Conductor and Conductance and Cell constant Questions in English

(C) The cell constant $G^* = \frac{\ell}{a} = \frac{120 \ cm}{1 \ cm^2} = 120 \ cm^{-1}$.
The conductivity $\kappa = G \times G^* = 5 \times 10^{-7} \ S \times 120 \ cm^{-1} = 6 \times 10^{-5} \ S \ cm^{-1}$.
The molar conductivity at concentration $C$ is $\Lambda_{m}^{c} = \frac{\kappa \times 1000}{C} = \frac{6 \times 10^{-5} \times 1000}{0.0015} = 40 \ S \ cm^2 \ mol^{-1}$.
For a weak acid,$pH = 4 \implies [H^+] = 10^{-4} \ M$. Since $[H^+] = C \alpha$,the degree of dissociation $\alpha = \frac{10^{-4}}{0.0015} = \frac{1}{15}$.
Using the relation $\alpha = \frac{\Lambda_{m}^{c}}{\Lambda_{m}^0}$,we get $\Lambda_{m}^0 = \frac{\Lambda_{m}^{c}}{\alpha} = \frac{40}{1/15} = 600 \ S \ cm^2 \ mol^{-1}$.
Given $\Lambda_{m}^0 = Z \times 10^2 \ S \ cm^2 \ mol^{-1}$,we have $Z \times 10^2 = 600$,which implies $Z = 6$.

(A) The reaction between $AgNO_3$ and $KCl$ is: $Ag^+(aq) + NO_3^-(aq) + K^+(aq) + Cl^-(aq) \rightarrow AgCl(s) + K^+(aq) + NO_3^-(aq)$.
Initially,the solution contains $K^+$ and $Cl^-$ ions. As $AgNO_3$ is added,$Ag^+$ ions react with $Cl^-$ ions to form $AgCl$ precipitate.
Since $AgCl$ is insoluble,the $Cl^-$ ions are replaced by $NO_3^-$ ions in the solution.
The ionic mobility of $Cl^-$ is higher than that of $NO_3^-$,so the conductance of the solution decreases until the equivalence point is reached.
After the equivalence point,adding more $AgNO_3$ increases the concentration of $Ag^+$ and $NO_3^-$ ions in the solution,which leads to an increase in the conductance.
Therefore,the graph shows a decrease followed by an increase,which corresponds to plot $(P)$.

(C) The degree of dissociation is given by $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$.
For the initial solution,$\alpha = \frac{y \times 10^2}{4 \times 10^2} = \frac{y}{4}$.
The dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1-\alpha} = \frac{C (\Lambda_m / \Lambda_m^\circ)^2}{1 - (\Lambda_m / \Lambda_m^\circ)} = \frac{C \Lambda_m^2}{\Lambda_m^\circ(\Lambda_m^\circ - \Lambda_m)}$.
For the initial solution: $K_a = \frac{C (y \times 10^2)^2}{4 \times 10^2 (4 \times 10^2 - y \times 10^2)} = \frac{C y^2}{4(4-y)}$.
After $20$ times dilution,the concentration becomes $C' = C/20$ and molar conductivity becomes $\Lambda_m' = 3y \times 10^2$.
$K_a = \frac{(C/20) (3y \times 10^2)^2}{4 \times 10^2 (4 \times 10^2 - 3y \times 10^2)} = \frac{C (9y^2)}{20 \times 4 (4-3y)} = \frac{9 C y^2}{80(4-3y)}$.
Equating the two expressions for $K_a$: $\frac{C y^2}{4(4-y)} = \frac{9 C y^2}{80(4-3y)}$.
$\frac{1}{4-y} = \frac{9}{20(4-3y)}$ $\Rightarrow 20(4-3y) = 9(4-y)$ $\Rightarrow 80 - 60y = 36 - 9y$.
$44 = 51y \Rightarrow y = \frac{44}{51} \approx 0.86$.
Then $\alpha = \frac{y}{4} = \frac{44/51}{4} = \frac{11}{51} \approx 0.2156 \approx 0.21$.

(A) $(P)$ $(C_2H_5)_3N + CH_3COOH \longrightarrow (C_2H_5)_3NH^+ + CH_3COO^-$. Since both reactants are weak,the formation of ions leads to an increase in conductivity initially. After the equivalence point,the addition of excess weak acid does not significantly change the ion concentration,so it remains nearly constant. Thus,$P-3$.
$(Q)$ $KI + AgNO_3 \longrightarrow AgI(s) + KNO_3$. Here,$Ag^+$ ions are replaced by $K^+$ ions in the solution. Since the mobility of $K^+$ is comparable to $Ag^+$,the conductivity does not change much initially. After the equivalence point,excess $AgNO_3$ adds $Ag^+$ and $NO_3^-$ ions,causing conductivity to increase. Thus,$Q-4$.
$(R)$ $CH_3COOH + KOH \longrightarrow CH_3COOK + H_2O$. $OH^-$ ions (high mobility) are replaced by $CH_3COO^-$ ions (low mobility),causing conductivity to decrease. After the equivalence point,excess $KOH$ adds $K^+$ and $OH^-$ ions,but the effect is minimal due to the common ion effect or dilution. Thus,$R-2$.
$(S)$ $NaOH + HI \longrightarrow NaI + H_2O$. $H^+$ ions (very high mobility) are replaced by $Na^+$ ions (low mobility),causing conductivity to decrease. After the equivalence point,excess $HI$ adds $H^+$ and $I^-$ ions,causing conductivity to increase. Thus,$S-1$.
The correct matching is $P-3, Q-4, R-2, S-1$,which corresponds to option $(A)$.

(A) From the graph,for the concentrated solution $(c)$: $\sqrt{C_c} = 0.15 \ (mol/L)^{1/2}$,so $C_c = (0.15)^2 = 0.0225 \ mol/L$. The molar conductivity $\Lambda_{m,c} = 100 \ S \ cm^2 \ mol^{-1}$.
For the dilute solution $(d)$: $\sqrt{C_d} = 0.1 \ (mol/L)^{1/2}$,so $C_d = (0.1)^2 = 0.01 \ mol/L$. The molar conductivity $\Lambda_{m,d} = 150 \ S \ cm^2 \ mol^{-1}$.
We know that $\kappa = \frac{\Lambda_m \cdot C}{1000}$ and $R = \frac{G^*}{\kappa}$,where $G^*$ is the cell constant.
Thus,$R = \frac{1000 \cdot G^*}{\Lambda_m \cdot C}$.
Taking the ratio of resistances: $\frac{R_d}{R_c} = \frac{\Lambda_{m,c} \cdot C_c}{\Lambda_{m,d} \cdot C_d}$.
Substituting the values: $\frac{R_d}{100} = \frac{100 \times 0.0225}{150 \times 0.01} = \frac{2.25}{1.5} = 1.5$.
Therefore,$R_d = 100 \times 1.5 = 150 \ \Omega$.

(D) For a weak electrolyte,the degree of dissociation increases as the concentration decreases (dilution increases).
At very low concentrations (near infinite dilution,where $\sqrt{C} \to 0$),the degree of dissociation increases rapidly,leading to a sharp increase in molar conductivity.
Conversely,as the concentration increases (moving away from the y-axis),the degree of dissociation decreases,causing a sharp decrease in molar conductivity.
Therefore,the graph shows that molar conductivity decreases sharply with an increase in concentration.

(B) Given: Concentration of $NaOH = 0.2 \% (w/v)$.
This means $0.2 \ g$ of $NaOH$ is present in $100 \ mL$ of solution.
Molarity $(M) = \frac{\text{moles of solute}}{\text{Volume of solution in } L} = \frac{0.2 / 40}{100 / 1000} = \frac{0.005}{0.1} = 0.05 \ M$.
Given resistivity $(\rho) = 870.0 \ m\Omega \ m = 870.0 \times 10^{-3} \ \Omega \ m = 0.87 \ \Omega \ m$.
Since $1 \ m = 10 \ dm$,$\rho = 0.87 \ \Omega \ m = 0.87 \ \Omega \ (10 \ dm) = 8.7 \ \Omega \ dm$.
Conductivity $(\kappa) = \frac{1}{\rho} = \frac{1}{8.7} \ S \ dm^{-1}$.
Molar conductivity $(\Lambda_m) = \frac{\kappa}{M} = \frac{1 / 8.7}{0.05} = \frac{1}{8.7 \times 0.05} = \frac{1}{0.435} \approx 2.2988 \ S \ dm^2 \ mol^{-1}$.
Converting to $mS \ dm^2 \ mol^{-1}$: $2.2988 \ S \ dm^2 \ mol^{-1} = 2298.8 \ mS \ dm^2 \ mol^{-1} = 22.988 \times 10^2 \ mS \ dm^2 \ mol^{-1}$.
Rounding to the nearest integer,we get $23 \times 10^2 \ mS \ dm^2 \ mol^{-1}$.

(B) The limiting molar conductivities $(\lambda^0)$ of the given ions in water at $298 \ K$ are as follows $:$
$H^{ } : 349.8 \ S \ cm^2 \ mol^{-1}$
$Ca^{2 } : 119.0 \ S \ cm^2 \ mol^{-1}$
$Mg^{2 } : 106.1 \ S \ cm^2 \ mol^{-1}$
$K^{ } : 73.5 \ S \ cm^2 \ mol^{-1}$
$Na^{ } : 50.1 \ S \ cm^2 \ mol^{-1}$
Comparing these values,the correct order of limiting molar conductivity is $H^{ } > Ca^{2 } > Mg^{2 } > K^{ } > Na^{ }$.

(D) Given: $\lambda_{m}^{\circ}(NH_4Cl) = 185 \ S \ cm^2 \ mol^{-1}$,$\lambda_{m}^{\circ}(OH^-) = 170 \ S \ cm^2 \ mol^{-1}$,$\lambda_{m}^{\circ}(Cl^-) = 70 \ S \ cm^2 \ mol^{-1}$.
Using Kohlrausch's law: $\lambda_{m}^{\circ}(NH_4Cl) = \lambda_{m}^{\circ}(NH_4^+) + \lambda_{m}^{\circ}(Cl^-) = 185$.
Therefore,$\lambda_{m}^{\circ}(NH_4^+) = 185 - 70 = 115 \ S \ cm^2 \ mol^{-1}$.
Now,$\lambda_{m}^{\circ}(NH_4OH) = \lambda_{m}^{\circ}(NH_4^+) + \lambda_{m}^{\circ}(OH^-) = 115 + 170 = 285 \ S \ cm^2 \ mol^{-1}$.
The degree of dissociation $(\alpha)$ is given by $\alpha = \frac{\lambda_{m}}{\lambda_{m}^{\circ}} = \frac{85.5}{285} = 0.3$.
Since $\alpha = x \times 10^{-1}$,we have $0.3 = x \times 10^{-1}$,which gives $x = 3$.

(C) Mohr's salt is $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$.
It dissociates into $Fe^{2+}$,$2NH_4^+$,and $2SO_4^{2-}$ ions. Thus,Statement $I$ is true.
According to Kohlrausch's law of independent migration of ions,the molar conductance at infinite dilution is the sum of the molar conductances of the constituent ions multiplied by their respective stoichiometric coefficients.
$\lambda_{m}^{\infty} (FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O) = \lambda^{\infty}(Fe^{2+}) + 2\lambda^{\infty}(NH_4^+) + 2\lambda^{\infty}(SO_4^{2-}) = x_1 + 2x_2 + 2x_3$.
Since the given expression in Statement $II$ is $x_1 + x_2 + 2x_3$,Statement $II$ is false.

(B) Conductivity (specific conductance) is defined as the conductance of a solution contained between two electrodes of unit area of cross-section separated by a unit distance.
With dilution,the number of ions per unit volume decreases.
Since conductivity depends on the number of ions present per unit volume,it decreases upon dilution.

(B) The variation of molar conductivity $(\lambda_m)$ with concentration $(C)$ for strong electrolytes is given by the Kohlrausch equation: $\lambda_m = \lambda_m^\circ - A\sqrt{C}$.
Here,$\lambda_m^\circ$ is the molar conductivity at infinite dilution,$A$ is a constant,and $\sqrt{C}$ is the square root of the concentration.
This equation represents a straight line equation of the form $y = mx + c$,where $y = \lambda_m$,$x = \sqrt{C}$,$m = -A$ (slope),and $c = \lambda_m^\circ$ (intercept).
Since the slope is negative $(-A)$,the graph of $\lambda_m$ versus $\sqrt{C}$ is a straight line with a negative slope.

(C) According to Kohlrausch's law,the equivalent conductivity at infinite dilution $(\lambda^{\infty})$ for the weak electrolyte $CH_3COOH$ is given by:
$\lambda^{\infty}_{CH_3COOH} = \lambda^{\infty}_{CH_3COONa} + \lambda^{\infty}_{HCl} - \lambda^{\infty}_{NaCl}$
Therefore,to calculate the value of $\lambda^{\infty}_{CH_3COOH}$,the value of $\lambda^{\infty}_{NaCl}$ must also be known.

(B) The formula for specific conductance $(\kappa)$ is given by: $\kappa = \frac{1}{R} \times \frac{\ell}{A}$
Given: Resistance $(R) = 50 \ \Omega$,distance $(\ell) = 0.02 \ m$,and area $(A) = 0.0004 \ m^2$.
Substituting the values: $\kappa = \frac{1}{50} \times \frac{0.02}{0.0004}$
$\kappa = \frac{0.02}{0.02} = 1 \ S \ m^{-1}$.

(A) According to Kohlrausch's law of independent migration of ions,the equivalent conductivity of an electrolyte at infinite dilution is the sum of the equivalent conductivities of its constituent ions.
For $BaCl_2$,the equivalent conductivity at infinite dilution is given by:
$\Lambda^{\infty}_{eq}(BaCl_2) = \lambda^{\infty}_{eq}(Ba^{2+}) + \lambda^{\infty}_{eq}(Cl^{-})$
Given that the ionic conductivities provided are already in terms of equivalent conductivities $(ohm^{-1} cm^2 eq^{-1})$,we simply add the values:
$\Lambda^{\infty}_{eq}(BaCl_2) = 127 + 76 = 203 \ ohm^{-1} cm^2 eq^{-1}$
Therefore,the correct option is $A$.

(A) The dissociation of $X_2Y$ is given by: $X_2Y \rightarrow 2X^{+} + Y^{-2}$.
According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.
$\Lambda_{m}^{\infty}(X_2Y) = 2 \times \lambda^{\infty}(X^{+}) + 1 \times \lambda^{\infty}(Y^{-2})$.
Substituting the given values: $\Lambda_{m}^{\infty}(X_2Y) = 2 \times 45 + 110 = 90 + 110 = 200 \ S \ cm^2 \ mol^{-1}$.

(C) For a strong electrolyte,the variation of molar conductance $(\Lambda_m)$ with concentration $(C)$ is given by the Kohlrausch equation: $\Lambda_m = \Lambda_m^0 - A\sqrt{C}$. This is a linear equation with a negative slope,which corresponds to graph $(ii)$.
For a weak electrolyte,molar conductance increases sharply as concentration decreases (i.e.,as $\sqrt{C}$ approaches zero) due to an increase in the degree of dissociation. This behavior is represented by graph $(i)$.

(A) The formula for molar conductance $(\Lambda_m)$ in $S \ m^2 \ mol^{-1}$ when concentration $(C)$ is in $mol \ m^{-3}$ is $\Lambda_m = \frac{\kappa}{C}$.
Given $\kappa = 9.2 \times 10^{-2} \ S \ m^{-1}$.
Concentration $C = 0.02 \ M = 0.02 \ mol \ L^{-1} = 0.02 \times 10^3 \ mol \ m^{-3} = 20 \ mol \ m^{-3}$.
$\Lambda_m = \frac{9.2 \times 10^{-2}}{20} = 0.46 \times 10^{-2} = 4.6 \times 10^{-3} \ S \ m^2 \ mol^{-1}$.

(B) According to Kohlrausch's law of independent migration of ions:
$\Lambda_{m}^{\circ}(CH_{3}COOH) = \Lambda_{m}^{\circ}(CH_{3}COONa) + \Lambda_{m}^{\circ}(HCl) - \Lambda_{m}^{\circ}(NaCl)$
Substituting the given values:
$\Lambda_{m}^{\circ}(CH_{3}COOH) = 91.0 + 425.9 - 126.4$
$= 516.9 - 126.4 = 390.5 \ S \ cm^{2} \ mol^{-1}$

(D) The degree of dissociation $(\alpha)$ is defined as the ratio of molar conductivity at a specific concentration $(\Lambda_c)$ to the limiting molar conductivity $(\Lambda_0)$.
$\alpha = \frac{\Lambda_c}{\Lambda_0}$
Given:
$\Lambda_c = 15.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\Lambda_0 = 300 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Substituting the values:
$\alpha = \frac{15.0}{300} = 0.05$

(B) The molar conductivity $(\Lambda_m)$ of a solution is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in a solution of volume $V \ mL$.
The relationship is given by the formula: $\Lambda_m = \frac{k \times 1000}{M}$,where $k$ is the conductivity in $S \ cm^{-1}$ and $M$ is the molarity in $mol \ L^{-1}$.
Therefore,option $B$ is the correct relationship.

(D) The chemical formula for aluminium sulphate is $Al_2(SO_4)_3$.
According to Kohlrausch's law of independent migration of ions,the molar conductivity at infinite dilution (zero concentration) is given by:
$\Lambda_m^0 (Al_2(SO_4)_3) = 2 \lambda_m^0 (Al^{3+}) + 3 \lambda_m^0 (SO_4^{2-})$
Given:
$\lambda_m^0 (Al^{3+}) = 189 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\lambda_m^0 (SO_4^{2-}) = 50.1 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Substituting the values:
$\Lambda_m^0 = 2(189) + 3(50.1)$
$\Lambda_m^0 = 378 + 150.3 = 528.3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$
Thus,the correct option is $D$.

(C) The cell constant $(G^*)$ is defined by the formula: $G^* = \kappa \times R$,where $\kappa$ is the conductivity and $R$ is the resistance.
Given:
Conductivity $(\kappa)$ = $0.001262 \ \Omega^{-1} \ cm^{-1}$
Resistance $(R)$ = $1440 \ \Omega$
Calculation:
$G^* = 0.001262 \ \Omega^{-1} \ cm^{-1} \times 1440 \ \Omega$
$G^* = 1.81728 \ cm^{-1}$
Rounding to three decimal places,we get $1.817 \ cm^{-1}$.

(A) The resistance $R$ of a conductor is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.
Rearranging for resistivity,we get $\rho = R \frac{A}{l}$.
The unit of resistance $R$ is $\Omega$ (ohm),the unit of area $A$ is $m^2$,and the unit of length $l$ is $m$.
Substituting these units,the unit of resistivity $\rho = \Omega \times \frac{m^2}{m} = \Omega \ m$.
Therefore,the $SI$ unit of resistivity is $\Omega \ m$.

(C) The conductivity $(\kappa)$ is given by the formula: $\kappa = \frac{\text{cell constant}}{R}$.
Given: Cell constant = $0.33 \ cm^{-1}$,Resistance $(R)$ = $30 \ \Omega$.
$\kappa = \frac{0.33 \ cm^{-1}}{30 \ \Omega} = 0.011 \ \Omega^{-1} \ cm^{-1}$.

(C) The relationship between conductivity $(k)$,cell constant $(G^*)$,and resistance $(R)$ is given by:
$k = \frac{G^*}{R}$
Therefore,the cell constant $(G^*)$ is:
$G^* = k \times R$
Substituting the given values:
$G^* = (1.64 \times 10^{-4} \ S \ cm^{-1}) \times (120 \ \Omega)$
$G^* = 0.01968 \ \text{cm}^{-1} \approx 0.0196 \ \text{cm}^{-1}$

(A) The formula for molar conductivity is $\Lambda_{m} = \frac{1000 \kappa}{C}$.
Given: $\Lambda_{m} = 410 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $C = 0.02 \ M$.
Rearranging the formula to find conductivity $(\kappa)$: $\kappa = \frac{\Lambda_{m} \times C}{1000}$.
Substituting the values: $\kappa = \frac{410 \times 0.02}{1000} \ \Omega^{-1} \ cm^{-1}$.
$\kappa = \frac{8.2}{1000} \ \Omega^{-1} \ cm^{-1} = 8.2 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.

(A) The cell constant $(G^*)$ is defined as the ratio of the distance between the electrodes $(l)$ to the area of cross-section $(A)$:
$G^* = \frac{l}{A}$
Given:
$l = 0.92 \ cm$
$A = 1.2 \ cm^2$
Calculation:
$G^* = \frac{0.92 \ cm}{1.2 \ cm^2} = 0.7666... \ cm^{-1} \approx 0.767 \ cm^{-1}$

(B) The relationship between conductivity $(k)$,resistance $(R)$,and cell constant $(G^*)$ is given by the formula:
$k = \frac{1}{R} \times G^*$
Rearranging the formula to solve for the cell constant:
$G^* = k \times R$
Given values:
$k = 0.0015 \ \Omega^{-1} \ cm^{-1}$
$R = 600 \ \Omega$
Substituting the values:
$G^* = 0.0015 \ \Omega^{-1} \ cm^{-1} \times 600 \ \Omega = 0.90 \ cm^{-1}$

(A) According to Kohlrausch's law of independent migration of ions,the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its constituent ions.
$\Lambda_{0(NaBr)} = \lambda^0_{Na^+} + \lambda^0_{Br^-}$
$\Lambda_{0(NaCl)} = \lambda^0_{Na^+} + \lambda^0_{Cl^-} = 126 \ S \ cm^2 \ mol^{-1}$
$\Lambda_{0(KBr)} = \lambda^0_{K^+} + \lambda^0_{Br^-} = 152 \ S \ cm^2 \ mol^{-1}$
$\Lambda_{0(KCl)} = \lambda^0_{K^+} + \lambda^0_{Cl^-} = 150 \ S \ cm^2 \ mol^{-1}$
To find $\Lambda_{0(NaBr)}$,we perform the operation:
$\Lambda_{0(NaBr)} = \Lambda_{0(NaCl)} + \Lambda_{0(KBr)} - \Lambda_{0(KCl)}$
$\Lambda_{0(NaBr)} = 126 + 152 - 150 = 128 \ S \ cm^2 \ mol^{-1}$

(A) The formula for molar conductivity is $\Lambda_{m} = \frac{1000 \times \kappa}{c}$.
Given,conductivity $\kappa = 6.07 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$ and concentration $c = 0.005 \ M$.
Substituting the values:
$\Lambda_{m} = \frac{1000 \times 6.07 \times 10^{-4}}{0.005} \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$\Lambda_{m} = \frac{0.607}{0.005} \ \Omega^{-1} \ cm^2 \ mol^{-1} = 121.4 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(B) Conductivity $(\kappa)$ is defined as the reciprocal of resistivity $(\varrho)$. Thus,$\kappa = \frac{1}{\varrho}$.
For a given electrolytic cell,the relationship between conductivity $(\kappa)$ and conductance $(G)$ is given by $\kappa = G \cdot G^*$,where $G^*$ is the cell constant,defined as $\frac{l}{A}$.
Evaluating the options:
$A$: $\kappa = \frac{1}{\varrho}$ is the correct definition of conductivity.
$B$: $\kappa = G \cdot \frac{1}{a}$ is incorrect because the correct relation is $\kappa = G \cdot a$ (assuming $a$ is the cell constant).
$C$: $\kappa = \frac{1}{R} \cdot \frac{l}{A}$ is correct because $G = \frac{1}{R}$ and $\frac{l}{A}$ is the cell constant.
$D$: $\kappa = \Lambda_m \cdot c$ is correct based on the definition of molar conductivity,$\Lambda_m = \frac{\kappa}{c}$.
Therefore,the incorrect expression is $B$.

(B) The conductivity $(\kappa)$ is calculated using the formula: $\kappa = \frac{\text{Cell constant}}{R}$
Substituting the given values: $\kappa = \frac{1.1 \ cm^{-1}}{94.5 \ \Omega}$
$\kappa = 0.0116 \ \Omega^{-1} \ cm^{-1}$

(C) The molar conductivity of a weak electrolyte like $CH_3COOH$ does not vary linearly with the square root of concentration,as shown in the graph.
Unlike strong electrolytes,the curve for a weak electrolyte does not intersect the y-axis upon extrapolation.
Therefore,the molar conductivity at infinite dilution $(\Lambda_0)$ for weak electrolytes cannot be determined by the graphical extrapolation method.

(D) The cell constant $\left(\frac{l}{A}\right)$ is defined by the relationship: $\frac{l}{A} = \kappa \times R$
where $\kappa$ is the conductivity and $R$ is the resistance.
Given:
$\kappa = 1.90 \times 10^{-6} \ S \ cm^{-1}$
$R = 115 \ \Omega$
Calculation:
$\frac{l}{A} = (1.90 \times 10^{-6} \ S \ cm^{-1}) \times (115 \ \Omega) = 0.2185 \ cm^{-1}$
Rounding to three decimal places,we get $0.218 \ cm^{-1}$.
Thus,the correct option is $D$.

(C) The conductivity $(\kappa)$ is calculated using the formula: $\kappa = \frac{\text{Cell constant}}{R}$.
Given: Cell constant = $1.32 \ cm^{-1}$,Resistance $(R)$ = $528 \ \Omega$.
$\kappa = \frac{1.32 \ cm^{-1}}{528 \ \Omega} = 0.0025 \ \Omega^{-1} \ cm^{-1}$.

(B) The relationship between conductivity $(k)$,resistance $(R)$,and cell constant $(G^*)$ is given by:
$k = \frac{G^*}{R}$
Therefore,the cell constant is:
$G^* = k \times R$
Given:
$k = 0.0012 \ \Omega^{-1} \ cm^{-1}$
$R = 600 \ \Omega$
Substituting the values:
$G^* = 0.0012 \ \Omega^{-1} \ cm^{-1} \times 600 \ \Omega = 0.72 \ cm^{-1}$

(D) $\because V = I R$ and $Q = I t$
Conductance $(G) = \frac{1}{R} = \frac{I}{V} = \frac{Q}{V t}$
Substituting with units,
$\therefore 1 \text{ Siemen} = \frac{1}{\Omega} = \Omega^{-1} = \frac{A}{V} = A V^{-1} = \frac{C}{V s} = C V^{-1} s^{-1}$
Since $\Omega$ is the unit of resistance,it is not equivalent to $1$ Siemen.

(A) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{1000 \times \kappa}{C}$
Given:
Conductivity $(\kappa)$ = $0.00250 \ \Omega^{-1} \ cm^{-1}$
Concentration $(C)$ = $0.02 \ M$
Substituting the values:
$\Lambda_m = \frac{1000 \times 0.00250}{0.02} \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\Lambda_m = \frac{2.5}{0.02} \ \Omega^{-1} \ cm^2 \ mol^{-1}$
$\Lambda_m = 125 \ \Omega^{-1} \ cm^2 \ mol^{-1}$

(B) The formula for cell constant $(G^*)$ is given by:
$G^* = \kappa \times R$
Where:
- $\kappa$ is the conductivity $(S \ cm^{-1})$,
- $R$ is the resistance $(\Omega)$.
Given:
- $\kappa = 1.70 \times 10^{-4} \ S \ cm^{-1}$,
- $R = 100 \ \Omega$.
Calculation:
$G^* = (1.70 \times 10^{-4} \ S \ cm^{-1}) \times (100 \ \Omega) = 0.017 \ cm^{-1}$.
Therefore,the cell constant is $0.017 \ cm^{-1}$.
Hence,the correct option is $B$.

(D) saturated $KCl$ solution is not typically used as a standard solution for conductivity calibration.
The reason is that a saturated $KCl$ solution may have variable composition due to temperature changes,partial crystallization,and the difficulty in ensuring a uniform,reproducible concentration.
Thus,it does not have a precisely known or easily reproducible conductivity value compared to the other standard concentrations.
Conclusion: Out of the given options,the saturated $KCl$ solution (Option $D$) is not suitable for use as a standard solution in determining the cell constant because its concentration and thus its conductivity are not stable or easily reproducible.

(C) The conductivity $(\kappa)$ is given by the formula: $\kappa = \frac{\text{Cell constant}}{R}$
Given: Cell constant $= 0.9 \ cm^{-1}$ and $R = 6530 \ \Omega$.
Substituting the values: $\kappa = \frac{0.9 \ cm^{-1}}{6530 \ \Omega} = 1.378 \times 10^{-4} \ \Omega^{-1} \ cm^{-1} \approx 1.38 \times 10^{-4} \ \Omega^{-1} \ cm^{-1}$.

(B) The formula for molar conductivity $(\Lambda_m)$ is given by: $\Lambda_m = \frac{1000 \times \kappa}{c}$
Given: $\Lambda_m = 101 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $\kappa = 1.01 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$
Substituting the values: $101 = \frac{1000 \times 1.01 \times 10^{-2}}{c}$
$101 = \frac{10.1}{c}$
$c = \frac{10.1}{101} = 0.1 \ M$

(C) The formula for molar conductivity is $\Lambda_m = \frac{1000 \times \kappa}{C}$.
Given: $\kappa = 0.00216 \ \Omega^{-1} \ cm^{-1}$ and $C = 0.02 \ M$.
Substituting the values: $\Lambda_m = \frac{1000 \times 0.00216}{0.02} = 108.0 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(D) The formula for conductivity $(k)$ is given by: $k = \frac{1}{R} \times \frac{l}{A}$
Given: Resistance $(R)$ = $31.5 \ \Omega$ and Cell constant $(\frac{l}{A})$ = $0.315 \ cm^{-1}$
Substituting the values: $k = \frac{1}{31.5 \ \Omega} \times 0.315 \ cm^{-1}$
$k = 0.01 \ \Omega^{-1} \ cm^{-1}$

(B) The relationship between conductivity $(\kappa)$,resistance $(R)$,and cell constant $(G^*)$ is given by:
$\kappa = \frac{1}{R} \times G^*$
Given:
Resistance $(R)$ = $60 \ \Omega$
Conductivity $(\kappa)$ = $0.014 \ \Omega^{-1} \ cm^{-1}$
Substituting the values:
$0.014 \ \Omega^{-1} \ cm^{-1} = \frac{1}{60 \ \Omega} \times G^*$
$G^* = 0.014 \ \Omega^{-1} \ cm^{-1} \times 60 \ \Omega$
$G^* = 0.84 \ cm^{-1}$

(A) The relationship between molar conductivity $(\Lambda_m)$ and conductivity $(k)$ is given by the formula: $\Lambda_m = \frac{1000 \times k}{C}$
Rearranging for conductivity $(k)$: $k = \frac{\Lambda_m \times C}{1000}$
Given: $\Lambda_m = 220 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $C = 0.05 \ M$
Substituting the values: $k = \frac{220 \times 0.05}{1000} \ \Omega^{-1} \ cm^{-1}$
$k = \frac{11}{1000} \ \Omega^{-1} \ cm^{-1} = 0.011 \ \Omega^{-1} \ cm^{-1}$

(A) The formula for molar conductivity is $\Lambda_m = \frac{1000 \times \kappa}{c}$.
Given,conductivity $\kappa = 1.26 \times 10^{-2} \ \Omega^{-1} \ cm^{-1}$ and concentration $c = 0.01 \ M$.
Substituting the values:
$\Lambda_m = \frac{1000 \times 1.26 \times 10^{-2}}{0.01} \ \Omega^{-1} \ cm^2 \ mol^{-1}$.
$\Lambda_m = \frac{12.6}{0.01} \ \Omega^{-1} \ cm^2 \ mol^{-1} = 1.26 \times 10^3 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(A) The relationship between molar conductivity $(\Lambda_{m})$ and conductivity $(k)$ is given by the formula: $\Lambda_{m} = \frac{1000 \times k}{c}$.
Rearranging the formula to solve for conductivity $(k)$: $k = \frac{\Lambda_{m} \times c}{1000}$.
Given values: $\Lambda_{m} = 407.2 \ \Omega^{-1} \ cm^2 \ mol^{-1}$ and $c = 0.02 \ M$.
Substituting the values: $k = \frac{407.2 \times 0.02}{1000}$.
$k = \frac{8.144}{1000} = 8.144 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.

(C) The molar conductivity $(\wedge_m)$ of an electrolyte increases upon dilution because the total volume of the solution containing one mole of electrolyte increases,leading to greater dissociation or increased ionic mobility.
Conversely,the conductivity $(k)$ (specific conductance) decreases upon dilution because the number of ions per unit volume of the solution decreases.