Explanation

Magnesium pyrophosphate has the formula:

\[ Mg_2P_2O_7 \]

First calculate its molar mass:

\[ = 2(24) + 2(31) + 7(16) \]

\[ = 48 + 62 + 112 \]

\[ = 222 \; g \, mol^{-1} \]

Mass of phosphorus present in 1 mole of Mg₂P₂O₇:

\[ = 2 \times 31 = 62 \; g \]

Thus, 222 g of Mg₂P₂O₇ contains 62 g of phosphorus.

Now, given 1.79 g of Mg₂P₂O₇ is obtained.

Phosphorus in 1.79 g:

\[ = \frac{62}{222} \times 1.79 \]

\[ = 0.50 \; g \; (approximately) \]

Original sample mass = 1.00 g

Mass of phosphorus = 0.50 g

Percentage of phosphorus:

\[ = \frac{0.50}{1.00} \times 100 \]

\[ = 50\% \]

Nearest integer = 50

Correct Option: D

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