1 L of 1 M CuSO solution is electrolysed.After passing 2F charge, the molarity of CuSOwill be (Assume no change in volume) :

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

M1V1 = M2V2

Given: Initial molarity (M1) = 1 M Initial volume (V1) = 1 L Final volume (V2) = 1 L (assuming no change in volume) Charge passed (Q) = 2F

Since 1 mole of CuSO4 corresponds to 2 moles of electrons (2F), we can say that the moles of CuSO4 formed after passing 2F charge is equal to the moles of electrons passed.

Now, we can calculate the moles of CuSO4 formed:

Moles of CuSO4 formed = Moles of electrons passed = Q / (2 * F)

Substituting the given values:

Moles of CuSO4 formed = 2F / (2 * F) = 1 mole

Since the final volume (V2) is the same as the initial volume (V1), the final molarity (M2) can be calculated as:

M2 = Moles of CuSO4 formed / Final volume (V2)

Substituting the values:

M2 = 1 mole / 1 L = 1 M

Therefore, the molarity of CuSO4 after passing 2F charge will be 1 M.