The amount of phosphorus present in (1.375, g) of (PCl\_3) i... | Chemistry | SolveeIt

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

$0.1 5\, g$

$1.37\,g$

$0.31\, g$

$2.37 \, g$

Answer

$0.31\, g$

Explanation

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 $P$ .

Chemistry Question on Mole concept and Molar Masses