
= 49.99 50
Answer: (4)