A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. Write the mass of copper deposited at the cathode by calculation.

**OR**

The conductivity of 0.001028 mol L^-1 acetic acid is 4.95×10^-5 S cm^-1. Calculate its dissociation constant if Λ°m for ace

Question

A solution of CuSO4 is electrolysed for 10 minutes with a current of 1.5 amperes. Write the mass of copper deposited at the cathode by calculation.

**OR**

The conductivity of 0.001028 mol L^-1 acetic acid is 4.95×10^-5 S cm^-1. Calculate its dissociation constant if Λ°m for acetic acid is 390.5 S cm^2 mol^-1.

Accepted Answer

>

> [!TLDR]
> Main: use Faraday's laws (Q = It, then relate charge to moles of Cu via the 2-electron reduction of Cu²⁺) to get ≈0.296 g of copper. OR: use molar conductivity and Ostwald's dilution law to get Ka ≈ 1.78×10⁻⁵.
>
> **Main — mass of copper deposited:**
> Charge passed: $Q = I 	imes t = 1.5\ 	ext{A} 	imes (10 	imes 60\ 	ext{s}) = 1.5 	imes 600 = 900\ 	ext{C}$.
> …

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