Question

112cm³ 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]


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35 points!!

Answer 1

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.

Answer 2

Answer:

Step-by-step explanation:

V=112cm3

Mass =0.63g

By expression :

Mass / Molecular mass = Volume /22400

then,

Molecular mass =126g

So,

PFn=126

31+19n=126

19n=126-31

n=5then,

Formula :-

PF5