Q.3. 112 cm3
(at STP ) of a gaseous fluoride of phosphorus has a mass of 0.63g. Calculate the
relative molecular mass of the fluoride. If the molecule of fluoride contains one atom of
phosphorus, then determine the formula of the phosphorus fluoride.
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Weight of 112 cm3 of gaseous fluoride is 0.63g
Weight of 22400 cm3 = 0.63/112 × 22400 g = 126 g.
Let PFn = 126
Or, 31 + 19n = 126
Or, 19n = 126 – 31 = 95
Or, n = 95/19 = 5
Hence, formula of Phosphorus fluoride is PF5.
Weight of 22400 cm3 = 0.63/112 × 22400 g = 126 g.
Let PFn = 126
Or, 31 + 19n = 126
Or, 19n = 126 – 31 = 95
Or, n = 95/19 = 5
Hence, formula of Phosphorus fluoride is PF5.
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