Find the n-factor of Cu in the following chemical changes. (3Cu+8HNO3(dil)→3Cu(NO3)2+4NO+2H2O)

The n-factor of an element in a chemical reaction is the total change in the oxidation number per molecule. In the given reaction, we need to find the n-factor of Cu (Copper).

Step 1: Identify the oxidation states of Cu in both the reactants and the products.

In the reactants, Cu is in its elemental form, so its oxidation state is 0.

In the products, Cu is in Cu(NO3)2. Here, the oxidation state of Cu is +2 (since Nitrate (NO3) has a charge of -1 and there are two of them, making the total charge -2. To balance this, Cu must have a charge of +2).

Step 2: Calculate the change in oxidation state.

The change in oxidation state is the final oxidation state minus the initial oxidation state. So, the change is +2 - 0 = +2.

Step 3: Multiply the change in oxidation state by the number of atoms of the element per molecule.

In this case, there is only one Cu atom per molecule of Cu(NO3)2. So, the total change in oxidation state per molecule is +2 * 1 = +2.

Therefore, the n-factor of Cu in the given reaction is +2.