(i)10 g of lead on heating gave 10.78 g of litharge, PbO. (ii) 9.775 g of red lead (Pb_(3)O_(4)) yielded on strong heating 9.545 g of litharge. (iii) 4.87 g of lead peroxide (PbO_(2)) gave on heating 4.545 g of litharge. Show that these results illustrate the law of multiple proportions.

(i)10 g of lead on heating gave 10.78 g of litharge, PbO. (ii) 9.775 g

1.	(i)10 g of lead on heating gave 10.78 g of litharge, PbO. (ii) 9.775 g of red lead (Pb_(3)O_(4)) yielded on strong heating 9.545 g of litharge. (iii) 4.87 g of lead peroxide (PbO_(2)) gave on heating 4.545 g of litharge. Show that these results illustrate the law of multiple proportions.
Answer» Solution :(i) 10 G of lead give litharage `(PBO)=10.78g` `therefore"Oxygen PRESENT"=10.78-10=0.78g` `therefore"1 g of lead combines with oxygen"=(0.78)/(10)g=0.078g` (ii) 10.78 g of LITHARGE contain Pb = 10 g `therefore "9.775 g of lithrage will contain "Pb=(10)/(10.78)xx9.545g=8.854g` Thus, 9.775 g of red lead `(Pb_(3)O_(3))` contain `Pb=8.854 g` `therefore"Oxygen present"=9.775-8.854g=0.92g` `rArr"8.854 g of lead combine with oxygen"=0.92 g` `"1 g of lead combines with oxygen "=(0.92)/(8.854)=0.104g` (iii) 10.78 g of litharge contain Pb = 10 g `therefore"4.545 g of litharge contain Pb"=(10)/(10.78)xx4.545g=4.216g` Thus, 4.87 of lead PEROXIDE `(PbO_(2))` contains `Pb=4.216g` `therefore"Oxygen present"=4.87-4.216=0.654g` `rArr"4.216 g of Pb combine with oxygen "=0.654g` `"1 g of Pb combines with oxygen "=(0.654)/(4.216)=0.155g` `therefore` Ratio of masses of oxygen which combine with fixed mass (1 g) of Pb are in the ratio `0.078:0.104:0.155=1:1.33:1.99=3:4:6.` which is a simple whole number ratio.