
When 1.375 g of cupric oxide is reduced on heating in a current of hydrogen, the weight of copper remains 1.098 g. In another experiment, 1.179 g of copper is dissolved in nitric acid and resulting copper nitrate converted into cupric oxide by ignition. The weight of cupric oxide formed is 1.476 g. This is in agreement with:
A.Law of definite proportions
B.Law of mass conservation
C.Law of momentum conservation
D.None of the above
Answer
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Hint: There are five laws which are the basis of chemical reactions. They are, law of conservation of mass, law of constant composition, law of multiple proportions, law of reciprocal proportions and Gay Lussac’s law of combining volumes.
Complete step by step answer:
Here, first we have to calculate the percentage composition of oxygen in both the experiments using the given quantities.
Given that in the first experiment, the mass of cupric oxide is 1.375 g and the weight of copper left is 1.098.
So, the mass of oxygen=1.375 g-1.098 g=0.277 g
Now, we have to calculate the percentage composition of oxygen in copper oxide (CuO).
${\rm{Percentage}}\,{\rm{of}}\,{\rm{O}} = \dfrac{{0.277}}{{1.375}} \times 100$
$ \Rightarrow \% \,{\rm{of}}\,{\rm{O}} = {\rm{20}}{\rm{.15\% }}$
In the second experiment, the weight of copper taken is 1.179 g and the mass of copper oxide (CuO) formed is 1.476 g.
Now, we have to calculate the amount of oxygen present.
Oxygen present=1.476 g-1.179 g=0.297 g
Now, we have to calculate the percentage composition of oxygen in CuO.
${\rm{Percentage}}\,{\rm{of}}\,{\rm{O}} = \dfrac{{0.297}}{{1.476}} \times 100$
$ \Rightarrow \% \,{\rm{of}}\,{\rm{O}} = {\rm{20}}{\rm{.12\% }}$
So, we get to know that in both the experiments the percentage composition of oxygen is approximately the same.
Law of definite proportions states that combination of a pure chemical compound is always of the same elements in a fixed proportion by mass.
Hence, law of definite proportions is applied here.
So, the correct answer is Option A.
Note: Law of conservation of mass says that in all chemical reactions and physical changes, there is equal mass of reactants and products. Also this law says that both creation and destruction of matter is impossible. Therefore, this law is known by the name 'indestructibility of matter’.
Complete step by step answer:
Here, first we have to calculate the percentage composition of oxygen in both the experiments using the given quantities.
Given that in the first experiment, the mass of cupric oxide is 1.375 g and the weight of copper left is 1.098.
So, the mass of oxygen=1.375 g-1.098 g=0.277 g
Now, we have to calculate the percentage composition of oxygen in copper oxide (CuO).
${\rm{Percentage}}\,{\rm{of}}\,{\rm{O}} = \dfrac{{0.277}}{{1.375}} \times 100$
$ \Rightarrow \% \,{\rm{of}}\,{\rm{O}} = {\rm{20}}{\rm{.15\% }}$
In the second experiment, the weight of copper taken is 1.179 g and the mass of copper oxide (CuO) formed is 1.476 g.
Now, we have to calculate the amount of oxygen present.
Oxygen present=1.476 g-1.179 g=0.297 g
Now, we have to calculate the percentage composition of oxygen in CuO.
${\rm{Percentage}}\,{\rm{of}}\,{\rm{O}} = \dfrac{{0.297}}{{1.476}} \times 100$
$ \Rightarrow \% \,{\rm{of}}\,{\rm{O}} = {\rm{20}}{\rm{.12\% }}$
So, we get to know that in both the experiments the percentage composition of oxygen is approximately the same.
Law of definite proportions states that combination of a pure chemical compound is always of the same elements in a fixed proportion by mass.
Hence, law of definite proportions is applied here.
So, the correct answer is Option A.
Note: Law of conservation of mass says that in all chemical reactions and physical changes, there is equal mass of reactants and products. Also this law says that both creation and destruction of matter is impossible. Therefore, this law is known by the name 'indestructibility of matter’.
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