Redox Reactions and Electrochemistry MCQs with Answers – Part 4 (NEET Chemistry)

Explanation: \(\textbf{Statement I: Correct.}\) The oxidising agent accepts electrons and undergoes reduction. Among the specified oxidised species, the silver(I)/silver couple has the highest standard reduction potential. Silver(I) therefore has the strongest standard reduction tendency among the three oxidised forms. The ranking is: \[ \mathrm{Ag^+\gt Fe^{3+}\gt Cu^{2+}} \] in oxidising tendency under the stated conditions. \(\textbf{Statement II: Correct.}\) The reducing agent releases electrons and undergoes oxidation. For the reduced forms of the supplied couples, a less positive reduction potential corresponds to a greater tendency to undergo oxidation. Copper has the lowest listed reduction potential. Therefore, copper metal is the strongest reducing agent among Cu, Fe2+ and Ag. The corresponding qualitative reducing-strength order is: \[ \mathrm{Cu\gt Fe^{2+}\gt Ag} \] \(\textbf{Statement III: Incorrect.}\) For silver to reduce iron(III), silver would have to undergo oxidation: \[ \mathrm{Ag\rightarrow Ag^++e^-} \] Iron(III) would undergo reduction: \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] The corresponding standard emf is: \[ E^\circ=0.77-0.80 \] \[ E^\circ=-0.03\ \mathrm V \] The proposed reaction is therefore not spontaneous under standard conditions. \(\textbf{Statement IV: Correct.}\) Copper oxidation is: \[ \mathrm{Cu\rightarrow Cu^{2+}+2e^-} \] Two iron(III) ions accept those electrons: \[ \mathrm{2Fe^{3+}+2e^-\rightarrow2Fe^{2+}} \] The standard emf is: \[ E^\circ=0.77-0.34 \] \[ E^\circ=+0.43\ \mathrm V \] The positive value favours the reaction: \[ \mathrm{Cu+2Fe^{3+}\rightarrow Cu^{2+}+2Fe^{2+}} \] \(\textbf{Interpret the redox-strength ranking:}\) The oxidising strength of the oxidised species and reducing strength of their corresponding reduced forms run in opposite directions along the reduction-potential scale. This distinction prevents the mistaken inference that a metal associated with a strongly oxidising ion must itself be a strong reducing agent. The two displacement tests further demonstrate that a difference of only \(0.03\ \mathrm V\) changes the standard thermodynamic direction between silver and iron(III). The correct statements are I, II and IV.

Explanation: \(\textbf{Identify the physical meaning of the graph:}\) The terminal voltage of a discharging cell is lower than its emf when current passes through the internal resistance. The stated model is: \[ V=E-Ir \] This is a straight-line relation between terminal voltage and current. Its vertical intercept is the open-circuit emf. Its slope is the negative of the internal resistance. \(\textbf{Determine the graph slope:}\) Using the two plotted observations: \[ (I_1,V_1)=(0.0500\ \mathrm A,\ 1.15\ \mathrm V) \] \[ (I_2,V_2)=(0.150\ \mathrm A,\ 0.950\ \mathrm V) \] The slope is: \[ m=\frac{V_2-V_1}{I_2-I_1} \] Substituting: \[ m=\frac{0.950-1.15}{0.150-0.0500} \] \[ m=\frac{-0.200}{0.100} \] \[ m=-2.00\ \mathrm{V\,A^{-1}} \] \(\textbf{Infer the internal resistance:}\) Comparing: \[ V=E-rI \] with a straight-line equation shows: \[ m=-r \] Therefore: \[ r=2.00\ \Omega \] The slope unit \(\mathrm{V\,A^{-1}}\) is equivalent to ohms. \(\textbf{Recover the open-circuit emf:}\) Rearrange the discharge equation: \[ E=V+Ir \] Using the first observation: \[ E=1.15+(0.0500)(2.00) \] \[ E=1.15+0.100 \] \[ E=1.25\ \mathrm V \] \(\textbf{Verify using the second observation:}\) \[ E=0.950+(0.150)(2.00) \] \[ E=0.950+0.300 \] \[ E=1.25\ \mathrm V \] Both observations give the same emf. \(\textbf{Interpret the intercept rather than extrapolated current:}\) The emf corresponds to the limiting terminal potential as current approaches zero. The internal resistance controls how rapidly terminal voltage falls with increasing discharge current. The downward slope does not mean that the cell's equilibrium emf becomes negative as soon as current flows. It reflects the growing internal resistive voltage drop under the assumed model. This method permits emf and internal resistance to be inferred from loaded measurements without treating either loaded terminal voltage as the emf itself. \(\textbf{Final answer:}\) \[ \boxed{(E,r)=(1.25\ \mathrm V,\ 2.00\ \Omega)} \]

Explanation: \(\textbf{Recognise the shared cathode electron supply:}\) Both iron(III) reduction and copper(II) reduction consume electrons at the same cathode. The total electron amount must be partitioned between these two reactions. It cannot be assigned independently in full to each product. \(\textbf{Infer copper deposition from the measured mass:}\) \[ n(\mathrm{Cu})=\frac{m(\mathrm{Cu})}{M(\mathrm{Cu})} \] Substituting: \[ n(\mathrm{Cu})=\frac{0.635}{63.5} \] \[ n(\mathrm{Cu})=0.0100\ \mathrm{mol} \] \(\textbf{Determine electrons used for copper reduction:}\) The cathode half-reaction is: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu} \] Therefore: \[ n(e^-)_{\mathrm{Cu}}=2(0.0100) \] \[ n(e^-)_{\mathrm{Cu}}=0.0200\ \mathrm{mol} \] \(\textbf{Calculate electrons available for iron(III) reduction:}\) The total electron amount is: \[ n(e^-)_{\mathrm{total}}=0.0600\ \mathrm{mol} \] The portion consumed by iron(III) must be: \[ n(e^-)_{\mathrm{Fe}}=0.0600-0.0200 \] \[ n(e^-)_{\mathrm{Fe}}=0.0400\ \mathrm{mol} \] \(\textbf{Determine iron(III) consumption:}\) The iron cathode reaction is: \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] One iron(III) ion accepts one electron. Therefore: \[ n(\mathrm{Fe^{3+}})_{\mathrm{reduced}}=0.0400\ \mathrm{mol} \] \(\textbf{Calculate iron(III) remaining:}\) Initially: \[ n(\mathrm{Fe^{3+}})_i=0.0500\ \mathrm{mol} \] After electrolysis: \[ n(\mathrm{Fe^{3+}})_f=0.0500-0.0400 \] \[ n(\mathrm{Fe^{3+}})_f=0.0100\ \mathrm{mol} \] \(\textbf{Determine oxygen evolution from total anode charge:}\) The same total electron amount is released at the anode: \[ n(e^-)_{\mathrm{anode}}=0.0600\ \mathrm{mol} \] For: \[ \mathrm{2H_2O\rightarrow O_2+4H^++4e^-} \] the oxygen amount is: \[ n(\mathrm{O_2})=\frac{0.0600}{4} \] \[ n(\mathrm{O_2})=0.0150\ \mathrm{mol} \] \(\textbf{Verify both electrode balances:}\) At the cathode: \[ 0.0400+2(0.0100)=0.0600\ \mathrm{mol\ of\ electrons} \] At the anode: \[ 4(0.0150)=0.0600\ \mathrm{mol\ of\ electrons} \] The electron amounts agree. \(\textbf{Interpret the measurements:}\) The copper mass identifies one part of the cathode electron consumption. The remaining cathode charge must reduce iron(III), according to the stipulated reaction set. The oxygen amount is determined by the complete cell charge, not only by the charge associated with copper deposition. The problem therefore requires distinguishing simultaneous cathode processes from their common complementary anode process. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{Fe^{3+}})_f,n(\mathrm{O_2})\right)=(0.0100,\ 0.0150)\ \mathrm{mol}} \]

Explanation: \(\textbf{Determine electron transfer during water electrolysis:}\) The first stage passes: \[ Q_1=3860\ \mathrm C \] The corresponding electron amount is: \[ n(e^-)_1=\frac{Q_1}{F} \] \[ n(e^-)_1=\frac{3860}{96500} \] \[ n(e^-)_1=0.0400\ \mathrm{mol} \] \(\textbf{Calculate hydrogen formation:}\) At the electrolytic cathode: \[ \mathrm{2H^++2e^-\rightarrow H_2} \] Two electrons produce one hydrogen molecule. Therefore: \[ n(\mathrm{H_2})=\frac{0.0400}{2} \] \[ n(\mathrm{H_2})=0.0200\ \mathrm{mol} \] \(\textbf{Identify the second cell's oxidation reaction:}\) The hydrogen collected from electrolysis is used as the fuel at the galvanic anode: \[ \mathrm{H_2\rightarrow2H^++2e^-} \] Each hydrogen molecule releases two electrons. For the available hydrogen: \[ n(e^-)_{2,\max}=2(0.0200) \] \[ n(e^-)_{2,\max}=0.0400\ \mathrm{mol} \] \(\textbf{Calculate the galvanic charge capacity:}\) \[ Q_{2,\max}=n(e^-)_{2,\max}F \] \[ Q_{2,\max}=0.0400(96500) \] \[ Q_{2,\max}=3860\ \mathrm C \] \(\textbf{Apply the copper cathode reaction:}\) Copper(II) ions accept the electrons supplied by hydrogen oxidation: \[ \mathrm{Cu^{2+}+2e^-\rightarrow Cu} \] Therefore: \[ n(\mathrm{Cu})=\frac{0.0400}{2} \] \[ n(\mathrm{Cu})=0.0200\ \mathrm{mol} \] \(\textbf{Calculate deposited copper mass:}\) \[ m(\mathrm{Cu})=nM \] \[ m(\mathrm{Cu})=0.0200(63.5) \] \[ m(\mathrm{Cu})=1.270\ \mathrm g \] \(\textbf{Verify the second cell's balanced reaction:}\) Combining hydrogen oxidation and copper(II) reduction: \[ \mathrm{H_2+Cu^{2+}\rightarrow2H^++Cu} \] The available \(0.0200\ \mathrm{mol}\) of hydrogen can therefore produce \(0.0200\ \mathrm{mol}\) of copper. This agrees with the electron-based calculation. \(\textbf{Interpret the energy-conversion sequence:}\) The first cell uses externally supplied electrical energy to generate hydrogen from water. The collected hydrogen then acts as a reducing agent in the second cell, which operates galvanically. The electron amount associated with hydrogen formation is recovered as the same electron amount during its subsequent oxidation under the stated ideal chemical conversion assumptions. Equal charge amounts do not imply that the electrical energy supplied in the first stage equals the electrical energy obtained from the second. Electrical energy also depends on the relevant operating voltage. The question establishes the charge and material balances without assuming equality of energy. \(\textbf{Final answer:}\) \[ \boxed{\left(Q_{2,\max},m(\mathrm{Cu})\right)=(3860\ \mathrm C,\ 1.270\ \mathrm g)} \]

Explanation: \(\textbf{Identify the signed measurement convention:}\) The voltmeter is specified to measure the potential of the right electrode relative to the left. Therefore: \[ E_{\mathrm{reading}}=E_{\mathrm{right}}-E_{\mathrm{left}} \] This signed quantity must be distinguished from the positive emf assigned to a cell written in its spontaneous galvanic direction. \(\textbf{Evaluate arrangement P:}\) Both electrodes are zinc/zinc(II) electrodes under identical standard conditions. Their electrode potentials are equal: \[ E^\circ_{\mathrm{right}}=E^\circ_{\mathrm{left}}=-0.76\ \mathrm V \] Therefore: \[ E^\circ_{\mathrm P}=-0.76-(-0.76) \] \[ E^\circ_{\mathrm P}=0 \] A complete ionic and electronic connection does not create a driving force when both electrode systems have identical potentials. \(\textbf{Evaluate arrangement Q:}\) The right electrode is copper, while the left electrode is zinc. Therefore: \[ E^\circ_{\mathrm Q}=0.34-(-0.76) \] \[ E^\circ_{\mathrm Q}=+1.10\ \mathrm V \] In this orientation, zinc is the galvanic anode and copper is the cathode. \(\textbf{Evaluate arrangement R:}\) The physical half-cells are the same as those in Q, but their positions are exchanged. The right electrode is now zinc. Therefore: \[ E^\circ_{\mathrm R}=-0.76-(+0.34) \] \[ E^\circ_{\mathrm R}=-1.10\ \mathrm V \] The negative reading reflects the selected right-minus-left measurement orientation. It does not imply that zinc oxidation coupled with copper(II) reduction has become non-spontaneous. \(\textbf{Evaluate arrangement S:}\) Both electrodes are copper/copper(II) electrodes under identical standard conditions. Thus: \[ E^\circ_{\mathrm S}=0.34-0.34 \] \[ E^\circ_{\mathrm S}=0 \] \(\textbf{Distinguish electrical continuity from driving potential:}\) All four arrangements have electronic and ionic conducting pathways. However, a conducting circuit alone does not ensure a sustained spontaneous current. Arrangements P and S have zero open-circuit potential differences because their corresponding electrode systems are identical. Arrangements Q and R have the same physical electrode-potential difference in magnitude but opposite measured signs because their positions are reversed. The correctly ordered sequence is \(0,\ +1.10,\ -1.10,\ 0\ \mathrm V\).

Explanation: \(\textbf{Observation P — dissolved iron redox couple:}\) At the platinum cathode: \[ \mathrm{Fe^{3+}+e^-\rightarrow Fe^{2+}} \] Iron(III) accepts an electron but remains in solution as iron(II). Platinum provides an electronically conducting surface without becoming part of the specified redox products. Consequently, the platinum mass need not increase even though reduction occurs at its surface. P therefore matches explanation 3. \(\textbf{Observation Q — chlorine gas cathode:}\) At the chlorine electrode: \[ \mathrm{Cl_2+2e^-\rightarrow2Cl^-} \] Chlorine gas accepts electrons and becomes dissolved chloride ions. The cathode reaction consumes a gaseous substance instead of depositing a solid metal. The platinum surface enables electron exchange but is not itself reduced to form another solid. Q therefore matches explanation 1. \(\textbf{Observation R — active copper anode:}\) At a copper anode: \[ \mathrm{Cu\rightarrow Cu^{2+}+2e^-} \] Copper atoms leave the metallic electrode as dissolved copper(II) ions. The electrode loses mass because its own atoms participate in the oxidation reaction. An anode does not have to be chemically inert. R therefore matches explanation 2. \(\textbf{Observation S — aqueous sodium chloride cathode:}\) Sodium ions migrate towards the negatively charged cathode. However, the specified aqueous operating conditions favour water reduction: \[ \mathrm{2H_2O+2e^-\rightarrow H_2+2OH^-} \] Hydrogen is consequently formed rather than sodium metal. Sodium ions can contribute to ionic conduction without being the species discharged at the electrode. S therefore matches explanation 4. \(\textbf{Integrate the electrode observations:}\) The terms anode and cathode identify oxidation and reduction sites. They do not, by themselves, identify whether an electrode gains mass, loses mass or remains unchanged. A cathode can reduce a dissolved ion without metal deposition, consume a gas, or deposit a metal when the appropriate reaction is favoured. Similarly, an anode can dissolve if its constituent metal is oxidised. The actual half-reaction and participating species determine the observed material changes. The correct matching is P–3, Q–1, R–2, S–4.

Explanation: \(\textbf{Recover the common electron amount:}\) One mole of silver ions accepts one mole of electrons. \[ n(\mathrm{Ag})=\frac{6.48}{108}=0.0600\ \mathrm{mol} \] Hence \(0.0600\ \mathrm{mol}\) of electrons passed through P and, because the cells are in series, through Q. \(\textbf{Apply the second electrode reaction:}\) Each copper(II) ion accepts two electrons, so: \[ n(\mathrm{Cu})=\frac{0.0600}{2}=0.0300\ \mathrm{mol} \] \[ m(\mathrm{Cu})=0.0300(63.5)=1.905\ \mathrm g \] Equal charge does not imply equal deposited mass: the electron requirement per ion and the molar mass both matter. \(\textbf{Final answer:}\) \[ \boxed{m(\mathrm{Cu})=1.905\ \mathrm g,\quad n(e^-)=0.0600\ \mathrm{mol}} \]

Explanation: \(\textbf{Add charge from both time intervals:}\) \[ Q_1=(1.50)(20.0\times60)=1800\ \mathrm C \] \[ Q_2=(0.500)(30.0\times60)=900\ \mathrm C \] \[ Q_{\mathrm{total}}=2700\ \mathrm C \] \(\textbf{Convert charge into copper deposition:}\) \[ n(e^-)=\frac{2700}{96500}=0.02798\ \mathrm{mol} \] Two electrons deposit one copper atom: \[ n(\mathrm{Cu})=\frac{0.02798}{2}=0.01399\ \mathrm{mol}\approx0.0140\ \mathrm{mol} \] \(\textbf{Use the initial copper-ion inventory:}\) \[ n(\mathrm{Cu^{2+}})_{\mathrm{remaining}}=0.0250-0.01399\approx0.0110\ \mathrm{mol} \] The lower second-stage current still contributes to the charge; neither interval can be omitted. \(\textbf{Final answer:}\) \[ \boxed{n(\mathrm{Cu})=0.0140\ \mathrm{mol},\quad n(\mathrm{Cu^{2+}})_{\mathrm{remaining}}=0.0110\ \mathrm{mol}} \]

Explanation: \(\textbf{Separate hydrogen pressure from total collected-gas pressure:}\) The gas is collected over water and therefore contains water vapour. The total pressure is the sum of the partial pressures: \[ P_{\mathrm{total}}=P_{\mathrm{H_2}}+P_{\mathrm{H_2O}} \] Rearranging: \[ P_{\mathrm{H_2}}=P_{\mathrm{total}}-P_{\mathrm{H_2O}} \] Substituting: \[ P_{\mathrm{H_2}}=1.00-0.0400 \] \[ P_{\mathrm{H_2}}=0.960\ \mathrm{atm} \] \(\textbf{Apply the ideal-gas relation to dry hydrogen:}\) For hydrogen: \[ P_{\mathrm{H_2}}V=n_{\mathrm{H_2}}RT \] Therefore: \[ n_{\mathrm{H_2}}=\frac{P_{\mathrm{H_2}}V}{RT} \] Substituting: \[ n_{\mathrm{H_2}}=\frac{(0.960)(1.230)}{(0.0820)(300)} \] \[ n_{\mathrm{H_2}}=\frac{1.1808}{24.60} \] \[ n_{\mathrm{H_2}}=0.0480\ \mathrm{mol} \] \(\textbf{Connect collected hydrogen with cathode electron transfer:}\) The cathode reduction is: \[ \mathrm{2H^++2e^-\rightarrow H_2} \] Each mole of hydrogen requires two moles of electrons. Thus: \[ n(e^-)=2n(\mathrm{H_2}) \] \[ n(e^-)=2(0.0480) \] \[ n(e^-)=0.0960\ \mathrm{mol} \] \(\textbf{Calculate the passed charge:}\) \[ Q=n(e^-)F \] \[ Q=(0.0960)(96500) \] \[ Q=9264\ \mathrm C \] \(\textbf{Check the consequences of ignoring water vapour:}\) Using the total moist-gas pressure as though it belonged entirely to hydrogen would give: \[ n_{\mathrm{incorrect}}=\frac{(1.00)(1.230)}{24.60} \] \[ n_{\mathrm{incorrect}}=0.0500\ \mathrm{mol} \] This overestimates the hydrogen amount and the required charge. \(\textbf{Interpret the two distinct measurements:}\) The gas-volume measurement determines the amount of hydrogen only after correcting its partial pressure. Faraday's law then connects that hydrogen amount to the number of electrons passed through the electrolyser. The gas collection condition and the electrode half-reaction are separate constraints. Neither can be omitted without changing the answer. \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{H_2}),Q\right)=(0.0480\ \mathrm{mol},\ 9264\ \mathrm C)} \]

Explanation: \(\textbf{Determine electron transfer:}\) \[ n(e^-)=\frac{3860}{96500}=0.0400\ \mathrm{mol} \] Both electrode half-reactions involve two electrons per copper atom: \[ n(\mathrm{Cu})=\frac{0.0400}{2}=0.0200\ \mathrm{mol} \] \[ m(\mathrm{Cu})=0.0200(63.5)=1.27\ \mathrm g \] At the anode, \(0.0200\ \mathrm{mol}\) of copper enters the solution as copper(II); at the cathode the same amount leaves the solution and forms metallic copper. \[ \Delta n(\mathrm{Cu^{2+}})=+0.0200-0.0200=0 \] Electrode mass loss and mass gain are equal in this idealised cell, even though oxidation and reduction occur at different electrodes. \(\textbf{Final answer:}\) The anode loses \(1.27\ \mathrm g\), the cathode gains \(1.27\ \mathrm g\), and the net change in dissolved copper(II) is zero.

Explanation: \(\textbf{Determine the total amount of iron dissolved:}\) The anode loses iron through oxidation to either iron(II) or iron(III). The total iron amount is: \[ n(\mathrm{Fe})_{\mathrm{total}}=\frac{m}{M} \] \[ n(\mathrm{Fe})_{\mathrm{total}}=\frac{4.1875}{55.85} \] \[ n(\mathrm{Fe})_{\mathrm{total}}=0.0750\ \mathrm{mol} \] \(\textbf{Introduce the two oxidation products:}\) Let \(x\) be the moles of iron(II) produced and \(y\) the moles of iron(III) produced. Iron-atom conservation gives: \[ x+y=0.0750 \] \(\textbf{Construct the independent electron balance:}\) Each mole of iron(II) formed releases two moles of electrons. Each mole of iron(III) formed releases three moles of electrons. Therefore: \[ 2x+3y=0.200 \] \(\textbf{Solve the two conservation relations:}\) From iron conservation: \[ x=0.0750-y \] Substituting into the electron balance: \[ 2(0.0750-y)+3y=0.200 \] \[ 0.150-2y+3y=0.200 \] \[ y=0.0500\ \mathrm{mol} \] Therefore: \[ x=0.0750-0.0500 \] \[ x=0.0250\ \mathrm{mol} \] \(\textbf{Verify the measured anode mass:}\) The total amount of iron ions formed is: \[ 0.0250+0.0500=0.0750\ \mathrm{mol} \] The corresponding mass is: \[ m=0.0750(55.85) \] \[ m=4.1875\ \mathrm g \] \(\textbf{Verify the electrical measurement:}\) Electrons released through iron(II) formation: \[ 2(0.0250)=0.0500\ \mathrm{mol} \] Electrons released through iron(III) formation: \[ 3(0.0500)=0.150\ \mathrm{mol} \] Their sum is: \[ 0.0500+0.150=0.200\ \mathrm{mol} \] \(\textbf{Interpret the reverse inference:}\) Anode mass loss measures the total number of iron atoms oxidised. The electrical charge independently measures the total electrons released. Together, these measurements distinguish the amounts of two products formed from the same metal in different oxidation states. The electron amount alone would not establish their individual amounts without the mass measurement. Likewise, the mass loss alone would not establish the distribution between iron(II) and iron(III). \(\textbf{Final answer:}\) \[ \boxed{\left(n(\mathrm{Fe^{2+}}),n(\mathrm{Fe^{3+}})\right)=(0.0250,\ 0.0500)\ \mathrm{mol}} \]

Explanation: \(\textbf{Interpret the graph intercept:}\) For the supplied strong-electrolyte relation: \[ \Lambda_m=\Lambda_m^0-A\sqrt c \] the vertical intercept represents limiting molar conductivity. The magnitude of the negative slope is \(A\). The two graph points can therefore determine both constants. \(\textbf{Calculate the graph slope:}\) \[ m=\frac{140-150}{0.200-0.100} \] \[ m=\frac{-10.0}{0.100} \] \[ m=-100 \] Therefore: \[ A=100 \] with the corresponding units implied by the stated concentration convention. \(\textbf{Determine the limiting molar conductivity:}\) Using the first graph point: \[ 150=\Lambda_m^0-100(0.100) \] \[ 150=\Lambda_m^0-10.0 \] \[ \Lambda_m^0=160\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Calculate the molar conductivity at the higher requested concentration:}\) For: \[ c_2=0.0900\ \mathrm{mol\,L^{-1}} \] \[ \sqrt{c_2}=0.300 \] Therefore: \[ \Lambda_{m,2}=160-100(0.300) \] \[ \Lambda_{m,2}=130\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Convert to solution conductivity:}\) Using: \[ \kappa=\frac{c\Lambda_m}{1000} \] \[ \kappa_2=\frac{0.0900(130)}{1000} \] \[ \kappa_2=0.01170\ \mathrm{S\,cm^{-1}} \] \(\textbf{Calculate the conductivity at the lower concentration:}\) At: \[ c_1=0.0100\ \mathrm{mol\,L^{-1}} \] \[ \sqrt{c_1}=0.100 \] The graph gives: \[ \Lambda_{m,1}=150\ \mathrm{S\,cm^2\,mol^{-1}} \] Therefore: \[ \kappa_1=\frac{0.0100(150)}{1000} \] \[ \kappa_1=0.00150\ \mathrm{S\,cm^{-1}} \] \(\textbf{Compare the two solution conductivities:}\) \[ \frac{\kappa_2}{\kappa_1}=\frac{0.01170}{0.00150} \] \[ \frac{\kappa_2}{\kappa_1}=7.80 \] \(\textbf{Interpret the competing concentration effects:}\) The analytical concentration increases by a factor of nine. However, molar conductivity decreases from \(150\) to \(130\ \mathrm{S\,cm^2\,mol^{-1}}\). Consequently, solution conductivity increases by less than a factor of nine. The graph's vertical intercept gives a limiting property per mole, while the requested conductivity ratio depends on both concentration and molar conductivity. The calculation must therefore use the concentration correction rather than equating conductivity directly with concentration. \(\textbf{Final answer:}\) \[ \boxed{\left(\Lambda_m^0,\frac{\kappa_2}{\kappa_1}\right)=\left(160\ \mathrm{S\,cm^2\,mol^{-1}},\ 7.80\right)} \]

Explanation: \(\textbf{Distinguish the three transport quantities:}\) Measured conductance depends on both the solution and the geometry of the conductivity cell. Conductivity corrects conductance for the cell geometry. Molar conductivity additionally relates conductivity to the analytical electrolyte concentration. Therefore, the three comparisons need not give identical rankings. \(\textbf{Calculate the measured conductance of P:}\) \[ G_{\mathrm P}=\frac1{R_{\mathrm P}} \] \[ G_{\mathrm P}=\frac1{120} \] \[ G_{\mathrm P}=8.33\times10^{-3}\ \mathrm S \] \(\textbf{Calculate the measured conductance of Q:}\) \[ G_{\mathrm Q}=\frac1{150} \] \[ G_{\mathrm Q}=6.67\times10^{-3}\ \mathrm S \] Therefore: \[ G_{\mathrm P}\gt G_{\mathrm Q} \] \(\textbf{Calculate conductivity for each solution:}\) The cell relation is: \[ \kappa=\frac{G^*}{R} \] For P: \[ \kappa_{\mathrm P}=\frac{1.20}{120} \] \[ \kappa_{\mathrm P}=0.0100\ \mathrm{S\,cm^{-1}} \] For Q: \[ \kappa_{\mathrm Q}=\frac{1.80}{150} \] \[ \kappa_{\mathrm Q}=0.0120\ \mathrm{S\,cm^{-1}} \] Thus: \[ \kappa_{\mathrm Q}\gt \kappa_{\mathrm P} \] \(\textbf{Calculate molar conductivity of P:}\) \[ \Lambda_{m,\mathrm P}=\frac{1000\kappa_{\mathrm P}}{c_{\mathrm P}} \] \[ \Lambda_{m,\mathrm P}=\frac{1000(0.0100)}{0.0500} \] \[ \Lambda_{m,\mathrm P}=200\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Calculate molar conductivity of Q:}\) \[ \Lambda_{m,\mathrm Q}=\frac{1000(0.0120)}{0.0800} \] \[ \Lambda_{m,\mathrm Q}=150\ \mathrm{S\,cm^2\,mol^{-1}} \] Therefore: \[ \Lambda_{m,\mathrm P}\gt \Lambda_{m,\mathrm Q} \] \(\textbf{Interpret the reversal of rankings:}\) P has lower resistance and hence greater conductance in the particular cells used. However, Q's larger cell constant changes the geometry-corrected comparison, giving Q the greater intrinsic solution conductivity. After conductivity is divided by electrolyte concentration, P has the greater molar conductivity. The result demonstrates why resistance, conductivity and molar conductivity cannot be ranked interchangeably when both cell geometry and concentration differ. The correct option is A.

Explanation: \(\textbf{Obtain the initial molar conductivity:}\) At the first concentration: \[ c_1=0.0100\ \mathrm{mol\,L^{-1}} \] Using: \[ \Lambda_m=\frac{1000\kappa}{c} \] \[ \Lambda_{m,1}=\frac{1000(8.00\times10^{-4})}{0.0100} \] \[ \Lambda_{m,1}=80.0\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Determine the first degree of dissociation:}\) \[ \alpha_1=\frac{\Lambda_{m,1}}{\Lambda_m^0} \] \[ \alpha_1=\frac{80.0}{400} \] \[ \alpha_1=0.200 \] \(\textbf{Calculate the dissociation constant without approximation:}\) \[ K_a=\frac{c_1\alpha_1^2}{1-\alpha_1} \] Substituting: \[ K_a=\frac{0.0100(0.200)^2}{1-0.200} \] \[ K_a=\frac{0.000400}{0.800} \] \[ K_a=5.00\times10^{-4} \] \(\textbf{Apply the unchanged equilibrium constant at higher concentration:}\) The temperature is unchanged, so the same \(K_a\) applies. At: \[ c_2=0.0400\ \mathrm{mol\,L^{-1}} \] \[ 5.00\times10^{-4}=\frac{0.0400\alpha_2^2}{1-\alpha_2} \] Rearranging: \[ \frac{\alpha_2^2}{1-\alpha_2}=0.0125 \] \[ \alpha_2^2+0.0125\alpha_2-0.0125=0 \] \(\textbf{Solve for the physically meaningful root:}\) \[ \alpha_2=\frac{-0.0125+\sqrt{(0.0125)^2+4(0.0125)}}{2} \] \[ \alpha_2=\frac{-0.0125+\sqrt{0.05015625}}{2} \] \[ \alpha_2\approx0.1057 \] \(\textbf{Determine the second molar conductivity:}\) \[ \Lambda_{m,2}=\alpha_2\Lambda_m^0 \] \[ \Lambda_{m,2}\approx0.1057(400) \] \[ \Lambda_{m,2}\approx42.3\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Calculate the conductivity at the higher concentration:}\) \[ \kappa_2=\frac{c_2\Lambda_{m,2}}{1000} \] \[ \kappa_2\approx\frac{0.0400(42.3)}{1000} \] \[ \kappa_2\approx1.69\times10^{-3}\ \mathrm{S\,cm^{-1}} \] \(\textbf{Interpret the non-proportional concentration change:}\) The acid concentration increases fourfold. Its degree of dissociation falls because a smaller fraction of the more concentrated weak acid ionises. Consequently, its molar conductivity decreases from \(80.0\) to approximately \(42.3\ \mathrm{S\,cm^2\,mol^{-1}}\). Nevertheless, the solution conductivity increases because the greater analytical concentration more than compensates for the decrease in ionised fraction. The calculation uses the exact dissociation expression because the initial ionisation fraction of \(0.200\) is not sufficiently small to discard its denominator without justification. \(\textbf{Final answer:}\) \[ \boxed{\left(\alpha_2,\kappa_2\right)=\left(0.106,\ 1.69\times10^{-3}\ \mathrm{S\,cm^{-1}}\right)} \]

Explanation: \(\textbf{Identify the common-ion constraint:}\) Both salts are assumed to dissociate completely: \[ \mathrm{NaCl\rightarrow Na^++Cl^-} \] \[ \mathrm{KCl\rightarrow K^++Cl^-} \] Each formula unit contributes one chloride ion. Consequently, the chloride concentration equals the sum of the two salt concentrations. \(\textbf{Express the unknown salt concentrations:}\) Let: \[ c_{\mathrm{KCl}}=x \] \[ c_{\mathrm{NaCl}}=y \] The stated total concentration gives: \[ x+y=0.100 \] Therefore: \[ [\mathrm{K^+}]=x,\quad [\mathrm{Na^+}]=y,\quad [\mathrm{Cl^-}]=0.100 \] with concentrations in \(\mathrm{mol\,L^{-1}}\). \(\textbf{Construct the conductivity balance:}\) For independent ionic contributions: \[ 1000\kappa=74.0x+50.0y+76.0(0.100) \] Substitute the measured conductivity: \[ 1000(1.356\times10^{-2})=74.0x+50.0y+7.60 \] \[ 13.56=74.0x+50.0y+7.60 \] Therefore: \[ 74.0x+50.0y=5.96 \] \(\textbf{Solve the mixture composition:}\) Use: \[ y=0.100-x \] Then: \[ 74.0x+50.0(0.100-x)=5.96 \] \[ 74.0x+5.00-50.0x=5.96 \] \[ 24.0x=0.960 \] \[ x=0.0400\ \mathrm{mol\,L^{-1}} \] Therefore: \[ y=0.100-0.0400 \] \[ y=0.0600\ \mathrm{mol\,L^{-1}} \] \(\textbf{Determine the salt concentration ratio:}\) \[ c_{\mathrm{KCl}}:c_{\mathrm{NaCl}}=0.0400:0.0600 \] \[ =2:3 \] \(\textbf{Calculate the chloride conductivity contribution:}\) The chloride concentration is \(0.100\ \mathrm{mol\,L^{-1}}\). Therefore: \[ \kappa_{\mathrm{Cl^-}}=\frac{0.100(76.0)}{1000} \] \[ \kappa_{\mathrm{Cl^-}}=0.00760\ \mathrm{S\,cm^{-1}} \] \(\textbf{Calculate its fraction of total conductivity:}\) \[ f_{\mathrm{Cl^-}}=\frac{\kappa_{\mathrm{Cl^-}}}{\kappa} \] \[ f_{\mathrm{Cl^-}}=\frac{0.00760}{0.01356} \] \[ f_{\mathrm{Cl^-}}\approx0.5605 \] Therefore: \[ 100f_{\mathrm{Cl^-}}\approx56.0\% \] \(\textbf{Verify the inferred composition:}\) Using the inferred salt concentrations: \[ 1000\kappa=0.0400(74.0)+0.0600(50.0)+0.100(76.0) \] \[ 1000\kappa=2.96+3.00+7.60 \] \[ 1000\kappa=13.56 \] This reproduces the measured conductivity. \(\textbf{Interpret the inverse measurement:}\) The total salt concentration fixes the chloride-ion concentration but not the relative amounts of sodium and potassium ions. The conductivity measurement supplies the second independent relation required to determine the mixture composition. Because the two cations have different ionic conductivity contributions, their relative amounts affect the observed conductivity. The chloride contribution depends on the combined salt concentration and can then be evaluated separately. \(\textbf{Final answer:}\) \[ \boxed{\left(c_{\mathrm{KCl}}:c_{\mathrm{NaCl}},f_{\mathrm{Cl^-}}\right)=\left(2:3,\ 56.0\%\right)} \]

Explanation: \(\textbf{Construct the limiting molar conductivity of AgCl:}\) Kohlrausch's law permits the ionic contributions of AgCl to be isolated using strong-electrolyte data. The required combination is: \[ \Lambda_m^0(\mathrm{AgCl})=\Lambda_m^0(\mathrm{AgNO_3})+\Lambda_m^0(\mathrm{NaCl})-\Lambda_m^0(\mathrm{NaNO_3}) \] The sodium and nitrate contributions cancel in this expression. \(\textbf{Calculate the limiting value:}\) \[ \Lambda_m^0(\mathrm{AgCl})=133+126-121 \] \[ \Lambda_m^0(\mathrm{AgCl})=138\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Correct the measured conductivity for the solvent:}\) The saturated solution contains the background conductivity of the water as well as the contribution from dissolved AgCl. Under the stated additive approximation: \[ \kappa_{\mathrm{AgCl}}=\kappa_{\mathrm{sat}}-\kappa_{\mathrm{water}} \] Substituting: \[ \kappa_{\mathrm{AgCl}}=(1.58-0.20)\times10^{-6} \] \[ \kappa_{\mathrm{AgCl}}=1.38\times10^{-6}\ \mathrm{S\,cm^{-1}} \] \(\textbf{Connect conductivity with molar solubility:}\) For the dissolved salt: \[ \mathrm{AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)} \] Let the molar solubility be \(s\). Then: \[ [\mathrm{Ag^+}]=[\mathrm{Cl^-}]=s \] At the stated extreme dilution: \[ \Lambda_m\approx\Lambda_m^0(\mathrm{AgCl}) \] Using: \[ \Lambda_m=\frac{1000\kappa}{s} \] rearrange: \[ s=\frac{1000\kappa_{\mathrm{AgCl}}}{\Lambda_m^0(\mathrm{AgCl})} \] \(\textbf{Calculate the molar solubility:}\) \[ s=\frac{1000(1.38\times10^{-6})}{138} \] \[ s=1.00\times10^{-5}\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate the solubility product:}\) For the stated ideal concentration model: \[ K_{sp}=[\mathrm{Ag^+}][\mathrm{Cl^-}] \] Therefore: \[ K_{sp}=s^2 \] \[ K_{sp}=(1.00\times10^{-5})^2 \] \[ K_{sp}=1.00\times10^{-10} \] \(\textbf{Interpret the experimental corrections:}\) The measured conductivity of an extremely dilute saturated salt solution need not arise exclusively from the dissolved salt. A solvent blank can therefore be significant. Failure to subtract the supplied water contribution would overestimate the salt's solubility. The limiting molar conductivity must also be constructed using the correct ionic combination; using the conductivity of an unrelated strong electrolyte directly would be invalid. The final inference combines an electrical measurement with the equilibrium expression for a sparingly soluble \(1:1\) salt. \(\textbf{Final answer:}\) \[ \boxed{\left(s,K_{sp}\right)=\left(1.00\times10^{-5}\ \mathrm{mol\,L^{-1}},\ 1.00\times10^{-10}\right)} \]

Explanation: \(\textbf{Calculate conductivity independently from both cells:}\) \[ \kappa=\frac{l/A}{R} \] For P: \[ \kappa_P=\frac{1.20}{80.0}=0.0150\ \mathrm{S\,cm^{-1}} \] For Q: \[ \kappa_Q=\frac{1.80}{120}=0.0150\ \mathrm{S\,cm^{-1}} \] The two measurements agree despite their different resistances because the cell constants differ. \(\textbf{Convert the unit:}\) \[ 1\ \mathrm{S\,cm^{-1}}=100\ \mathrm{S\,m^{-1}} \] \[ \kappa=0.0150\times100=1.50\ \mathrm{S\,m^{-1}} \] Conductivity is a property of the solution at the stated temperature, not of the dimensions of the measurement cell. \(\textbf{Final answer:}\) \[ \boxed{\kappa=0.0150\ \mathrm{S\,cm^{-1}}=1.50\ \mathrm{S\,m^{-1}}} \]

Explanation: \(\textbf{Determine the original solubility product:}\) In pure water, dissolved AgCl produces equal concentrations of silver and chloride ions: \[ [\mathrm{Ag^+}]=[\mathrm{Cl^-}]=s_0 \] Therefore: \[ K_{sp}=s_0^2 \] \[ K_{sp}=(1.00\times10^{-5})^2 \] \[ K_{sp}=1.00\times10^{-10} \] \(\textbf{Apply the added common-ion condition:}\) NaCl supplies a chloride-ion concentration of approximately: \[ [\mathrm{Cl^-}]=0.0100\ \mathrm{mol\,L^{-1}} \] Let the new AgCl solubility be \(s\). The equilibrium expression gives: \[ K_{sp}=[\mathrm{Ag^+}][\mathrm{Cl^-}] \] Under the stated approximation: \[ 1.00\times10^{-10}\approx s(0.0100) \] Therefore: \[ s=\frac{1.00\times10^{-10}}{0.0100} \] \[ s=1.00\times10^{-8}\ \mathrm{mol\,L^{-1}} \] \(\textbf{Compare the two AgCl solubilities:}\) \[ \frac{s_0}{s}=\frac{1.00\times10^{-5}}{1.00\times10^{-8}} \] \[ \frac{s_0}{s}=1000 \] The common chloride ion therefore suppresses AgCl dissolution by a factor of \(1000\). \(\textbf{Calculate conductivity of the original saturated solution:}\) Using the limiting-contribution approximation: \[ \kappa_0=\frac{s_0\Lambda_m^0(\mathrm{AgCl})}{1000} \] \[ \kappa_0=\frac{(1.00\times10^{-5})(138)}{1000} \] \[ \kappa_0=1.38\times10^{-6}\ \mathrm{S\,cm^{-1}} \] \(\textbf{Estimate the conductivity of the salt-containing solution:}\) The dissolved NaCl concentration is much greater than the AgCl solubility. Its contribution is approximately: \[ \kappa_{\mathrm{NaCl}}=\frac{0.0100(126)}{1000} \] \[ \kappa_{\mathrm{NaCl}}=1.26\times10^{-3}\ \mathrm{S\,cm^{-1}} \] The much smaller silver-ion contribution does not change the conclusion. \(\textbf{Interpret the apparently opposing trends:}\) The added chloride ions reduce AgCl solubility through the common-ion effect. Nevertheless, the added NaCl supplies a large concentration of mobile sodium and chloride ions. Therefore, the total number of conducting ions in the salt-containing solution is much greater than in the saturated AgCl solution prepared using pure water. The solution can accordingly become much more conducting even though less AgCl dissolves. \(\textbf{Identify the experimental misconception:}\) Total conductivity of a mixture cannot automatically be treated as conductivity due only to the sparingly soluble salt. Inferring AgCl solubility directly from the uncorrected conductivity of the NaCl-containing sample would be invalid. The result combines a chemical-equilibrium suppression of dissolution with an independent increase in ionic charge carriers. Both effects must be considered to understand the observations.

Explanation: \(\textbf{Separate material properties from cell geometry:}\) Resistivity is a property of the electrolyte under specified conditions. The measured resistance also depends on the dimensions of the conducting path: \[ R=\rho\frac{l}{A} \] Molar conductivity further depends on electrolyte concentration. These quantities must be calculated separately. \(\textbf{Convert the final cell dimensions into SI units:}\) The final electrode separation is: \[ l_2=1.00\ \mathrm{cm}=0.0100\ \mathrm m \] The final area is: \[ A_2=8.00\ \mathrm{cm^2} \] Since: \[ 1\ \mathrm{cm^2}=10^{-4}\ \mathrm{m^2} \] the final area is: \[ A_2=8.00\times10^{-4}\ \mathrm{m^2} \] \(\textbf{Calculate the final resistance:}\) \[ R_2=\rho_2\frac{l_2}{A_2} \] Substituting: \[ R_2=3.20\frac{0.0100}{8.00\times10^{-4}} \] \[ R_2=\frac{0.0320}{0.000800} \] \[ R_2=40.0\ \Omega \] \(\textbf{Convert final resistivity into conductivity:}\) Conductivity and resistivity are reciprocals: \[ \kappa_2=\frac1{\rho_2} \] \[ \kappa_2=\frac1{3.20} \] \[ \kappa_2=0.3125\ \mathrm{S\,m^{-1}} \] \(\textbf{Convert the concentration into SI units:}\) The diluted concentration is: \[ c_2=0.0125\ \mathrm{mol\,L^{-1}} \] Since one cubic metre contains \(1000\) litres: \[ c_2=12.5\ \mathrm{mol\,m^{-3}} \] \(\textbf{Calculate molar conductivity in SI units:}\) \[ \Lambda_{m,2}=\frac{\kappa_2}{c_2} \] \[ \Lambda_{m,2}=\frac{0.3125}{12.5} \] \[ \Lambda_{m,2}=0.0250\ \mathrm{S\,m^2\,mol^{-1}} \] \(\textbf{Convert to conventional chemistry units:}\) Using: \[ 1\ \mathrm{S\,m^2\,mol^{-1}}=10^4\ \mathrm{S\,cm^2\,mol^{-1}} \] \[ \Lambda_{m,2}=0.0250(10^4) \] \[ \Lambda_{m,2}=250\ \mathrm{S\,cm^2\,mol^{-1}} \] \(\textbf{Check the apparent resistance change:}\) The initial resistance was: \[ R_1=2.00\frac{0.0200}{4.00\times10^{-4}} \] \[ R_1=100\ \Omega \] Although the diluted electrolyte has greater resistivity, the changed cell geometry reduces its measured resistance from \(100\ \Omega\) to \(40.0\ \Omega\). \(\textbf{Interpret the result:}\) Dilution changes the material's conductivity and molar conductivity. Changing electrode separation and area changes the measured resistance independently. The final resistance decreases because the geometrical reduction outweighs the increase in resistivity. The calculation therefore requires distinguishing intrinsic electrolyte transport properties from the dimensions of the measurement cell. \(\textbf{Final answer:}\) \[ \boxed{\left(R_2,\Lambda_{m,2}\right)=\left(40.0\ \Omega,\ 250\ \mathrm{S\,cm^2\,mol^{-1}}\right)} \]

Explanation: \(\textbf{Determine the initial barium chloride inventory:}\) The first solution contains: \[ n(\mathrm{BaCl_2})=cV \] \[ n(\mathrm{BaCl_2})=(0.100)(0.100) \] \[ n(\mathrm{BaCl_2})=0.0100\ \mathrm{mol} \] Complete dissociation gives: \[ n(\mathrm{Ba^{2+}})=0.0100\ \mathrm{mol} \] \[ n(\mathrm{Cl^-})=0.0200\ \mathrm{mol} \] \(\textbf{Determine the initial sodium sulphate inventory:}\) Similarly: \[ n(\mathrm{Na_2SO_4})=(0.100)(0.100) \] \[ n(\mathrm{Na_2SO_4})=0.0100\ \mathrm{mol} \] Complete dissociation gives: \[ n(\mathrm{Na^+})=0.0200\ \mathrm{mol} \] \[ n(\mathrm{SO_4^{2-}})=0.0100\ \mathrm{mol} \] \(\textbf{Apply the precipitation reaction:}\) The net ionic equation is: \[ \mathrm{Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4(s)} \] Barium and sulphate are present in equal mole amounts. They are both completely removed as the precipitate under the stated conditions. The remaining dissolved ions are sodium and chloride. \(\textbf{Calculate the total final solution volume:}\) \[ V_f=0.100+0.100 \] \[ V_f=0.200\ \mathrm L \] \(\textbf{Determine final ionic concentrations:}\) The sodium amount remains \(0.0200\ \mathrm{mol}\): \[ [\mathrm{Na^+}]_f=\frac{0.0200}{0.200} \] \[ [\mathrm{Na^+}]_f=0.100\ \mathrm{mol\,L^{-1}} \] Similarly: \[ [\mathrm{Cl^-}]_f=\frac{0.0200}{0.200} \] \[ [\mathrm{Cl^-}]_f=0.100\ \mathrm{mol\,L^{-1}} \] \(\textbf{Calculate the final conductivity:}\) \[ \kappa_f=\frac{[\mathrm{Na^+}]\lambda_{\mathrm{Na^+}}+[\mathrm{Cl^-}]\lambda_{\mathrm{Cl^-}}}{1000} \] Substituting: \[ \kappa_f=\frac{0.100(50.0)+0.100(76.0)}{1000} \] \[ \kappa_f=\frac{5.00+7.60}{1000} \] \[ \kappa_f=1.26\times10^{-2}\ \mathrm{S\,cm^{-1}} \] \(\textbf{Count the ions before precipitation:}\) The combined initial dissolved-ion amount is: \[ n_{\mathrm{ions},i}=0.0100+0.0200+0.0200+0.0100 \] \[ n_{\mathrm{ions},i}=0.0600\ \mathrm{mol} \] \(\textbf{Count the remaining dissolved ions:}\) After precipitation: \[ n_{\mathrm{ions},f}=0.0200+0.0200 \] \[ n_{\mathrm{ions},f}=0.0400\ \mathrm{mol} \] \(\textbf{Calculate the fractional decrease:}\) \[ f_{\mathrm{decrease}}=\frac{0.0600-0.0400}{0.0600} \] \[ f_{\mathrm{decrease}}=\frac{0.0200}{0.0600} \] \[ f_{\mathrm{decrease}}=\frac13 \] \(\textbf{Interpret the chemical and transport changes:}\) Precipitation removes barium and sulphate ions from the conducting solution. Sodium and chloride remain as mobile spectator ions. The final conductivity must therefore be calculated from the remaining ionic species, not from the original formula-unit concentrations. The reduction in total dissolved-ion amount is determined by the precipitation stoichiometry. The conductivity is determined by the concentrations and transport contributions of the ions remaining after mixing. \(\textbf{Final answer:}\) \[ \boxed{\left(\kappa_f,f_{\mathrm{decrease}}\right)=\left(1.26\times10^{-2}\ \mathrm{S\,cm^{-1}},\frac13\right)} \]