By usual analysis, 1.00 g of compow1d (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is: (nearest integer)
(Given, molar mass in $$gmol^{-1}$$ : 0 = 16, Mg = 24, P = 31 )

1.00 g of compound X gives 1.79 g of $$Mg_2P_2O_7$$ (magnesium pyrophosphate).

Molar mass of $$Mg_2P_2O_7 = 2(24) + 2(31) + 7(16) = 48 + 62 + 112 = 222$$ g/mol.

Each mole of $$Mg_2P_2O_7$$ contains 2 moles of P.

Moles of $$Mg_2P_2O_7 = \frac{1.79}{222}$$

Moles of P = $$\frac{2 \times 1.79}{222} = \frac{3.58}{222}$$

Mass of P = $$\frac{3.58}{222} \times 31 = \frac{110.98}{222} = 0.4999 \approx 0.50$$ g

% P = $$\frac{0.50}{1.00} \times 100 = 50\%$$

The answer is Option 4: 50.