Question

112 cm³ of a gaseous fluoride of phosphorus has a mass of 0.63 g. Calculate the relative molecular mass of the fluoride. If the molecule of the fluoride contains only one atom of phosphorus, then determine the formula of the phosphorus fluoride. [ F = 19, P = 31].

Answer 1

As we know, 22400 cm³ at S.T. P. of all gases weigh equal to the relative molecular weights.

Now, 112 cm³ of phosphorus fluoride at S.T.P. weighs 0.63 g

∴ 22400 cm³ of phosphorus fluoride at STP weights = `(0.63 xx 22400)/112`

= 126 g

Hence, the relative molecular mass of fluoride = [126] amu

As it t is given that the fluoride has only one atom of phosphorus, its formula can be assumed as PF_n.

Molecular mass of PF_n = 31 + n × 19

∴ 31+ 19n = 126   ...[As mass of P = 31, F = 19]

n = 5

∴ The formula of phosphorus fluoride = PF₅.