$2.16$ grams of Cu, on reaction with $H N O_{3}$ , followed by ignition of the nitrate, gave $2.7$ g of copper oxide. In another experiment $1.15$ g of copper oxide, upon reaction with hydrogen, gave $0.92$ g of copper. This data illustrate the law of:

multiple proportions

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definite proportions

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reciprocal proportions

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conservation of mass

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Solution

The correct option is B definite proportions
In the first sample of copper oxide (obtained by the action of nitric acid on Cu), the mass of Cu is 2.16 g and the mass of oxygen is $2.7 - 2.16 = 0.54$ g.
The ratio of the mass of Cu to the mass of oxygen is $\frac{2.16}{0.54} = 4 : 1$
In the second sample of copper oxide (which reacts with hydrogen) , the mass of Cu is 0.92 g and the mass of oxygen is $1.15 - 0.92 = 0.23$ g.
The ratio of the mass of Cu to the mass of oxygen is $\frac{0.92}{0.23} = 4 : 1$
Hence, this illustrates the law of definite Proportions.

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2.16 grams of Cu, on reaction with HNO3, followed by ignition of the nitrate, gave 2.7 g of copper oxide. In another experiment 1.15 g of copper oxide, upon reaction with hydrogen, gave 0.92 g of copper. This data illustrate the law of:

$2.16$ grams of Cu, on reaction with $H N O_{3}$ , followed by ignition of the nitrate, gave $2.7$ g of copper oxide. In another experiment $1.15$ g of copper oxide, upon reaction with hydrogen, gave $0.92$ g of copper. This data illustrate the law of: