lead (pb) forms three oxide A,B and C. The quantity of oxygen in each of the oxides A, B and C is 7.143% , 10.345% and 13.133% respectively, show that the law of multiple proportion is obeyed.
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Answer:
Answer: -
Mass of lead = 207
Mass of oxygen= 16
For the first oxide A
Oxide = 7.143
Lead = 100 - 7.143 = 88.57
Ratio of atoms = \frac{88.57}{207}
207
88.57
: \frac{7.143}{16}
16
7.143
= 0.43 : 0.44
= 1 : 1
For the first oxide B
Oxide = 10.345
Lead = 100 - 10.345 = 89.655
Ratio of atoms = \frac{89.655}{207}
207
89.655
: \frac{710.345}{16}
16
710.345
= 0.43 : 0.646
= 2:3
For the first oxide C
Oxide = 13.133
Lead = 100 - 13.133 = 86.867
Ratio of atoms = \frac{86.867}{207}
207
86.867
: \frac{13.133}{16}
16
13.133
= 0.41 : 0.82
= 1 : 2
Law of multiple proportions, statement that when two elements combine with each other to form more than one compound, the weights of one element that combine with a fixed weight of the other are in a ratio of small whole numbers.
From above we see the valencies of lead are 2+, 3+ and 4+. Hence they are whole numbers and the law of multiple proportions is observed.