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

Question

The amount of phosphorus present in 1.375 g of PCl3 is (atomic weight of P = 31, Cl = 35.5) (A) 0.1 5 g (B) 1.37 g (C) 0.31 g (D) 2.37 g

Tardigrade Admin · Accepted Answer

Molecular weight of PCl3 = 137.5 137.5 g PCl3 contains 31 g of P, Thus, 1.375 g PCl5 contains 0.31 g of P.

Q. The amount of phosphorus present in $1.375 g$ of $PC l_{3}$ is (atomic weight of $P = 31$ , $Cl = 35.5$ )