A divalent metal oxide contains 60% of metal. What is atomic weight of metal?
wt of metal= 60 g
wt of oxygen= (100-60) g= 40 g
Eq. wt. of metal = wt. of metal/ wt of oxygen* 8
= 12
As metal is divalent, At. wt= Eq. wt* valency= 12*2 = 24
In an experiment 0.936 g of zinc was converted into the oxide and the weight of oxide formed was
1.165 g. In a second experiment 1.236 g of the metal was converted into the oxide and the weight of
oxide formed was 1.538 g. Show that these results illustrate the law of constant composition.
.
Ans: Solution:
Given,
Experiment 1,
Wt. of zinc = 0.936 g
Wt. of oxide = 1.165 g
Wt. of oxygen = (1.165 – 0.936) g = 0.229 g
The ratio of weight of oxygen is to zinc in zinc oxide is
Experiment 2,
Wt. of zinc = 1.236 g
Wt. of oxide = 1.538 g
Wt. of oxygen = (1.538 – 1.236) g = 0.302 g
The ratio of weight of oxygen is to zinc in this oxide is
The ratio of weight of oxygen to weight of zinc in both the oxides is the same. Hence, given data
illustrate the law of definite proportions.
Alternative Method
In Experiment 1
In Experiment 2
It is observed that % composition of Zn in both experiments is nearly same. Hence, given data illustrate the law of definite proportions