Q: A metal oxide has the formula Z2O3. It can be reduced by hydrogen to give free metal and water. 0.1596g 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

Solution: Valency of metal in Z_{2}O_{3} = 3

Z_{2}O_{3} + 3H_{2} → 2Z + 3H_{2}O

0.1596g of Z_{2}O_{3} reacts with 6 mg of H_{2}

1g H_{2} will react with g of Z_{2}O_{3}

equivalent weight of Z_{2}O_{3} = 0.1596/0.006 = 26.6g

equivalent weight of Z_{2}O_{3} = 26.6

= eq.wt. of Z + eq. wt. of = E + 8

E + 8 = 26.6

E = 18.6

Atomic weight of Z = 18.6 × 3 = 55.8