Mole Concept & Stoichiometry Miscellaneous Exercises — Question 30

Given,

112 cm³ of gaseous fluoride has mass = 0.63 g

so, 22400 cm³ of gaseous fluoride will have mass = $\dfrac{0.63}{112}$ x 22400

= 126 g

∴ Relative molecular mass of fluoride = 126 g

The molecular mass = At mass P + At. mass of F

126 = 31 + At. Mass of F

∴ At. Mass of F = 126 - 31 = 95 g

However, At. mass of F = 19

∴ $\dfrac{95}{19}$ = 5

So, 5 atoms of F, hence, the molecular formula = PF₅