10 g of lead on heating gave 10.78 g of litharge, PbO. (ii) 9.775 g of red lead (Pb O4) yielded on strong heating 9-545 g of litharge. (i) 4-87 g of lead peroxide (PbO2) gave on heating 4.545 g of litharge. Show that these results illustrate the law of multiple proportions.
Answers
Answer:
(i) 1g of Pb will combine with 0.078grams of O2.
(ii) 1g of lead will get combined with 0.11g of O2
(iii) 0.155grams is the correct answer.
Explanation:
Let us consider the first case
Given data
(i) Weight of litharge = 10.78grams
Weight of lead= 10grams
Weight of O2 = 10.78-10 = 0.78grams
Hence, 1g of Pb will combine with 0.078grams of O2.
(ii)
Let us consider the second case:
Given data:
Weight of litharge = 9.775 grams
Weight of Pb for 10.78grams lithrage = 10grams
Hence,9.545grams of litharge will contain
= 10/10.78 * 9.545
= 8.854grams
Also,
Weight of red lead = 9.775grams
Weight of Pb = 8.854grams
Weight of O2 = 9.775-8.854 = 0.921g
Thus, 1g of lead will get combined with 0.11g of O2.
(iii)
Let us consider the third case
Weight of litharge in PbO2 = 4.545grams
Weight of Pb in 4.545grams of litharge = 10/10.78 * 4.545 = 4.216grams
Weight of PbO2 = 4.87 grams
Weight of Pb = 4.216 grams
Weight of O2 = 4.87 - 4.216 = 0.654grams
Weight of O2 will combine with 1g of Pb = 0.654/4.216 = 0.155grams
The law of multiple proportion state that when there are two elements which combine to form another compounds in different proportions, the weight of let's say the first element combining with other is in the ratio of whole numbers.
The weights of oxygen with its oxides is different which is calculated above 0.078/0.11/0.155
Therefore law of multiple proportions are applied for the above elements.
(i) 1g of Pb will combine with 0.078grams of O2.
(ii) 1g of lead will get combined with 0.11g of O2
(iii) 0.155grams is the correct answer.