By usual analysis, 1.00 g of compound (X) ga | vedclass

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(A) The molar mass of magnesium pyrophosphate $(Mg_2P_2O_7)$ is: $2 	imes 24 + 2 	imes 31 + 7 	imes 16 = 48 + 62 + 112 = 222 \ g \ mol^{-1}$.The mo

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(A) The molar mass of magnesium pyrophosphate $(Mg_2P_2O_7)$ is: $2 	imes 24 + 2 	imes 31 + 7 	imes 16 = 48 + 62 + 112 = 222 \ g \ mol^{-1}$.The moles of $Mg_2P_2O_7$ formed = $\frac{1.79 \ g}{222 \ g \ mol^{-1}} \approx 0.008063 \ mol$.Since each mole of $Mg_2P_2O_7$ contains $2 \ mol$ of phosphorus atoms, the moles of $P$ = $2 	imes 0.008063 = 0.016126 \ mol$.Mass of phosphorus = $0.016126 \ mol 	imes 31 \ g \ mol^{-1} \approx 0.4999 \ g$.Percentage of phosphorus = $\frac{	ext{Mass of } P}{	ext{Mass of compound } X} 	imes 100 = \frac{0.4999}{1.00} 	imes 100 = 49.99\%$.Rounding to the nearest integer, we get $50\%$.

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 $g \ mol^{-1}$; $O=16, Mg=24, P=31$)

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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 $g \ mol^{-1}$; $O=16, Mg=24, P=31$)