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First method: <br> Since, `0.8 g` of higher oxide is obtained from `0.72 g` of lower oxide, therefore, the weight of lower oxide that would produce `100 g` higher oxide on oxidation <br> `= (0.72 xx 100)/(0.80) = 90 g` <br> Thus, `90 g` of lower oxide contains as much metal as `100 g` of higher oxide, i.e., `80 g` (given). <br> Hence, `80 g` of metal combines with `10 g` of oxygen in the lower oxide and `20 g` of oxygen in the higher oxide. The weights of oxygen that combine with the same weight of the metal in the two oxides are in the ratio of `10 : 20` or `1 : 2`, The ratio, being simple, proves the law of multiple proportions. <br> Second method : Lower oixde `+ O_(2) rarr` Higher oxide <br> Lower oxide : `0.72 g` <br> Weight of oxide `= 0.72 g` <br> Weight of `M` (fixed weight) `= 0.64 g` <br> Weight of `O_(2) = 0.72 - 0.64 = 0.08 g` of `O_(2)` <br> Higher oxide : `0.2 g` <br> `100 g` of oxide contains `implies 80 g of M` <br> `0.8 g` of oxide contains `implies (80)/(100) xx 0.8 implies 0.64 g of M` <br> weight of `O_(2) = 0.8 - 0.64 = 0.16 g of O_(2)` <br> Ratio of oxygen lower and higher oxide <br> `= 0.08 : 0.16 implies 1 : 2`