1 litre of 1M CuSO4 solution is electrolysed. After passing 2F charge, the molarity of CuSO4 will be:

To find the molarity of CuSO4 after passing 2F charge, we need to use the formula:

M1V1 = M2V2

Where: M1 = initial molarity of CuSO4 solution V1 = initial volume of CuSO4 solution M2 = final molarity of CuSO4 solution V2 = final volume of CuSO4 solution

Given that we have 1 litre (1000 mL) of 1M CuSO4 solution, we can substitute the values into the formula:

1M * 1000 mL = M2 * V2

Since we are passing 2F charge, we know that 1 mole of CuSO4 requires 2 moles of electrons (2F) for electrolysis. Therefore, the number of moles of CuSO4 that will be electrolyzed is:

2F / (96500 C/mol) = 2/96500 mol

Now, we need to determine the final volume of the CuSO4 solution. Since the electrolysis process does not change the volume of the solution, the final volume will also be 1000 mL.

Substituting the values into the formula:

1M * 1000 mL = M2 * 1000 mL

Simplifying the equation:

1M = M2

Therefore, the molarity of CuSO4 after passing 2F charge will remain the same as the initial molarity, which is 1M.