Answer the following: When a current of is passed through a solution for , of copper is deposited…

Formula (Faraday's first law of electrolysis): $w = \frac{\text{Atomic weight} \times I \times t}{n \times F}$ $\text{Atomic weight} = \frac{w \times n \times F}{I \times t}$ For copper deposition from $\text{CuSO}_4$: $\text{Cu}^{2+} + 2e^- \longrightarrow \text{Cu} \quad (n = 2)$ $t = 25 \times 60 = 1500\text{ seconds}$ $F = 96,500\text{ C mol}^{-1}$ Substitution: $\text{Atomic weight} = \frac{0\cdot370\text{ g} \times 2 \times 96,500\text{ C mol}^{-1}}{0\cdot75\text{ A} \times 1500\text{ s}}$ $= \frac{71,410}{1125} = 63\cdot48\text{ g mol}^{-1}$ Final answer: 63.48 g mol^-1

Metal A will liberate $\text{H}_2$ gas from dil. $\text{H}_2\text{SO}_4$. Reason: The standard reduction potential of hydrogen is $E^\circ_{\text{H}^+/\text{H}_2} = 0\cdot00\text{ V}$. Metal A has a negative reduction potential ($E^\circ = -0\cdot76\text{ V}$), which is lower than that of hydrogen, making it a stronger reducing agent capable of spontaneously reducing $\text{H}^+$ ions to $\text{H}_2$ gas ($E^\circ_{\text{cell}} = +0\cdot76\text{ V} > 0$). Metal B has a positive reduction potential ($+0\cdot34\text{ V}$), so it cannot reduce $\text{H}^+$ ions.

Specific conductivity ($\kappa$) is the conductance of ions present in a unit volume ($1\text{ cm}^3$) of the solution. On dilution, although the total volume increases and the degree of dissociation may increase, the number of current-carrying ions per unit volume decreases, leading to a decrease in specific conductivity.