Given below is the plot of the molar conductivity vs sqrttextconcentration$\sqrt{\text{concentration}}$ for KCl in aqueous solution.

If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100Omega$100\Omega$ then the resistance of the same cell with the dilute solution is mathrmxOmega$\mathrm{x}\Omega$ The value of mathbfx$\mathbf{x}$ is (Nearest integer)

### Related Formula E = E^circ - frac2.303RTnF log Q$$E = E^\circ - \frac{2.303RT}{nF} \log Q$$ For the Standard Hydrogen Electrode half-reaction: 2H^+ + 2e^- rightarrow H_2$2H^+ + 2e^- \rightarrow H_2$ E_H^+/H_2 = E^circ_H^+/H_2 - frac0.062 log fracP_H_2[H^+]^2$$E_{H^+/H_2} = E^\circ_{H^+/H_2} - \frac{0.06}{2} \log \frac{P_{H_2}}{[H^+]^2}$$ ### Step 1: Substitute the given values E^circ_H^+/H_2 = 0.00 mathrm~V$E^\circ_{H^+/H_2} = 0.00 \mathrm{~V}$ (by definition) [H^+] = 1 mathrm~M$[H^+] = 1 \mathrm{~M}$ P_H_2 = 2 mathrm~atm$P_{H_2} = 2 \mathrm{~atm}$ n = 2$n = 2$ electrons E = 0.00 - frac0.062 log left( frac21^2 right)$$E = 0.00 - \frac{0.06}{2} \log \left( \frac{2}{1^2} \right)$$ ### Step 2: Solve the calculation E = -0.03 log 2$E = -0.03 \log 2$ Given log 2 = 0.3$\log 2 = 0.3$ E = -0.03 times 0.3$E = -0.03 \times 0.3$ E = -0.009 mathrm~V$E = -0.009 \mathrm{~V}$ E = -0.9 times 10^-2 mathrm~V$E = -0.9 \times 10^{-2} \mathrm{~V}$ ### Step 3: Match the requested format The question asks for (-) dots times 10^-2 mathrm~V$(-) \dots \times 10^{-2} \mathrm{~V}$. This gives exactly 0.9$0.9$. For NAT type with integer expected, 0.9$0.9$ can be rounded to 1$1$. However, exact calculation yields 0.9$0.9$. According to official JEE rounding, 0.9 approx 1$0.9 \approx 1$. ### Pattern Recognition Hydrogen electrode non-standard potential depends strictly on pressure of H_2$H_2$ and concentration of H^+$H^+$. If [H^+]=1$[H^+]=1$, increasing H_2$H_2$ pressure lowers the potential below zero (makes it negative). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula textNumber of equivalents deposited = fracW, E = fracQ, F$$\text{Number of equivalents deposited} = \frac{W}, {E} = \frac{Q}, {F}$$ textEquivalent Weight (E) = fractextAtomic Mass, ntext-factor$$\text{Equivalent Weight } (E) = \frac{\text{Atomic Mass}}, {n\text{-factor}}$$ ### Core Logic In the reduction of gold from the tetrachloroaurate(III$III$) complex anion: mathrmAuCl_4^- + 3e^- rightarrow mathrmAu(s) + 4mathrmCl^- implies ntext-factor = 3$$\mathrm{AuCl}_4^- + 3e^- \rightarrow \mathrm{Au}(s) + 4\mathrm{Cl}^- \implies n\text{-factor} = 3$$ Calculate the equivalent weight (E$E$) of Gold: E = frac197, 3$$E = \frac{197}, {3}$$ Set up the Faraday equivalence relation to solve for charge (Q$Q$ in Faradays): frac1.314, left(frac197, 3right) = Q$$\frac{1.314}, {\left(\frac{197}, {3}\right)} = Q$$ ### Step 1: Arithmetic Resolution Q = frac1.314 times 3, 197 = frac3.942, 197 = 0.02text F = 2 times 10^-2text F$$Q = \frac{1.314 \times 3}, {197} = \frac{3.942}, {197} = 0.02\text{ F} = 2 \times 10^{-2}\text{ F}$$ Thus, the required integer value is **2**. ### Pattern Recognition Always determine the correct change in oxidation state (+3$+3$ to 0$0$) to establish the proper n$n$-factor value for calculations using Faraday's laws. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Related Formula By Faraday's Second Law of Electrolysis: textEquivalents of Ag = textEquivalents of O_2$$\text{Equivalents of Ag} = \text{Equivalents of O}_2$$ textEquivalents = fractextMasstextEquivalent Mass = textMoles times ntext-factor$$\text{Equivalents} = \frac{\text{Mass}}{\text{Equivalent Mass}} = \text{Moles} \times n\text{-factor}$$ ### Step 1: Calculate equivalents using standard metrics Let x$x$ grams of Silver be displaced. Using the older STP molar volume baseline (22.4text L$22.4\text{ L}$ or 22400text mL$22400\text{ mL}$): textMoles of O_2 = frac560022400 = 0.25text moles$$\text{Moles of O}_2 = \frac{5600}{22400} = 0.25\text{ moles}$$ Since the n$n$-factor of O_2$O_2$ is 4 (2textO^2- rightarrow textO_2 + 4texte^-$2\text{O}^{2-} \rightarrow \text{O}_2 + 4\text{e}^-$): textEquivalents of O_2 = 0.25 times 4 = 1$$\text{Equivalents of O}_2 = 0.25 \times 4 = 1$$ ### Step 2: Equating equivalents for silver mass textEquivalents of Ag = fracx108 times 1 = 1 implies x = 108text g$$\text{Equivalents of Ag} = \frac{x}{108} \times 1 = 1 \implies x = 108\text{ g}$$ ### Step 3: Alternative calculation using current STP metric Using modern STP volume metrics (22.7text L$22.7\text{ L}$): fracx times 1108 = frac5.622.7 times 4 implies x approx 106.57text g rightarrow 107text g$$\frac{x \times 1}{108} = \frac{5.6}{22.7} \times 4 \implies x \approx 106.57\text{ g} \rightarrow 107\text{ g}$$ ### Pattern Recognition Equivalents equations bypass complex current/time measurements. Always link volume fractions directly to n$n$-factor equivalents. ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 11 Chemistry: Some Basic Concepts of Chemistry

### Related Formula W = Z cdot I cdot t = fracMn cdot F cdot I cdot t$$W = Z \cdot I \cdot t = \frac{M}{n \cdot F} \cdot I \cdot t$$ where, W$W$ = mass deposited Z$Z$ = electrochemical equivalent I$I$ = current in amperes t$t$ = time in seconds M$M$ = molar mass n$n$ = n-factor (electrons exchanged) F$F$ = Faraday's constant (96500 text C/mol$96500 \text{ C/mol}$) ### Core Logic The electrolysis of zinc sulphate (ZnSO_4$ZnSO_4$) involves the reduction of zinc ions at the cathode: Zn^+2 + 2e^- rightarrow Zn$$Zn^{+2} + 2e^- \rightarrow Zn$$ Here, the n-factor (n$n$) is 2. ### Step 1: Calculation Given values: I = 0.015text A$I = 0.015\text{ A}$ t = 15text minutes = 15 times 60text seconds = 900text s$t = 15\text{ minutes} = 15 \times 60\text{ seconds} = 900\text{ s}$ M = 65.4text g/mol$M = 65.4\text{ g/mol}$ F approx 96500text C$F \approx 96500\text{ C}$ Plugging the values into Faraday's First Law: W = frac65.42 times 96500 times 0.015 times 15 times 60$$W = \frac{65.4}{2 \times 96500} \times 0.015 \times 15 \times 60$$ W = frac65.4193000 times 13.5$$W = \frac{65.4}{193000} \times 13.5$$ W = 3.3886 times 10^-4 times 13.5$$W = 3.3886 \times 10^{-4} \times 13.5$$ W = 45.746 times 10^-4text g$$W = 45.746 \times 10^{-4}\text{ g}$$ Rounding to two decimal places (or nearest integer depending on convention), we get 45.75 times 10^-4text g$45.75 \times 10^{-4}\text{ g}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry

### Core Logic The Standard Reduction Potential (E^circ$E^\circ$) measures a species' tendency to undergo reduction (gain electrons). A higher, more positive E^circ$E^\circ$ value means the species has a stronger tendency to be reduced, which in turn makes it a stronger oxidizing agent. Comparing the given E^circ$E^\circ$ values: mathrmBrO_4^-$\mathrm{BrO}_4^-$: 1.74mathrmV$1.74\mathrm{V}$ mathrmIO_4^-$\mathrm{IO}_4^-$: 1.65mathrmV$1.65\mathrm{V}$ mathrmClO_4^-$\mathrm{ClO}_4^-$: 1.19mathrmV$1.19\mathrm{V}$ The order of oxidizing power follows the magnitude of the reduction potential: mathrmBrO_4^- > mathrmIO_4^- > mathrmClO_4^-$\mathrm{BrO}_4^- > \mathrm{IO}_4^- > \mathrm{ClO}_4^-$ ### Pattern Recognition Higher +ve Standard Reduction Potential (SRP) = Stronger Oxidising Agent. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: Electrochemistry Class 12 Chemistry: The p Block Elements