Question

A metallic fluoride contains 67.78% fluorine. One mole of that compound contains 1.5 times the number of atoms in one mole of dry fluorine gas. Find the atomic weight of the metal.

• A. 27 gm
• B. 54 gm
• C. 13.5 gm
• D. none of these

Answer 1

The correct option is A 27 gm

Given,
67.8 g fluorine is present in 100 g of metallic fluoride
1 mole of fluorine gas - 2 N fluorine atoms
$Numberoffluorineionsinonemoleofcompound=1.5\times 2N=3N$
$Theformulaofthecompound\to M{F}_3$
$57gfluorineispresentin1moleofM{F}_3$
$67.8gfluorine100gM{F}_3$
$57gfluorine⟶\frac{5700}{67.8}=84gm$
Mass of $M{F}_3=84gm=x+19\ast 3$
$Atomicweightofmetal=27g$