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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].
Solution
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 PFn.
Molecular mass of PFn = 31 + n × 19
∴ 31+ 19n = 126 ...[As mass of P = 31, F = 19]
n = 5
∴ The formula of phosphorus fluoride = PF5.
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