Find the n-factor of Cu in the following chemical changes. (3Cu+8HNO3(dil)→3Cu(NO3)2+4NO+2H2O)
A
2
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B
3
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C
1
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D
4
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Solution
The correct option is A 2 Steps for finding valency factor/n-factor: 1. Find the oxidation state (O.S.) of metal in reactant side. 2. Find the oxidation state of metal in product side. 3. Take the magnitude of difference of two oxidation state |O.S.Product−O.S.Reactant|×number of atom for the valency factor (Cu→Cu(NO3)2) O.S. of Cu in product side =x+2×(−1)=0 x=+2 O.S. of Cu in reactant side =2
valency factor=number of Cu atom×|O.S.Product−O.S.Reactant| valency factor=1×|+2−0| valency factor=2