i) Molar conductivity
Given : C = 0.01 M;
\(\kappa=1.5 \times 10^{-4} \mathrm{~S} \mathrm{~cm}^{-1} \)
\(=1.5 \times 10^{-2} \mathrm{~S} \mathrm{~m}^{-1} \)
\(\lambda_{\text {cation }}^{0}=248.2 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\lambda_{\text {anion }}^{0}=51.8 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1} \)
\(\Lambda_{m}^{0}=\frac{\kappa \times 10^{-3}}{C} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1} \)
\(=\frac{1.5 \times 10^{-2} \times 10^{-3}}{0.01} \)
\(=1.5 \times 10^{-3} \mathrm{~S} \mathrm{~m}^{2} \mathrm{~mol}^{-1}\)
ii) \(\alpha=\frac{\Lambda_{m}}{\Lambda_{m}^{0}}\)
\(\Lambda_{\mathrm{m}}^{0}=\lambda_{\text {cation }}^{0}+\lambda_{\text {anion }}^{0} \)
\(=(248.2+51.8) \mathrm{S} \mathrm{cm}^{2} \mathrm{~mol}^{-1} \)
\(=300 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\)
\(=300 \times 10^{-4} \mathrm{Sm}^{2} \mathrm{~mol}^{-1}  \)
\(\alpha =\frac{1.5 \times 10^{-3}}{300 \times 10^{-4}}=0.05\)
iii) \(\mathrm{K}_{\mathrm{a}} =\frac{\alpha^{2} \mathrm{C}}{1-\alpha} \)
\(\mathrm{K}_{\mathrm{a}} =\frac{(0.05)^{2} \times(0.01)}{1-0.05}=2.6 \times 10^{-5} \)
(or) 
\(\mathrm{K}_{\mathrm{a}} =\alpha^{2} \mathrm{C}  \)
\(=(0.05)^{2} \times(0.01)  \)
\(\mathrm{K}_{\mathrm{a}} =2.5 \times 10^{-5}\)

Given : I = 1.608A;
t = 50 min = 50 \(\times\) 60 = 3000sec ; S = 250 mL;
C = 0.5 M ; \(\eta=100%\)%
i) Q = It     (I = Q / t)
= 1.608 x 3000 = 4824 Columb
No. of Faradays of electricity \(=\frac{4824}{96500}=0.04 \mathrm{~F}\)
ii) Electrolysis of CuSO₄
\(\mathrm{Cu}_{(\mathrm{aq})}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}_{(\mathrm{s})}\)
\(\therefore\) 2 F electricity will deposit 1 mole of Cu²⁺ to Cu ​​​​
\(\therefore\) 0.05 F electricity will deposit 
\(x=\frac{1 \times 0.05}{2}=0.025\) mole
Initial No. of moles of Cu²⁺ in 250 ml of solution \(=\frac{0.5}{1000} \times 250 \) mole = 0.125 moles
\(\therefore\) No. of moles of Cu²⁺ after elettrolysis = 0.125 - 0.025 = 0.1 mole
\(\therefore\) concentration of Cu²⁺
\(=\frac{0.1}{250} \times 1000=0.4 \mathrm{M}\)

(a) If we apply DC current through the conductivity cell, it will lead to the electrolysis of the solution taken in the cell.
(b) So, AC current is used for this measurement to prevent electrolysis.

The oxidation potential of M₁ is more +ve than the oxidation potential of Fe which indicates that it will prevent iron from rusting.

\((E^{0}_{ox})_{Fe^{2+}|Fe} = -0.44\)  and 
\((E^{0}_{red})_{Cu^{2+}|Cu} = 0.34\)
These +ve emf values shows that iron will oxidise and copper will get reduced i.e., the vessel will dissolve. Hence it is not possible to store copper sulphate in an iron vessel.