Redox Reactions and Electrochemistry MCQs with Answers – Part 6 (NEET Chemistry)

Explanation: \(\textbf{Identify the concentration-cell driving force:}\) Both electrodes involve: \[ \mathrm{Cu^{2+}+2e^-\rightleftharpoons Cu} \] The standard reduction potentials are identical. The more concentrated compartment has the greater actual reduction potential and acts as the cathode. The lower-concentration compartment acts as the anode. \(\textbf{Calculate the reversible concentration-cell emf:}\) For the copper(II)/copper couple: \[ n=2 \] Therefore: \[ E_{\mathrm{rev}}=\frac{0.0591}{2}\log\left(\frac{a_{\mathrm{high}}}{a_{\mathrm{low}}}\right) \] Substituting: \[ E_{\mathrm{rev}}=\frac{0.0591}{2}\log\left(\frac{0.100}{0.00100}\right) \] \[ E_{\mathrm{rev}}=\frac{0.0591}{2}\log100 \] \[ E_{\mathrm{rev}}=0.0591\ \mathrm V \] \(\textbf{Relate bridge conductivity to its resistance:}\) For a uniform conducting bridge: \[ R=\rho\frac{\ell}{A} \] Conductivity and resistivity satisfy: \[ \rho=\frac1\kappa \] Therefore: \[ R=\frac{\ell}{\kappa A} \] \(\textbf{Calculate the bridge resistance:}\) \[ R=\frac{3.00}{(2.00\times10^{-2})(1.00)} \] \[ R=150\ \Omega \] The units are consistent because \(\kappa\) is expressed in \(\mathrm{S\,cm^{-1}}\), length in centimetres and area in square centimetres. \(\textbf{Convert the operating current:}\) \[ I=0.200\ \mathrm{mA} \] \[ I=2.00\times10^{-4}\ \mathrm A \] \(\textbf{Calculate the internal potential drop:}\) \[ V_{\mathrm{loss}}=IR \] \[ V_{\mathrm{loss}}=(2.00\times10^{-4})(150) \] \[ V_{\mathrm{loss}}=0.0300\ \mathrm V \] \(\textbf{Determine the terminal voltage:}\) For the stipulated discharging-cell model: \[ V_{\mathrm{terminal}}=E_{\mathrm{rev}}-IR \] \[ V_{\mathrm{terminal}}=0.0591-0.0300 \] \[ V_{\mathrm{terminal}}=0.0291\ \mathrm V \] \(\textbf{Check the direction of energy delivery:}\) The terminal voltage remains positive: \[ 0.0291\gt 0 \] The cell can therefore deliver electrical power to the external load under the specified conditions. The terminal voltage is smaller than the reversible emf because part of the potential is lost across the bridge resistance. \(\textbf{Interpret the connection between thermodynamics and ionic transport:}\) The reversible emf is determined by the copper-ion activity ratio. The bridge resistance is determined by its conductivity and geometry. These quantities enter different parts of the operating-cell calculation. A highly resistive ionic pathway can substantially reduce the useful terminal voltage even though the electrode reaction quotient remains unchanged. The problem therefore requires a Nernst calculation, a conductivity-to-resistance conversion and an operating-voltage calculation. The reversible emf should not be confused with the voltage delivered to the load while current passes through the electrolyte bridge. \(\textbf{Final answer:}\) \[ \boxed{(E_{\mathrm{rev}},V_{\mathrm{terminal}})=(0.0591\ \mathrm V,\ 0.0291\ \mathrm V)} \]

Explanation: \(\textbf{Isolate the conductivity caused by dissolved AgCl:}\) The measured conductivity includes the background contribution of water. Under the specified additive model, this contribution must be subtracted before determining the solubility of the salt. \[ \kappa_{\mathrm{AgCl}}=\kappa_{\mathrm{sat}}-\kappa_{\mathrm{water}} \] \[ \kappa_{\mathrm{AgCl}}=(1.88-0.50)\times10^{-6} \] \[ \kappa_{\mathrm{AgCl}}=1.38\times10^{-6}\ \mathrm{S\,cm^{-1}} \] \(\textbf{Construct the limiting molar conductivity:}\) Dissolved AgCl supplies one silver ion and one chloride ion: \[ \mathrm{AgCl(s)\rightleftharpoons Ag^++Cl^-} \] By Kohlrausch's law: \[ \Lambda_m^\circ(\mathrm{AgCl})=\lambda^\circ_{\mathrm{Ag^+}}+\lambda^\circ_{\mathrm{Cl^-}} \] \[ \Lambda_m^\circ(\mathrm{AgCl})=62.0+76.0 \] \[ \Lambda_m^\circ(\mathrm{AgCl})=138.0\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Infer the solubility in pure water:}\) At the very low dissolved concentration, the molar conductivity is approximated by its limiting value. The conductivity relation is: \[ \Lambda_m=\frac{1000\kappa}{c} \] Therefore: \[ s=\frac{1000\kappa_{\mathrm{AgCl}}}{\Lambda_m^\circ} \] \[ s=\frac{1000(1.38\times10^{-6})}{138.0} \] \[ s=1.00\times10^{-5}\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate the solubility product:}\) In pure water, dissolution supplies equal concentrations of both ions: \[ [\mathrm{Ag^+}]=[\mathrm{Cl^-}]=s \] Thus: \[ K_{sp}=[\mathrm{Ag^+}][\mathrm{Cl^-}] \] \[ K_{sp}=s^2 \] \[ K_{sp}=(1.00\times10^{-5})^2 \] \[ K_{sp}=1.00\times10^{-10} \] \(\textbf{Apply the common-ion condition:}\) The second solution contains a much larger chloride concentration supplied by NaCl. Its total free chloride concentration is approximately: \[ [\mathrm{Cl^-}]\approx0.0100\ \mathrm{mol\,L^{-1}} \] The additional chloride contributed by dissolved AgCl is negligible compared with this amount. \(\textbf{Calculate the new solubility:}\) \[ K_{sp}=s_{\mathrm{NaCl}}[\mathrm{Cl^-}] \] \[ s_{\mathrm{NaCl}}=\frac{1.00\times10^{-10}}{0.0100} \] \[ s_{\mathrm{NaCl}}=1.00\times10^{-8}\ \mathrm{mol\,L^{-1}} \] \(\textbf{Interpret the coupled measurements:}\) The conductivity measurement provides the extremely small concentration of dissolved AgCl without requiring a direct mass measurement. Kohlrausch's law connects that conductivity to the ionic contributions at high dilution. The resulting solubility determines the solubility product. Adding a common chloride ion then suppresses dissolution without changing the equilibrium constant at the same temperature under the ideal model. Using the uncorrected conductivity would overestimate the solubility, while equating solubility in NaCl solution with solubility in pure water would ignore the common-ion effect. \(\textbf{Final answer:}\) \[ \boxed{\left(K_{sp},s_{\mathrm{NaCl}}\right)=\left(1.00\times10^{-10},\ 1.00\times10^{-8}\ \mathrm{mol\,L^{-1}}\right)} \]

Explanation: \(\textbf{Identify the appropriate concentration variable:}\) The strong-electrolyte relation is linear in the square root of concentration, not concentration itself. The two measured square-root concentrations are: \[ \sqrt{0.0100}=0.100 \] \[ \sqrt{0.0400}=0.200 \] \(\textbf{Use the first experimental point:}\) \[ 140=\Lambda_m^\circ-A(0.100) \] \(\textbf{Use the second experimental point:}\) \[ 130=\Lambda_m^\circ-A(0.200) \] \(\textbf{Eliminate the limiting molar conductivity:}\) Subtract the two equations: \[ 140-130=A(0.200-0.100) \] \[ 10=0.100A \] \[ A=100 \] The value of \(A\) carries the units appropriate to the chosen concentration convention. \(\textbf{Determine the limiting molar conductivity:}\) Substituting into the first equation: \[ 140=\Lambda_m^\circ-100(0.100) \] \[ 140=\Lambda_m^\circ-10 \] \[ \Lambda_m^\circ=150\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Predict molar conductivity at the third concentration:}\) For: \[ c=0.0900\ \mathrm{mol\,L^{-1}} \] \[ \sqrt c=0.300 \] Therefore: \[ \Lambda_{m,3}=150-100(0.300) \] \[ \Lambda_{m,3}=120\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Convert molar conductivity into conductivity:}\) The relation is: \[ \Lambda_m=\frac{1000\kappa}{c} \] Rearranging: \[ \kappa=\frac{c\Lambda_m}{1000} \] Substituting: \[ \kappa_3=\frac{0.0900(120)}{1000} \] \[ \kappa_3=0.0108\ \mathrm{S\,cm^{-1}} \] \(\textbf{Check the apparently opposing concentration trends:}\) The molar conductivity falls as concentration increases because ionic interactions become more important. However, the total number of charge carriers per unit volume increases. Under the supplied data, this increase is sufficiently large that conductivity rises. \(\textbf{Interpret the extrapolation:}\) The intercept of the straight-line plot of \(\Lambda_m\) against \(\sqrt c\) represents \(\Lambda_m^\circ\). Its slope gives \(-A\). The predicted molar conductivity at the third concentration must then be converted using the concentration-dependent relation between \(\kappa\) and \(\Lambda_m\). Directly multiplying the limiting value by concentration would ignore the measured decrease of molar conductivity with concentration. \(\textbf{Final answer:}\) \[ \boxed{\left(\kappa_3,\Lambda_m^\circ\right)=\left(0.0108\ \mathrm{S\,cm^{-1}},\ 150\ \mathrm{S\,cm^2\,mol^{-1}}\right)} \]

Explanation: \(\textbf{Calibrate the cell geometry:}\) A conductivity cell has cell constant: \[ G^*=\frac{\ell}{A} \] Conductivity is related to resistance through: \[ \kappa=G^*\frac1R \] Therefore: \[ G^*=\kappa R \] \(\textbf{Calculate the cell constant from KCl:}\) \[ G^*=(0.0120)(125) \] \[ G^*=1.50\ \mathrm{cm^{-1}} \] The same cell constant applies to both measurements of X. \(\textbf{Calculate conductivity before dilution:}\) \[ \kappa_1=\frac{G^*}{R_1} \] \[ \kappa_1=\frac{1.50}{300} \] \[ \kappa_1=0.00500\ \mathrm{S\,cm^{-1}} \] \(\textbf{Determine initial molar conductivity:}\) \[ \Lambda_{m,1}=\frac{1000\kappa_1}{c_1} \] \[ \Lambda_{m,1}=\frac{1000(0.00500)}{0.0250} \] \[ \Lambda_{m,1}=200\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Calculate conductivity after dilution:}\) \[ \kappa_2=\frac{1.50}{550} \] \[ \kappa_2=2.727\times10^{-3}\ \mathrm{S\,cm^{-1}} \] The conductivity decreases because the resistance increases at fixed cell geometry. \(\textbf{Determine the diluted solution's molar conductivity:}\) \[ \Lambda_{m,2}=\frac{1000(2.727\times10^{-3})}{0.0125} \] \[ \Lambda_{m,2}\approx218.2\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Compare the two measurements:}\) The conductivity changes from: \[ 0.00500\rightarrow0.002727\ \mathrm{S\,cm^{-1}} \] The molar conductivity changes from: \[ 200\rightarrow218.2\ \mathrm{S\,cm^2\,mol^{-1}} \] Thus, the two quantities change in opposite directions. \(\textbf{Interpret the distinction:}\) Conductivity measures the conducting ability of a unit volume of solution. Dilution decreases the number of electrolyte formula units per unit volume. Molar conductivity refers to conducting ability per mole of electrolyte and can increase because ions experience less interaction and, for weak electrolytes, may become more extensively dissociated. The resistance measurement alone does not reveal which concentration-normalised trend occurs. The calibrated cell constant and both concentration values are required to establish the actual behaviour of X. The observed measurements support option B without requiring an unsupported assumption about whether X is strong or weak.

Explanation: \(\textbf{Determine the molar conductivity of P:}\) For concentration expressed in \(\mathrm{mol\,L^{-1}}\): \[ \Lambda_m=\frac{1000\kappa}{c} \] Therefore: \[ \Lambda_{m,P}=\frac{1000(8.00\times10^{-4})}{0.0100} \] \[ \Lambda_{m,P}=80.0\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Calculate the initial dissociation fraction:}\) \[ \alpha_P=\frac{80.0}{400} \] \[ \alpha_P=0.200 \] \(\textbf{Obtain the dissociation constant from P:}\) \[ K_{a,P}=\frac{0.0100(0.200)^2}{1-0.200} \] \[ K_{a,P}=\frac{4.00\times10^{-4}}{0.800} \] \[ K_{a,P}=5.00\times10^{-4} \] \(\textbf{Determine the molar conductivity of Q:}\) \[ \Lambda_{m,Q}=\frac{1000(4.00\times10^{-4})}{0.00300} \] \[ \Lambda_{m,Q}=133.33\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Calculate the diluted dissociation fraction:}\) \[ \alpha_Q=\frac{133.33}{400} \] \[ \alpha_Q\approx0.3333 \] \(\textbf{Check the dissociation constant at the new concentration:}\) \[ K_{a,Q}=\frac{0.00300(1/3)^2}{1-1/3} \] \[ K_{a,Q}=\frac{0.00300/9}{2/3} \] \[ K_{a,Q}=5.00\times10^{-4} \] \(\textbf{Compare the ionised fractions:}\) \[ \alpha_P=0.200,\qquad\alpha_Q\approx0.3333 \] The acid is more extensively dissociated after dilution. However: \[ \kappa_P=8.00\times10^{-4} \] \[ \kappa_Q=4.00\times10^{-4}\ \mathrm{S\,cm^{-1}} \] The conductivity decreases because the number of charge carriers per unit volume is lower overall. \(\textbf{Check the hydrogen-ion concentration trend:}\) For P: \[ [\mathrm{H^+}]_P=c_P\alpha_P=(0.0100)(0.200) \] \[ [\mathrm{H^+}]_P=0.00200\ \mathrm{mol\,L^{-1}} \] For Q: \[ [\mathrm{H^+}]_Q=(0.00300)(1/3) \] \[ [\mathrm{H^+}]_Q=0.00100\ \mathrm{mol\,L^{-1}} \] Thus, dilution raises pH despite increasing the dissociation fraction. \(\textbf{Interpret the experimentally consistent behaviour:}\) The dissociation constant is a characteristic equilibrium quantity at fixed temperature under the ideal model. It must not be identified with conductivity or degree of dissociation. Dilution favours a greater fractional dissociation of a weak electrolyte, but the reduced analytical concentration can still lower the total ion concentration and specific conductivity. The two measurements independently give the same value of \(K_a\), providing a consistency check on the stipulated model. The correct conclusion is option D.

Explanation: \(\textbf{Represent each limiting conductivity as ionic contributions:}\) At infinite dilution, individual ionic contributions are additive. For MgCl2: \[ \Lambda_m^\circ(\mathrm{MgCl_2})=\lambda^\circ_{\mathrm{Mg^{2+}}}+2\lambda^\circ_{\mathrm{Cl^-}} \] For Na2SO4: \[ \Lambda_m^\circ(\mathrm{Na_2SO_4})=2\lambda^\circ_{\mathrm{Na^+}}+\lambda^\circ_{\mathrm{SO_4^{2-}}} \] For NaCl: \[ \Lambda_m^\circ(\mathrm{NaCl})=\lambda^\circ_{\mathrm{Na^+}}+\lambda^\circ_{\mathrm{Cl^-}} \] \(\textbf{Add the magnesium and sulphate-containing salts:}\) \[ \Lambda_m^\circ(\mathrm{MgCl_2})+\Lambda_m^\circ(\mathrm{Na_2SO_4}) \] \[ =\lambda^\circ_{\mathrm{Mg^{2+}}}+\lambda^\circ_{\mathrm{SO_4^{2-}}}+2\lambda^\circ_{\mathrm{Na^+}}+2\lambda^\circ_{\mathrm{Cl^-}} \] The desired MgSO4 contribution is accompanied by two sodium and two chloride contributions. \(\textbf{Cancel the unwanted ions correctly:}\) Two NaCl formula units supply: \[ 2\Lambda_m^\circ(\mathrm{NaCl})=2\lambda^\circ_{\mathrm{Na^+}}+2\lambda^\circ_{\mathrm{Cl^-}} \] Subtracting twice the NaCl value cancels all unwanted contributions. Therefore: \[ \Lambda_m^\circ(\mathrm{MgSO_4})=\Lambda_m^\circ(\mathrm{MgCl_2})+\Lambda_m^\circ(\mathrm{Na_2SO_4})-2\Lambda_m^\circ(\mathrm{NaCl}) \] \(\textbf{Evaluate the correct limiting conductivity:}\) \[ \Lambda_m^\circ(\mathrm{MgSO_4})=258+260-2(126) \] \[ \Lambda_m^\circ(\mathrm{MgSO_4})=518-252 \] \[ \Lambda_m^\circ(\mathrm{MgSO_4})=266\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Diagnose the proposed error:}\) Subtracting only one NaCl contribution would leave: \[ \lambda^\circ_{\mathrm{Na^+}}+\lambda^\circ_{\mathrm{Cl^-}} \] in the expression. The result would therefore describe the desired ionic sum plus an unwanted NaCl contribution. The numerical error arises from incorrect ionic stoichiometry. \(\textbf{Distinguish ionic charge from electrode-electron transfer:}\) Kohlrausch's law concerns independently migrating ions at infinite dilution. The coefficients two in MgCl2 and Na2SO4 count chloride and sodium ions per electrolyte formula unit. They do not represent an electron-transfer number. No redox reaction is required to combine the limiting conductivity values. \(\textbf{Interpret the structural check:}\) Writing each electrolyte in terms of ionic contributions makes the correct combination transparent. Numerical subtraction without tracking the ion counts can produce a plausible-looking but incorrect answer. The result is \(266\ \mathrm{S\,cm^2\,mol^{-1}}\), for the stoichiometric reason given in option B.

Explanation: \(\textbf{Relate electron transfer to the discharge reaction:}\) The balanced reaction transfers two electrons. It consumes two moles of sulphuric acid and produces one mole of lead sulphate at each electrode per two moles of electrons. \(\textbf{Find sulphuric acid consumption:}\) \[ n(\mathrm{H_2SO_4})_{\mathrm{consumed}}=\frac{2}{2}(0.0800)=0.0800\ \mathrm{mol} \] The final acid amount is: \[ n(\mathrm{H_2SO_4})_{\mathrm{final}}=0.200-0.0800=0.120\ \mathrm{mol} \] At a fixed total volume of \(1.00\ \mathrm L\): \[ c_{\mathrm{final}}=\frac{0.120}{1.00}=0.120\ \mathrm{mol\,L^{-1}} \] \(\textbf{Distinguish a single electrode from both electrodes combined:}\) The total amount of PbSO4 formed at both electrodes is \(0.0800\ \mathrm{mol}\), but the positive electrode produces only half of that amount: \[ n(\mathrm{PbSO_4})_{\mathrm{positive}}=\frac{0.0800}{2}=0.0400\ \mathrm{mol} \] Counting PbSO4 from both electrodes when only the positive electrode is requested would double the answer. \(\textbf{Final answer:}\) \[ \boxed{c_{\mathrm{H_2SO_4,final}}=0.120\ \mathrm{mol\,L^{-1}},\quad n(\mathrm{PbSO_4})_{\mathrm{positive}}=0.0400\ \mathrm{mol}} \]

Explanation: \(\textbf{Statement I — Correct:}\) Silver ions react with chloride ions: \[ \mathrm{Ag^++Cl^-\rightarrow AgCl(s)} \] If chloride enters the silver-ion compartment, this precipitation reaction can occur. The free silver-ion activity may consequently decrease. The precise extent depends on the chloride amount and relevant equilibrium conditions. \(\textbf{Statement II — Correct:}\) For the silver reduction: \[ \mathrm{Ag^++e^-\rightarrow Ag} \] the Nernst equation is: \[ E_{\mathrm{Ag}}=E^\circ_{\mathrm{Ag}}+0.0591\log a_{\mathrm{Ag^+}} \] at \(298\ \mathrm K\). Reducing free silver-ion activity makes its logarithm smaller. The silver reduction potential therefore decreases. This chemical change can alter the measured cell emf even if the salt-bridge resistance is unchanged. \(\textbf{Statement III — Incorrect:}\) The electrical charges carried by ions do not fully determine their chemical suitability in a salt bridge. Chloride can react with silver ions, whereas nitrate is not expected to cause the same AgCl precipitation. Equal bridge geometry and equal electrical resistance therefore do not guarantee chemically equivalent cell operation. A precipitation reaction changes the half-cell composition, not merely its ionic conductivity. \(\textbf{Statement IV — Correct:}\) A salt bridge provides a pathway for ionic charge transport and helps maintain electrical neutrality in the half-cells. Its electrolyte should also avoid reactions that substantially change the intended electrode solutions. Otherwise, the bridge can become a source of chemical interference rather than a purely supporting component. \(\textbf{Integrate the experimental diagnosis:}\) The zinc–silver cell relies on the activities of zinc(II) and silver(I) to establish its reversible electrode potentials. The salt bridge is intended to support charge balance without materially altering those activities. A chloride-containing bridge can violate that assumption in a silver-ion half-cell. The potential change would arise from unintended precipitation even though chloride is not part of the intended overall redox reaction. The correct statements are I, II and IV.

Explanation: \(\textbf{Determine the initial series resistance:}\) For two resistances connected in series: \[ R_{\mathrm{total},i}=R_{\mathrm{metal},i}+R_{\mathrm{electrolyte},i} \] \[ R_{\mathrm{total},i}=20.0+30.0 \] \[ R_{\mathrm{total},i}=50.0\ \Omega \] \(\textbf{Calculate the initial current:}\) \[ I_i=\frac VR \] \[ I_i=\frac{1.00}{50.0} \] \[ I_i=0.0200\ \mathrm A=20.0\ \mathrm{mA} \] \(\textbf{Apply the temperature change to metallic conduction:}\) The metallic resistance increases by \(20.0\%\). Therefore: \[ R_{\mathrm{metal},f}=20.0(1.20) \] \[ R_{\mathrm{metal},f}=24.0\ \Omega \] The increase is consistent with greater scattering of conducting electrons in an ordinary metal at higher temperature. \(\textbf{Apply the temperature change to electrolytic conduction:}\) For fixed geometry: \[ R=\frac{\ell}{\kappa A} \] Thus: \[ R\propto\frac1\kappa \] The electrolyte conductivity increases by a factor of \(1.20\). Consequently: \[ R_{\mathrm{electrolyte},f}=\frac{30.0}{1.20} \] \[ R_{\mathrm{electrolyte},f}=25.0\ \Omega \] \(\textbf{Determine the new total resistance:}\) \[ R_{\mathrm{total},f}=24.0+25.0 \] \[ R_{\mathrm{total},f}=49.0\ \Omega \] \(\textbf{Calculate the new current:}\) \[ I_f=\frac{1.00}{49.0} \] \[ I_f=0.020408\ \mathrm A \] \[ I_f\approx20.4\ \mathrm{mA} \] \(\textbf{Interpret the competing temperature effects:}\) Metallic conduction occurs predominantly through electrons. Heating an ordinary metal generally increases its resistance. Electrolytic conduction occurs through moving ions; its conductivity commonly increases with temperature because ionic mobility increases. The same percentage increase in metallic resistance and electrolyte conductivity does not produce exactly cancelling resistance changes. The electrolyte resistance is divided by \(1.20\), not multiplied by \(0.80\). Under the supplied initial resistances, the electrolyte's resistance decrease slightly exceeds the metallic resistance increase. The net circuit resistance therefore falls, and the current rises. \(\textbf{Final answer:}\) \[ \boxed{(R_{\mathrm{total},f},I_f)=(49.0\ \Omega,\ 20.4\ \mathrm{mA})} \]

Explanation: \(\textbf{Determine the total electron transfer:}\) Faraday's law gives: \[ n(e^-)=\frac QF \] \[ n(e^-)=\frac{9650}{96500} \] \[ n(e^-)=0.100\ \mathrm{mol} \] \(\textbf{Calculate copper dissolved at the anode:}\) The anode reaction is: \[ \mathrm{Cu\rightarrow Cu^{2+}+2e^-} \] Two electrons are released per copper atom. Therefore: \[ n(\mathrm{Cu^{2+}})_{\mathrm{anode,formed}}=\frac{0.100}{2} \] \[ n(\mathrm{Cu^{2+}})_{\mathrm{anode,formed}}=0.0500\ \mathrm{mol} \] \(\textbf{Determine copper deposition at the cathode:}\) The cathodic metal reaction is: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu} \] Using the measured deposited mass: \[ n(\mathrm{Cu})_{\mathrm{deposited}}=\frac{2.54}{63.5} \] \[ n(\mathrm{Cu})_{\mathrm{deposited}}=0.0400\ \mathrm{mol} \] \(\textbf{Calculate the electrons used for copper deposition:}\) \[ n(e^-)_{\mathrm{Cu}}=2(0.0400) \] \[ n(e^-)_{\mathrm{Cu}}=0.0800\ \mathrm{mol} \] \(\textbf{Identify the remaining cathodic electron transfer:}\) The total cathodic electron amount is \(0.100\ \mathrm{mol}\). Therefore: \[ n(e^-)_{\mathrm{H_2}}=0.100-0.0800 \] \[ n(e^-)_{\mathrm{H_2}}=0.0200\ \mathrm{mol} \] \(\textbf{Calculate hydrogen evolved:}\) Hydrogen evolution requires two electrons per hydrogen molecule: \[ \mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-} \] Thus: \[ n(\mathrm{H_2})=\frac{0.0200}{2} \] \[ n(\mathrm{H_2})=0.0100\ \mathrm{mol} \] \(\textbf{Track the dissolved copper inventory:}\) Copper(II) is produced at the active anode and consumed during copper deposition. The net change is: \[ \Delta n(\mathrm{Cu^{2+}})=0.0500-0.0400 \] \[ \Delta n(\mathrm{Cu^{2+}})=+0.0100\ \mathrm{mol} \] \(\textbf{Verify cathodic charge conservation:}\) Electrons used for metal deposition: \[ 0.0800\ \mathrm{mol} \] Electrons used for hydrogen formation: \[ 0.0200\ \mathrm{mol} \] Their sum is \(0.100\ \mathrm{mol}\), equal to the electron amount released at the anode. \(\textbf{Interpret the unexpected concentration increase:}\) An active copper anode replenishes copper(II) ions, but this does not guarantee an unchanged copper-ion concentration if cathodic current is shared by another reaction. Here, some cathodic electrons reduce water to hydrogen instead of depositing copper. Consequently, the copper anode produces more copper(II) than the cathode removes. The net copper(II) inventory increases even though copper metal is visibly deposited at the cathode. The calculation therefore requires separate anode and cathode material balances in addition to total electron conservation. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{H_2}),\Delta n(\mathrm{Cu^{2+}})\right)=(0.0100,\ +0.0100)\ \mathrm{mol}} \]

Explanation: \(\textbf{Determine the electron amount shared by the two experiments:}\) Both electrolytic cells receive the same electrical charge. Using: \[ n(e^-)=\frac QF \] \[ n(e^-)=\frac{19300}{96500} \] \[ n(e^-)=0.200\ \mathrm{mol} \] \(\textbf{Identify the cathode reaction in molten NaCl:}\) Molten sodium chloride contains mobile sodium and chloride ions but no water. The stipulated cathodic reduction is: \[ \mathrm{Na^++e^-\rightarrow Na} \] Each sodium atom requires one electron. Therefore: \[ n(\mathrm{Na})_P=0.200\ \mathrm{mol} \] \(\textbf{Identify the cathode reaction in aqueous NaCl:}\) Water is present in the aqueous electrolyte. Under the specified conditions, hydrogen is produced at the cathode: \[ \mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-} \] Two electrons are required for every hydrogen molecule formed. \(\textbf{Calculate hydrogen formation:}\) \[ n(\mathrm{H_2})_Q=\frac{n(e^-)}2 \] \[ n(\mathrm{H_2})_Q=\frac{0.200}{2} \] \[ n(\mathrm{H_2})_Q=0.100\ \mathrm{mol} \] \(\textbf{Check the anode chemistry:}\) Under the specified conditions, chloride ions undergo oxidation at the anodes: \[ \mathrm{2Cl^-\rightarrow Cl_2+2e^-} \] Therefore, each experiment produces: \[ n(\mathrm{Cl_2})=\frac{0.200}{2} \] \[ n(\mathrm{Cl_2})=0.100\ \mathrm{mol} \] \(\textbf{Compare the complete transformations:}\) For molten NaCl: \[ \mathrm{2NaCl(l)\rightarrow2Na(l)+Cl_2(g)} \] For concentrated aqueous NaCl under the specified conditions: \[ \mathrm{2NaCl+2H_2O\rightarrow H_2+Cl_2+2NaOH} \] These reactions produce different cathodic products because the chemical environments differ. \(\textbf{Interpret the role of the solvent:}\) The presence of water introduces another reducible species and changes the cathode chemistry. Equal electrical charges correspond to equal total electron transfer, not equal mole amounts of every electrode product. Sodium deposition is a one-electron reduction, whereas hydrogen formation is a two-electron reduction per gas molecule. The reaction products and stoichiometric electron requirements must therefore both be identified before applying Faraday's law. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{Na})_P,n(\mathrm{H_2})_Q\right)=(0.200,\ 0.100)\ \mathrm{mol}} \]

Explanation: \(\textbf{Statement I — Correct:}\) Both electrodes involve: \[ \mathrm{M^{2+}+2e^-\rightleftharpoons M(s)} \] Their standard reduction potentials are identical. The Nernst equation is: \[ E=E^\circ+\frac{0.0591}{2}\log a_{\mathrm{M^{2+}}} \] Q has the greater metal-ion activity and therefore the greater reduction potential. \(\textbf{Statement II — Incorrect:}\) Bulk conductivity and electrode potential are not interchangeable quantities. The reversible metal-ion reduction potential depends on the specified metal-ion activity. The question explicitly keeps that activity unchanged when the inert supporting electrolyte is adjusted. Under these assumptions, improving bulk conductivity does not change the Nernst reaction quotient of the metal-ion electrode. It therefore does not necessarily increase the reversible emf. \(\textbf{Statement III — Correct:}\) A more conducting electrolyte pathway generally produces a smaller ohmic potential drop at a given current. The terminal voltage of a discharging cell can consequently increase under load. However, an ideal reversible open-circuit emf is calculated in the absence of sustained current. It is fixed by the electrochemical activities and thermodynamic properties rather than by a resistive voltage loss. \(\textbf{Statement IV — Correct:}\) The lower-activity metal-ion compartment P is the anode: \[ \mathrm{M(s)\rightarrow M^{2+}+2e^-} \] The higher-activity compartment Q is the cathode: \[ \mathrm{M^{2+}+2e^-\rightarrow M(s)} \] The resulting electrode processes tend to reduce the activity difference. \(\textbf{Check the reversible potential difference:}\) \[ E_{\mathrm{cell}}=\frac{0.0591}{2}\log\left(\frac{0.100}{0.00100}\right) \] \[ E_{\mathrm{cell}}=0.0591\ \mathrm V \] This potential is independent of the stated differences in inert supporting-electrolyte conductivity. \(\textbf{Interpret the distinction:}\) An ionic activity controls the chemical potential of an electroactive species. Bulk conductivity reflects the ease with which the entire solution carries ionic current. The two can respond differently to addition of inert supporting electrolyte. A performance improvement under load therefore need not represent an increase in the thermodynamic driving force of the cell. Statements I, III and IV are correct.

Explanation: \(\textbf{Establish the initial conducting-ion inventory:}\) KOH is a strong electrolyte: \[ \mathrm{KOH\rightarrow K^++OH^-} \] Initially: \[ n(\mathrm{K^+})=0.0200\ \mathrm{mol} \] \[ n(\mathrm{OH^-})=0.0200\ \mathrm{mol} \] \(\textbf{Calculate the initial conductivity:}\) With volume \(1.00\ \mathrm L\): \[ \kappa_i=\frac{0.0200(73.0)+0.0200(200)}{1000} \] \[ \kappa_i=\frac{1.460+4.000}{1000} \] \[ \kappa_i=5.46\times10^{-3}\ \mathrm{S\,cm^{-1}} \] \(\textbf{Determine the limiting reagent in neutralisation:}\) The added HCl provides: \[ n(\mathrm{H^+})=0.0100\ \mathrm{mol} \] Hydrogen ions and hydroxide ions react in a \(1:1\) mole ratio. The initially available hydroxide amount is \(0.0200\ \mathrm{mol}\). Consequently, the added acid is completely consumed, leaving excess hydroxide. \(\textbf{Calculate the final hydroxide amount:}\) \[ n(\mathrm{OH^-})_f=0.0200-0.0100 \] \[ n(\mathrm{OH^-})_f=0.0100\ \mathrm{mol} \] \(\textbf{Track the spectator ions:}\) Potassium ions remain dissolved: \[ n(\mathrm{K^+})_f=0.0200\ \mathrm{mol} \] Chloride ions introduced with HCl also remain dissolved: \[ n(\mathrm{Cl^-})_f=0.0100\ \mathrm{mol} \] \(\textbf{Verify ionic charge balance:}\) The positive charge equivalents are: \[ 0.0200\ \mathrm{mol} \] The final negative charge equivalents are: \[ 0.0100+0.0100=0.0200\ \mathrm{mol} \] Thus, the final electrolyte composition is electrically neutral. \(\textbf{Calculate the final conductivity:}\) \[ \kappa_f=\frac{0.0200(73.0)+0.0100(200)+0.0100(76.0)}{1000} \] \[ \kappa_f=\frac{1.460+2.000+0.760}{1000} \] \[ \kappa_f=4.22\times10^{-3}\ \mathrm{S\,cm^{-1}} \] \(\textbf{Determine the final pH:}\) The hydroxide concentration is: \[ [\mathrm{OH^-}]_f=\frac{0.0100}{1.00} \] \[ [\mathrm{OH^-}]_f=0.0100\ \mathrm{mol\,L^{-1}} \] Therefore: \[ \mathrm{pOH}=-\log(0.0100) \] \[ \mathrm{pOH}=2.00 \] At \(298\ \mathrm K\): \[ \mathrm{pH}=14.00-\mathrm{pOH} \] \[ \mathrm{pH}=12.00 \] \(\textbf{Interpret the change in conducting species:}\) Neutralisation removes some highly conducting hydroxide ions. The added chloride ions remain dissolved and contribute to conductivity, but their stipulated limiting ionic contribution is smaller than that of hydroxide. The potassium concentration remains unchanged. Consequently, the total conductivity decreases even though the solution still contains substantial amounts of mobile ions. The final solution remains alkaline because only half the original hydroxide is neutralised. The conductivity calculation requires tracking all surviving ions, while the pH calculation requires identifying the excess acid–base reactant. \(\textbf{Final answer:}\) \[ \boxed{\left(\kappa_f,\mathrm{pH}_f\right)=\left(4.22\times10^{-3}\ \mathrm{S\,cm^{-1}},\ 12.00\right)} \]

Explanation: \(\textbf{Distinguish the material property from the measurement:}\) Conductivity characterises the conducting ability of a solution at a specified temperature and composition. Measured conductance also depends on the geometry of the conducting path. The given cell dimensions must therefore be included before comparing conductances. \(\textbf{Calculate the resistance of X:}\) \[ R_X=\frac{\ell_X}{\kappa_XA_X} \] \[ R_X=\frac{2.00}{(0.0120)(4.00)} \] \[ R_X=41.67\ \Omega \] \(\textbf{Determine the conductance of X:}\) \[ G_X=\frac1{R_X} \] \[ G_X=\frac1{41.67} \] \[ G_X=0.0240\ \mathrm S \] \(\textbf{Calculate the resistance of Y:}\) \[ R_Y=\frac{3.00}{(0.0200)(1.00)} \] \[ R_Y=150\ \Omega \] \(\textbf{Determine the conductance of Y:}\) \[ G_Y=\frac1{150} \] \[ G_Y=0.00667\ \mathrm S \] \(\textbf{Compare the two results:}\) The conductivity ordering is: \[ \kappa_X\lt \kappa_Y \] However, the measured conductance ordering is: \[ G_X\gt G_Y \] The larger effective area and smaller separation-to-area ratio favour current flow through cell X. \(\textbf{Check the geometry effect independently:}\) Conductance may be written as: \[ G=\kappa\frac A\ell \] The geometry factor for X is: \[ \frac{A_X}{\ell_X}=\frac{4.00}{2.00}=2.00\ \mathrm{cm} \] For Y: \[ \frac{A_Y}{\ell_Y}=\frac{1.00}{3.00}=0.333\ \mathrm{cm} \] The geometry factor for X is six times that for Y. This outweighs its smaller material conductivity. \(\textbf{Interpret the experimental error:}\) The student compares a material property with a geometry-dependent measured quantity. A solution with lower conductivity can show greater conductance when placed in a more favourable conducting geometry. The two measurements must be normalised using the respective cell constants before inferring relative conductivities. The correct interpretation is given in option C.

Explanation: \(\textbf{Use the unchanged cell geometry:}\) The resistance of the conducting electrolyte is: \[ R=\frac{\ell}{\kappa A} \] The electrode separation and effective area remain constant. Therefore: \[ R\propto\frac1\kappa \] \(\textbf{Infer conductivity from the resistance change:}\) The resistance triples: \[ R_2=3R_1 \] Consequently: \[ \frac{\kappa_2}{\kappa_1}=\frac{R_1}{R_2} \] \[ \frac{\kappa_2}{\kappa_1}=\frac13 \] Therefore: \[ \kappa_2=\frac{2.50\times10^{-2}}3 \] \[ \kappa_2=8.33\times10^{-3}\ \mathrm{S\,m^{-1}} \] \(\textbf{Convert the conductivity unit:}\) Because: \[ 1\ \mathrm{S\,m^{-1}}=10^{-2}\ \mathrm{S\,cm^{-1}} \] \[ \kappa_2=(8.33\times10^{-3})(10^{-2}) \] \[ \kappa_2=8.33\times10^{-5}\ \mathrm{S\,cm^{-1}} \] \(\textbf{Compare the molar conductivities:}\) Molar conductivity satisfies: \[ \Lambda_m=\frac{\kappa}{c} \] when coherent units are used. Therefore: \[ \frac{\Lambda_{m,2}}{\Lambda_{m,1}}=\frac{\kappa_2}{\kappa_1}\frac{c_1}{c_2} \] \(\textbf{Substitute both changed-condition ratios:}\) \[ \frac{\Lambda_{m,2}}{\Lambda_{m,1}}=\frac13\left(\frac{0.0100}{0.00250}\right) \] \[ \frac{\Lambda_{m,2}}{\Lambda_{m,1}}=\frac13(4) \] \[ \frac{\Lambda_{m,2}}{\Lambda_{m,1}}=\frac43 \] \(\textbf{Verify with coherent SI concentrations:}\) The initial concentration is: \[ c_1=10.0\ \mathrm{mol\,m^{-3}} \] Thus: \[ \Lambda_{m,1}=\frac{0.0250}{10.0} \] \[ \Lambda_{m,1}=0.00250\ \mathrm{S\,m^2\,mol^{-1}} \] The final concentration is: \[ c_2=2.50\ \mathrm{mol\,m^{-3}} \] Therefore: \[ \Lambda_{m,2}=\frac{8.33\times10^{-3}}{2.50} \] \[ \Lambda_{m,2}=0.00333\ \mathrm{S\,m^2\,mol^{-1}} \] Their ratio is again approximately \(4:3\). \(\textbf{Interpret the two dilution effects:}\) The resistance measurement establishes that the bulk conductivity falls to one third of its initial value. The electrolyte concentration, however, falls to one quarter. Consequently, conductivity per mole of electrolyte increases. A decrease in specific conductivity does not imply a decrease in molar conductivity. The correct conclusion requires combining the measured resistance ratio with the separately specified concentration change and maintaining consistent units. \(\textbf{Final answer:}\) \[ \boxed{\left(\kappa_2,\frac{\Lambda_{m,2}}{\Lambda_{m,1}}\right)=\left(8.33\times10^{-5}\ \mathrm{S\,cm^{-1}},\ \frac43\right)} \]

Explanation: \(\textbf{Statement I — Correct:}\) For a dilute strong electrolyte, the approximate relation is: \[ \Lambda_m=\Lambda_m^\circ-A\sqrt c \] with a suitable positive coefficient \(A\). This predicts a straight line with negative slope when molar conductivity is plotted against \(\sqrt c\). Curve P is consistent with that behaviour. \(\textbf{Statement II — Correct:}\) The limiting molar conductivity is the value approached as concentration tends to zero. For the stipulated straight-line relation: \[ \lim_{c\rightarrow0}\Lambda_m=\Lambda_m^\circ \] The vertical intercept therefore provides the limiting molar conductivity. Extrapolation should nevertheless be based on measurements within the range where the dilute strong-electrolyte approximation is applicable. \(\textbf{Statement III — Incorrect:}\) A weak electrolyte can show a much sharper increase in molar conductivity on dilution. Its degree of ionisation rises as dilution shifts its ionisation equilibrium. The increase in the number of ions per mole of weak electrolyte contributes strongly to its molar-conductivity change. A steeply curved plot is therefore not evidence that the electrolyte is more extensively ionised at its original concentration. \(\textbf{Statement IV — Correct:}\) Weak electrolytes do not generally follow the same simple linear variation of molar conductivity with \(\sqrt c\). Their limiting molar conductivity cannot be obtained reliably by a straight-line extrapolation of that nonlinear curve. Kohlrausch's law instead permits the limiting value to be constructed from appropriate strong electrolytes containing the required constituent ions. \(\textbf{Integrate the graph interpretation:}\) The graph's shape supplies information about the dependence of ion transport on dilution. For a strong electrolyte, decreasing interionic interactions commonly increases molar conductivity towards its limiting value. For a weak electrolyte, increasing ionisation introduces an additional strong concentration dependence. Both types can have increasing molar conductivity on dilution while their ordinary conductivity decreases. The distinguishing feature is the appropriate limiting-conductivity model, not merely the direction of the trend. Statements I, II and IV are correct.

Explanation: \(\textbf{Statement I — Correct:}\) Metallic conductors transport charge predominantly through mobile electrons. Electrons carry negative charge, so their drift direction is opposite to conventional current. The external source determines the direction in which electrons enter or leave each electrode connection. This electronic conduction differs from conduction through the aqueous electrolyte. \(\textbf{Statement II — Correct:}\) Electrode P is connected to the positive terminal and functions as the anode. Electrode Q is connected to the negative terminal and functions as the cathode. Cations migrate towards the cathode, while anions migrate towards the anode. Thus, copper(II) ions move towards Q and sulphate ions towards P under the applied field. \(\textbf{Statement III — Incorrect:}\) Ordinary aqueous CuSO4 conducts electricity through the movement of dissolved ions. It does not principally conduct by transporting free electrons across the solution. Electrons are transferred at electrode interfaces during the oxidation and reduction reactions. They are transported through the external metallic connections, whereas mobile ions carry charge through the solution. \(\textbf{Statement IV — Correct:}\) At the positive copper anode: \[ \mathrm{Cu(s)\rightarrow Cu^{2+}+2e^-} \] At the negative copper cathode: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu(s)} \] The same total current passes through both electrodes. With these exclusive half-reactions, two electrons correspond to one copper atom dissolved and one copper atom deposited. Therefore, the copper amounts are equal. \(\textbf{Integrate transport and electrode chemistry:}\) Ionic migration and electrode conversion are different but coupled requirements of an electrolytic circuit. The electric field moves oppositely charged ions in opposite directions. The electrode reactions exchange electrons between the external wires and the participating ions. The dissolution of copper at P can replenish copper(II) removed by deposition at Q. Equal electrode reaction extents do not imply that dissolved ions remain stationary. The correct statements are I, II and IV.

Explanation: Acetic acid is a weak electrolyte whose degree of ionisation changes appreciably upon dilution. Its measured molar conductivity therefore generally exhibits pronounced nonlinear concentration dependence. A simple straight-line extrapolation of finite-concentration measurements is not the appropriate method for obtaining its limiting molar conductivity. Kohlrausch's law provides an alternative based on independent ionic contributions at infinite dilution. For HCl: \[ \Lambda_m^\circ(\mathrm{HCl})=\lambda^\circ_{\mathrm{H^+}}+\lambda^\circ_{\mathrm{Cl^-}} \] For sodium acetate: \[ \Lambda_m^\circ(\mathrm{CH_3COONa})=\lambda^\circ_{\mathrm{Na^+}}+\lambda^\circ_{\mathrm{CH_3COO^-}} \] For NaCl: \[ \Lambda_m^\circ(\mathrm{NaCl})=\lambda^\circ_{\mathrm{Na^+}}+\lambda^\circ_{\mathrm{Cl^-}} \] Adding the first two expressions and subtracting the third cancels both unwanted ionic contributions. The remaining expression is: \[ \lambda^\circ_{\mathrm{H^+}}+\lambda^\circ_{\mathrm{CH_3COO^-}} \] This is precisely the limiting molar conductivity of acetic acid. The law uses limiting ionic contributions, so the finite-concentration measurements cannot be inserted indiscriminately in place of the limiting values. After determining \(\Lambda_m^\circ\), a separate finite-concentration measurement can be used to estimate acetic acid's degree of dissociation under the appropriate weak-electrolyte model. Thus, the Reason is true and directly explains the Assertion.

Explanation: \(\textbf{Identify the meaning of equal conductivity:}\) Conductivity depends on both electrolyte concentration and molar conductivity. Equal conductivities do not require equal analytical concentrations. The two solutions contain different cations with different limiting ionic conductivity contributions. \(\textbf{Calculate the KCl conductivity:}\) Under the supplied approximation: \[ \kappa=\frac{c\Lambda_m^\circ}{1000} \] For KCl: \[ \kappa_{\mathrm{KCl}}=\frac{0.0200(150)}{1000} \] \[ \kappa_{\mathrm{KCl}}=3.00\times10^{-3}\ \mathrm{S\,cm^{-1}} \] \(\textbf{Set the required HCl conductivity equal to this value:}\) \[ \kappa_{\mathrm{HCl}}=\kappa_{\mathrm{KCl}} \] Therefore: \[ \frac{c_{\mathrm{HCl}}(426)}{1000}=3.00\times10^{-3} \] \(\textbf{Calculate the HCl concentration:}\) \[ c_{\mathrm{HCl}}=\frac{1000(3.00\times10^{-3})}{426} \] \[ c_{\mathrm{HCl}}=0.007042\ \mathrm{mol\,L^{-1}} \] \[ c_{\mathrm{HCl}}\approx0.00704\ \mathrm{mol\,L^{-1}} \] \(\textbf{Extract the ionic contribution difference:}\) At infinite dilution: \[ \Lambda_m^\circ(\mathrm{HCl})=\lambda^\circ_{\mathrm{H^+}}+\lambda^\circ_{\mathrm{Cl^-}} \] \[ \Lambda_m^\circ(\mathrm{KCl})=\lambda^\circ_{\mathrm{K^+}}+\lambda^\circ_{\mathrm{Cl^-}} \] Subtracting: \[ \Lambda_m^\circ(\mathrm{HCl})-\Lambda_m^\circ(\mathrm{KCl})=\lambda^\circ_{\mathrm{H^+}}-\lambda^\circ_{\mathrm{K^+}} \] \(\textbf{Evaluate the difference:}\) \[ \lambda^\circ_{\mathrm{H^+}}-\lambda^\circ_{\mathrm{K^+}}=426-150 \] \[ \lambda^\circ_{\mathrm{H^+}}-\lambda^\circ_{\mathrm{K^+}}=276\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Check against the sodium reference:}\) The NaCl value additionally shows: \[ \lambda^\circ_{\mathrm{K^+}}-\lambda^\circ_{\mathrm{Na^+}}=150-126 \] \[ \lambda^\circ_{\mathrm{K^+}}-\lambda^\circ_{\mathrm{Na^+}}=24\ \mathrm{S\,cm^2\,mol^{-1}} \] This is consistent with the relative limiting contributions inferred through the common chloride ion. \(\textbf{Interpret the concentration difference:}\) HCl has a substantially greater limiting molar conductivity than KCl. Consequently, a lower concentration of HCl can produce the same conductivity as the specified KCl solution under the given model. Kohlrausch's law allows the difference between ionic contributions to be extracted without knowing the chloride contribution separately. The chloride cancellation is a statement about limiting molar conductivities, not a claim that chloride ions stop carrying current. The calculation integrates ionic mobility contributions with conductivity and concentration. \(\textbf{Final answer:}\) \[ \boxed{\left(c_{\mathrm{HCl}},\lambda^\circ_{\mathrm{H^+}}-\lambda^\circ_{\mathrm{K^+}}\right)=\left(0.00704\ \mathrm{mol\,L^{-1}},\ 276\ \mathrm{S\,cm^2\,mol^{-1}}\right)} \]

Explanation: \(\textbf{Represent the unknown mixture composition:}\) Let: \[ c_{\mathrm{NaCl}}=x \] \[ c_{\mathrm{MgCl_2}}=y \] The analytical salt balance gives: \[ x+y=0.0300 \] \(\textbf{Use chloride stoichiometry:}\) One mole of NaCl supplies one mole of chloride ions. One mole of MgCl2 supplies two moles of chloride ions. Therefore: \[ [\mathrm{Cl^-}]=x+2y \] The measured chloride concentration gives: \[ x+2y=0.0450 \] \(\textbf{Solve the simultaneous balances:}\) Subtracting the analytical-salt equation: \[ (x+2y)-(x+y)=0.0450-0.0300 \] \[ y=0.0150\ \mathrm{mol\,L^{-1}} \] Then: \[ x=0.0300-0.0150 \] \[ x=0.0150\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the individual ionic concentrations:}\) Complete dissociation gives: \[ [\mathrm{Na^+}]=0.0150\ \mathrm{mol\,L^{-1}} \] \[ [\mathrm{Mg^{2+}}]=0.0150\ \mathrm{mol\,L^{-1}} \] \[ [\mathrm{Cl^-}]=0.0450\ \mathrm{mol\,L^{-1}} \] \(\textbf{Construct the conductivity sum:}\) With the supplied ionic contributions: \[ \kappa=\frac{[\mathrm{Na^+}]\lambda_{\mathrm{Na^+}}+[\mathrm{Mg^{2+}}]\lambda_{\mathrm{Mg^{2+}}}+[\mathrm{Cl^-}]\lambda_{\mathrm{Cl^-}}}{1000} \] \(\textbf{Substitute the inferred composition:}\) \[ \kappa=\frac{0.0150(50.0)+0.0150(106)+0.0450(76.0)}{1000} \] \[ \kappa=\frac{0.750+1.590+3.420}{1000} \] \[ \kappa=5.76\times10^{-3}\ \mathrm{S\,cm^{-1}} \] \(\textbf{Determine the requested ion ratio:}\) \[ [\mathrm{Mg^{2+}}]:[\mathrm{Na^+}]=0.0150:0.0150 \] \[ [\mathrm{Mg^{2+}}]:[\mathrm{Na^+}]=1:1 \] \(\textbf{Verify electrical neutrality:}\) Positive ionic charge equivalents: \[ [\mathrm{Na^+}]+2[\mathrm{Mg^{2+}}] \] \[ =0.0150+2(0.0150) \] \[ =0.0450\ \mathrm{mol\,L^{-1}} \] This equals the chloride concentration and confirms charge balance. \(\textbf{Interpret the multivalent-ion effect:}\) The chloride concentration exceeds the total salt concentration because MgCl2 supplies two chloride ions per formula unit. The magnesium and sodium ion concentrations happen to be equal in this mixture, but their charge contributions are not equal. Conductivity must be calculated from each ionic species and its supplied conductivity contribution. The analytical salt balance alone would not reveal the composition; the chloride measurement provides the second independent constraint. The final calculation combines formula-unit stoichiometry, charge neutrality and ionic conductance. \(\textbf{Final answer:}\) \[ \boxed{\left(\kappa,[\mathrm{Mg^{2+}}]:[\mathrm{Na^+}]\right)=\left(5.76\times10^{-3}\ \mathrm{S\,cm^{-1}},\ 1:1\right)} \]