4.4 g of an oxide of nitrogen gives 2.24 L of nitrogen. 60 g of another oxide of nitrogen gives 22.4 L of nitrogen at STP. This data illustrates the law of:
A
conservation of mass
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B
multiple proportions
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C
constant proportions
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D
gaseous volume
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Solution
The correct option is B multiple proportions For the first oxide, the ratio of the mass of the oxide to the volume of the nitrogen liberated is 4.42.24=1.96
For the second oxide, the ratio of the mass of the oxide to the volume of the nitrogen liberated is 6022.4=2.67 These two are in the ratio 1.962.67=11.33=34 or 3:4
Hence, this illustrates the Law of multiple proportions.
Note: In the above example, we have used volume of nitrogen which is proportional to the mass of nitrogen.