Question

A metal oxide has the formula Z₂O₃. It can be reduced by hydrogen to give free metal and water. 0.1596 g of the metal oxide requires 6 mg of hydrogen for complete reduction. The atomic weight of the metal is

(a) 27.9

(b) 159.6

(c) 79.8

(d) 55.8

Answer 1

Correct option is (d) 55.8

Weight of oxygen present in 0.1596 g of metal oxide = 0.006 x 8 = 0.048 g

Weight of element combined with 0.048 g oxygen = 0.1596 - 0.048 = 0.1116g

Eq wt. of element = \(\frac{0.1116 \times 8}{0.048}\) = 18.6

Valency of element Z in Z₂O₃ = 3

Hence Atomic Weight of Z = 3 x 18.6 = 55.8