1 liter of 8M aqueous solution of copper sulphate is electrolysed with copper electrodes using a current of 200 amperes for 965 seconds . The final cocentration (in M) of the electrolyte is _________

To find the final concentration of the electrolyte, we can use the formula:

M1V1 = M2V2

Where: M1 = initial concentration of the electrolyte V1 = initial volume of the electrolyte M2 = final concentration of the electrolyte V2 = final volume of the electrolyte

Given: Initial concentration (M1) = 8M Initial volume (V1) = 1 liter Current (I) = 200 amperes Time (t) = 965 seconds

First, we need to calculate the amount of copper deposited during electrolysis using Faraday's law:

n = (I * t) / F

Where: n = number of moles of copper deposited I = current in amperes t = time in seconds F = Faraday's constant (96500 C/mol)

Substituting the given values: n = (200 A * 965 s) / 96500 C/mol n = 2 mol

Since copper sulfate (CuSO4) is a 1:1 electrolyte, the number of moles of copper sulfate consumed is also 2 mol.

Next, we need to calculate the final volume (V2) of the electrolyte. Since copper is deposited on the cathode, the volume of the solution decreases. The molar volume of copper is 7.1 cm^3/mol. Therefore, the decrease in volume is:

ΔV = n * molar volume ΔV = 2 mol * 7.1 cm^3/mol ΔV = 14.2 cm^3

Converting the volume to liters: ΔV = 14.2 cm^3 * (1 L / 1000 cm^3) ΔV = 0.0142 L

The final volume (V2) is: V2 = V1 - ΔV V2 = 1 L - 0.0142 L V2 = 0.9858 L

Finally, we can calculate the final concentration (M2) using the formula:

M2 = (M1 * V1) / V2 M2 = (8 M * 1 L) / 0.9858 L M2 ≈ 8.12 M

Therefore, the final concentration of the electrolyte is approximately 8.12 M.