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  4. The amount of phosphorus present in 1.375 g of PCl3 is (atomic weight of P = 31, Cl = 35.5)

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Q. The amount of phosphorus present in $1.375\, g$ of $PCl_3$ is (atomic weight of $P = 31$, $Cl = 35.5$)

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Solution:

Molecular weight of $PCl_3 = 137.5$
$137.5\, g \,PCl_3$ contains $31\, g$ of $P$,
Thus, $1.375\, g\, PCl_5$ contains $0.31\, g$ of $P$.