Conductor and Conductance and Cell constant Questions in English

(B) According to Kohlrausch's law of independent migration of ions,the equivalent conductance at infinite dilution is the sum of the ionic conductances of the cation and the anion.
For $NaF$,$\Lambda^{\infty}_{NaF} = \lambda^{\infty}_{Na^+} + \lambda^{\infty}_{F^-} = 90.1 \, \Omega^{-1} \, cm^2$.
For $KF$,$\Lambda^{\infty}_{KF} = \lambda^{\infty}_{K^+} + \lambda^{\infty}_{F^-}$.
Since the ionic mobility of $K^+$ is higher than $Na^+$,the equivalent conductance of $KF$ will be higher than $NaF$.
However,in many simplified textbook contexts,if the question implies a comparison where the values are expected to be similar or if it is a trick question regarding the nature of the electrolyte,we must note that $KF$ has a higher value than $90.1$.
Given the options provided,$111.2$ is the only value greater than $90.1$ that fits the expected trend for the substitution of $Na^+$ with $K^+$.

(B) Given: Resistance $R = 31.6\,\Omega$,Cell constant $G^* = 0.367\,cm^{-1}$,Molarity $M = 0.5\,M$.
Conductance $C = \frac{1}{R} = \frac{1}{31.6} \approx 0.0316\,S$.
Specific conductance $\kappa = C \times G^* = 0.0316\,S \times 0.367\,cm^{-1} \approx 0.0116\,S\,cm^{-1}$.
Molar conductance $\Lambda_m = \frac{\kappa \times 1000}{M} = \frac{0.0116 \times 1000}{0.5} = \frac{11.6}{0.5} = 23.2\,S\,cm^2\,mol^{-1}$.

(A) According to Kohlrausch's law,the molar conductance at infinite dilution is the sum of the ionic conductances of the constituent ions.
$\lambda_m^\infty (BaSO_4) = \lambda_{Ba^{2+}}^\infty + \lambda_{SO_4^{2-}}^\infty$
We can express this using the given values:
$\lambda_m^\infty (BaSO_4) = \lambda_m^\infty (BaCl_2) + \lambda_m^\infty (H_2SO_4) - 2\lambda_m^\infty (HCl)$
$\lambda_m^\infty (BaSO_4) = x_1 + x_2 - 2x_3$
Equivalent conductance $(\lambda_e^\infty)$ is related to molar conductance $(\lambda_m^\infty)$ by the formula $\lambda_e^\infty = \frac{\lambda_m^\infty}{n}$,where $n$ is the valency factor. For $BaSO_4$,$n = 2$.
Therefore,$\lambda_e^\infty (BaSO_4) = \frac{x_1 + x_2 - 2x_3}{2}$.

(C) The formula for conductivity $(\kappa)$ is given by: $\kappa = \frac{1}{R} \times \text{Cell constant}$.
Given: $\text{Cell constant} = 0.47 \, cm^{-1}$,$\text{Resistance } (R) = 31.6 \, \Omega$.
Substituting the values: $\kappa = \frac{0.47}{31.6} \, S \, cm^{-1}$.
$\kappa \approx 0.01487 \, S \, cm^{-1} \approx 0.015 \, S \, cm^{-1}$.

(D) The Assertion is incorrect because the electrical conductivity of metals like copper decreases with an increase in temperature. This happens because an increase in temperature causes increased vibrations of metal ions,which increases the resistance to the flow of electrons.
The Reason is correct because the electrical conductivity of metals is indeed due to the motion of free electrons.
Therefore,the Assertion is incorrect,but the Reason is correct.

(D) Specific conductance (conductivity,$\kappa$) is defined as the conductance of $1 \ cm^3$ of the solution.
On increasing dilution,the number of ions per unit volume decreases,which leads to a decrease in specific conductance.
Therefore,the Assertion is incorrect.
Although the degree of ionization and ionic mobility increase with dilution,the Assertion itself is false,making the correct choice $D$.

(N/A) The cell constant $(G^*)$ is calculated as:
$G^* = \text{conductivity} \times \text{resistance} = 1.29 \, S \, m^{-1} \times 100 \, \Omega = 129 \, m^{-1}$.
Conductivity $(\kappa)$ of $0.02 \, mol \, L^{-1}$ $KCl$ solution:
$\kappa = \frac{G^*}{R} = \frac{129 \, m^{-1}}{520 \, \Omega} = 0.248 \, S \, m^{-1}$.
Molar conductivity $(\Lambda_m)$ is given by $\Lambda_m = \frac{\kappa}{c}$.
Concentration $c = 0.02 \, mol \, L^{-1} = 20 \, mol \, m^{-3}$.
$\Lambda_m = \frac{0.248 \, S \, m^{-1}}{20 \, mol \, m^{-3}} = 0.0124 \, S \, m^2 \, mol^{-1} = 124 \times 10^{-4} \, S \, m^2 \, mol^{-1}$.

$A = \pi r^{2} = 3.14 \times (0.5 \ cm)^{2} = 0.785 \ cm^{2} = 0.785 \times 10^{-4} \ m^{2}$
$l = 50 \ cm = 0.5 \ m$
$\rho = \frac{R A}{l} = \frac{5.55 \times 10^{3} \ \Omega \times 0.785 \ cm^{2}}{50 \ cm} = 87.135 \ \Omega \ cm$
$\kappa = \frac{1}{\rho} = \frac{1}{87.135} \ S \ cm^{-1} = 0.01148 \ S \ cm^{-1}$
$\Lambda_{m} = \frac{\kappa \times 1000}{c} = \frac{0.01148 \ S \ cm^{-1} \times 1000 \ cm^{3} \ L^{-1}}{0.05 \ mol \ L^{-1}} = 229.6 \ S \ cm^{2} \ mol^{-1}$

(N/A) The Kohlrausch equation is given by $\Lambda_m = \Lambda_m^o - A c^{1/2}$.
To verify this,we calculate the square root of the concentration $(c^{1/2})$:

$c^{1/2} / (mol \, L^{-1})^{1/2}$	$\Lambda_m / S \, cm^2 \, mol^{-1}$
$0.01407$	$148.61$
$0.01758$	$148.29$
$0.02283$	$147.81$
$0.03145$	$147.09$

$A$ plot of $\Lambda_m$ ($y$-axis) versus $c^{1/2}$ ($x$-axis) yields a straight line.
By extrapolating the line to $c^{1/2} = 0$,we find the intercept $\Lambda_m^o = 150.0 \, S \, cm^2 \, mol^{-1}$.
The slope of the line is given by $A = - \text{slope} = - \frac{147.09 - 148.61}{0.03145 - 0.01407} \approx 87.46 \, S \, cm^2 \, mol^{-1} / (mol \, L^{-1})^{1/2}$.

According to Kohlrausch law of independent migration of ions:
$\Lambda _{m(CaCl_2)}^o = \lambda _{Ca^{2+}}^o + 2\lambda _{Cl^{-}}^o$
$= 119.0 \ S \ cm^2 \ mol^{-1} + 2(76.3 \ S \ cm^2 \ mol^{-1})$
$= 119.0 + 152.6 = 271.6 \ S \ cm^2 \ mol^{-1}$
$\Lambda _{m(MgSO_4)}^o = \lambda _{Mg^{2+}}^o + \lambda _{SO_4^{2-}}^o$
$= 106.0 \ S \ cm^2 \ mol^{-1} + 160.0 \ S \ cm^2 \ mol^{-1}$
$= 266.0 \ S \ cm^2 \ mol^{-1}$

(A) According to Kohlrausch's law of independent migration of ions:
$\Lambda _{m(HAc)}^\circ = \lambda _{H^{+}}^\circ + \lambda _{Ac^{-}}^\circ$
We can express this in terms of the given values:
$\Lambda _{m(HAc)}^\circ = (\lambda _{H^{+}}^\circ + \lambda _{Cl^{-}}^\circ) + (\lambda _{Na^{+}}^\circ + \lambda _{Ac^{-}}^\circ) - (\lambda _{Na^{+}}^\circ + \lambda _{Cl^{-}}^\circ)$
$\Lambda _{m(HAc)}^\circ = \Lambda _{m(HCl)}^\circ + \Lambda _{m(NaAc)}^\circ - \Lambda _{m(NaCl)}^\circ$
Substituting the given values:
$\Lambda _{m(HAc)}^\circ = 425.9 + 91.0 - 126.4$
$\Lambda _{m(HAc)}^\circ = 390.5 \ S \ cm^2 \ mol^{-1}$

(N/A) The conductivity $(\kappa)$ of a solution is defined as the conductance of ions present in a unit volume $(1 \ cm^3)$ of the solution.
Upon dilution,the total number of ions remains the same,but the number of ions per unit volume decreases.
Since the conductivity depends directly on the concentration of ions in a unit volume,the conductivity of the solution decreases with dilution.

(N/A) Applying Kohlrausch's law of independent migration of ions,the $\Lambda _{m}^{o}$ value of water can be determined as follows:
$\Lambda _{m(H_2O)}^o = \lambda _{H^{+}}^o + \lambda _{OH^{-}}^o$
$= (\lambda _{H^{+}}^o + \lambda _{Cl^{-}}^o) + (\lambda _{Na^{+}}^o + \lambda _{OH^{-}}^o) - (\lambda _{Na^{+}}^o + \lambda _{Cl^{-}}^o)$
$= \Lambda _{m(HCl)}^o + \Lambda _{m(NaOH)}^o - \Lambda _{m(NaCl)}^o$
Hence,by knowing the values of $\Lambda _m^o$ for $HCl, NaOH,$ and $NaCl$,the $\Lambda _m^o$ value of water can be determined.

(N/A) Conductivity of a solution is defined as the conductance of a solution of $1 \, cm$ in length and area of cross-section $1 \, cm^2$. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol $\kappa$. If $\rho$ is resistivity,then we can write:
$\kappa = \frac{1}{\rho}$
The conductivity of a solution at any given concentration is the conductance $(G)$ of one unit volume of solution kept between two platinum electrodes with the unit area of cross-section and at a distance of unit length.
i.e.,$G = \kappa \frac{a}{l} = \kappa \cdot 1 = \kappa$
(since $a = 1, l = 1$)
Conductivity always decreases with a decrease in concentration,both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.
Molar conductivity:
Molar conductivity of a solution at a given concentration is the conductance of volume $V$ of a solution containing $1 \, mole$ of the electrolyte kept between two electrodes with the area of cross-section $A$ and distance of unit length.
$\Lambda_m = \kappa \cdot \frac{A}{l}$
Now,$l = 1$ and $A = V$ (volume containing $1 \, mole$ of the electrolyte).
$\therefore \Lambda_m = \kappa \cdot V$
Molar conductivity increases with a decrease in concentration. This is because the total volume $V$ of the solution containing one mole of the electrolyte increases on dilution.
The variation of $\Lambda_m$ with $\sqrt{c}$ for strong and weak electrolytes is shown in the following plot:

$(124 S CM^2 MOL^-1)$ Given,conductivity $\kappa = 0.0248 \, S \, cm^{-1}$.
Concentration $c = 0.20 \, M$.
Formula for molar conductivity is $\Lambda_{m} = \frac{\kappa \times 1000}{c}$.
Substituting the values: $\Lambda_{m} = \frac{0.0248 \times 1000}{0.20}$.
$\Lambda_{m} = \frac{24.8}{0.20} = 124 \, S \, cm^{2} \, mol^{-1}$.

(N/A) Given:
Conductivity,$\kappa = 0.146 \times 10^{-3} \ S \ cm^{-1}$
Resistance,$R = 1500 \ \Omega$
The formula for cell constant $(G^*)$ is:
$G^* = \kappa \times R$
Substituting the values:
$G^* = (0.146 \times 10^{-3} \ S \ cm^{-1}) \times (1500 \ \Omega)$
$G^* = 0.219 \ cm^{-1}$
Thus,the cell constant is $0.219 \ cm^{-1}$.

(A) The molar conductivity ${\Lambda _m}$ is calculated using the formula: ${\Lambda _m} = \frac{\kappa \times 1000}{c}$ where $\kappa$ is in $S \, cm^{-1}$ and $c$ is in $mol \, L^{-1}$.
Given $\kappa$ values are in $S \, m^{-1}$,we convert them to $S \, cm^{-1}$ by multiplying by $10^{-2}$.
$1$. For $c = 0.001 \, M$: $\kappa = 1.237 \times 10^{-4} \, S \, cm^{-1} \implies {\Lambda _m} = \frac{1.237 \times 10^{-4} \times 1000}{0.001} = 123.7 \, S \, cm^2 \, mol^{-1}$. $c^{1/2} = 0.0316 \, M^{1/2}$.
$2$. For $c = 0.010 \, M$: $\kappa = 11.85 \times 10^{-4} \, S \, cm^{-1} \implies {\Lambda _m} = \frac{11.85 \times 10^{-4} \times 1000}{0.010} = 118.5 \, S \, cm^2 \, mol^{-1}$. $c^{1/2} = 0.1000 \, M^{1/2}$.
$3$. For $c = 0.020 \, M$: $\kappa = 23.15 \times 10^{-4} \, S \, cm^{-1} \implies {\Lambda _m} = \frac{23.15 \times 10^{-4} \times 1000}{0.020} = 115.8 \, S \, cm^2 \, mol^{-1}$. $c^{1/2} = 0.1414 \, M^{1/2}$.
$4$. For $c = 0.050 \, M$: $\kappa = 55.53 \times 10^{-4} \, S \, cm^{-1} \implies {\Lambda _m} = \frac{55.53 \times 10^{-4} \times 1000}{0.050} = 111.1 \, S \, cm^2 \, mol^{-1}$. $c^{1/2} = 0.2236 \, M^{1/2}$.
$5$. For $c = 0.100 \, M$: $\kappa = 106.74 \times 10^{-4} \, S \, cm^{-1} \implies {\Lambda _m} = \frac{106.74 \times 10^{-4} \times 1000}{0.100} = 106.7 \, S \, cm^2 \, mol^{-1}$. $c^{1/2} = 0.3162 \, M^{1/2}$.
Plotting ${\Lambda _m}$ vs $c^{1/2}$ and extrapolating to $c^{1/2} = 0$,we find the intercept $\Lambda _m^o \approx 124.0 \, S \, cm^2 \, mol^{-1}$.

(A) $Al$ (Aluminum) has approximately double the electrical conductivity compared to $Cu$ (Copper) when compared on an equal mass basis.

(N/A) The conduction of electricity in conductors is due to the movement of electrons or ions. Conductors are classified into metallic conductors and electrolytic conductors.
- Metals conduct electricity in both solid and molten states. The conductivity of metals depends on the number of valence electrons available per atom.
- The atomic orbitals of metal atoms form molecular orbitals,which are so close in energy that they form bands.
- If these bands are partially filled or overlap with a higher energy unoccupied conduction band,electrons can flow easily under an electric field,and the metal shows conductivity.
- If the gap between the filled valence band and the next higher unoccupied band (conduction band) is large,electrons cannot jump into it. Such substances have very low conductivity and behave as insulators.

(N/A) Electrical resistance is a measure of the opposition to current flow in an electrical circuit. It is defined as the ratio of potential difference across a conductor to the current flowing through it.
The electrical resistance $R$ of a conductor is directly proportional to its length $l$ and inversely proportional to its area of cross-section $A$. That is,$R \propto \frac{l}{A}$.
$R = \rho \left( \frac{l}{A} \right)$,where $\rho$ is the resistivity constant.
The electrical resistance is represented by the symbol $R$ and is measured in ohm $(\Omega)$.
Its $SI$ base unit is $(kg \cdot m^{2}) / (s^{3} \cdot A^{2})$. It can be measured using a Wheatstone bridge.
$1 \, \Omega = 1 \, (kg \cdot m^{2}) / (s^{3} \cdot A^{2})$.
Conversion: $1 \, \Omega \cdot m = 100 \, \Omega \cdot cm$ or $1 \, \Omega \cdot cm = 0.01 \, \Omega \cdot m$.

(N/A) The electrical resistance of any object is directly proportional to its length $l$ and inversely proportional to its area of cross-section $A$.
$R \propto \frac{l}{A}$
$\therefore R = \rho \left( \frac{l}{A} \right)$ and $\rho = R \left( \frac{A}{l} \right)$
Here,$\rho$ (rho) is the specific resistance or resistivity.
Unit: The $SI$ unit of resistivity is ohm meter $(\Omega \ m)$,and sometimes it is expressed as ohm centimetre $(\Omega \ cm)$.
Note: $IUPAC$ recommends the use of the term resistivity over specific resistance.
If $l = 1 \ m$ and $A = 1 \ m^2$,then $R = \rho$.
$SI \ unit \ of \ \rho = \frac{(\Omega)(m)^2}{m} = \Omega \ m$.
Conversion: $1 \ \Omega \ m = 100 \ \Omega \ cm$ and $1 \ \Omega \ cm = 0.01 \ \Omega \ m$.

(N/A) Conductivity $(k)$,also known as specific conductance,is the reciprocal of resistivity $( ho)$.
Mathematically,$k = \frac{1}{\rho}$.
Conductivity depends on the following factors:
$1$. Nature of the material: It depends on the nature of the electrolyte and the solvent.
$2$. Concentration of ions: It depends on the concentration of ions present in the solution.
$3$. Temperature: It increases with an increase in temperature because the kinetic energy of ions increases,leading to higher mobility.

(N/A) The inverse of resistivity is called conductivity (specific conductance) and is represented by the symbol $k$ (Greek,kappa).
$k = \frac{1}{\rho}$
$IUPAC$ has recommended the use of the term conductivity over specific conductance.
The $SI$ units of conductivity are $S \ m^{-1}$,but quite often,$k$ is expressed in $S \ cm^{-1}$.
Conductivity of a material in $S \ m^{-1}$ is its conductance when it is $1 \ m$ long and its area of cross-section is $1 \ m^{2}$.
It may be noted that $1 \ S \ cm^{-1} = 100 \ S \ m^{-1}$.
Unit of kappa $(k): S \ m^{-1}$ or $S \ cm^{-1}$ or $\Omega^{-1} \ m^{-1}$ or $\Omega^{-1} \ cm^{-1}$.
Since $\rho = \frac{RA}{l}$,substituting this into $k = \frac{1}{\rho}$ gives:
$k = \left( \frac{l}{A} \right) \frac{1}{R}$
Since $G = \frac{1}{R}$ (conductance) and $G^{*} = \frac{l}{A}$ (cell constant):
$k = G \times G^{*}$

(N/A) Definition: The flow of electric current through metals due to the movement of electrons is known as metallic or electronic conductivity.
Factors affecting metallic conductivity:
$i$. Nature and structure of the metal.
$ii$. Number of valence electrons per atom.
$iii$. Temperature: As the temperature increases,the conductivity decreases because of the increased vibration of metal ions which hinders the flow of electrons.
Key characteristic: Since electrons enter at one end and exit through the other,the chemical composition of the metallic conductor remains unchanged.

(N/A) Definition: When electrolytes are dissolved in water,they furnish their own ions in the solution,which increases the conductivity of the solution. The conductance of electricity by ions present in the solution is called electrolytic or ionic conductance. Pure water has very low conductivity,approximately $3.5 \times 10^{-5} \ S \ m^{-1}$.
Ionic conductivity depends on the following factors:
$(i)$ The nature of the electrolyte added.
$(ii)$ The size of the ions produced and their degree of solvation.
$(iii)$ The nature of the solvent and its viscosity.
$(iv)$ The concentration of the electrolyte.
$(v)$ Temperature: It increases with an increase in temperature.
Note: The passage of direct current $(DC)$ through an ionic solution over a prolonged period can lead to a change in its composition due to electrochemical reactions.

(N/A) The inverse of the resistance of an ionic solution is known as the conductivity of the ionic solution,given by $k = \frac{G^{*}}{R}$.
Accurate measurement of an unknown resistance can be performed using a Wheatstone bridge.
However,for measuring the resistance of an ionic solution,we face two main problems:
$1$. Firstly,passing direct current $(DC)$ through the solution changes its chemical composition due to electrolysis.
This difficulty is resolved by using an alternating current $(AC)$ source of power.
$2$. Secondly,an ionic solution cannot be connected to the bridge like a metallic wire or other solid conductor.
This second problem is solved by using a specially designed vessel called a conductivity cell.

(N/A) The inverse of the resistance of an ionic solution is known as its conductance. The relationship is given by $k = \frac{G^{*}}{R}$.
Accurate measurement of an unknown resistance is typically performed using a Wheatstone bridge. However,measuring the resistance of an ionic solution presents two primary difficulties:
$1$. Passing direct current $(DC)$ through the solution causes electrolysis,which changes the chemical composition of the solution.
This difficulty is resolved by using an alternating current $(AC)$ source of power,which prevents electrolysis.
$2$. An ionic solution cannot be connected to the Wheatstone bridge like a metallic wire or other solid conductor.
This problem is solved by using a specially designed vessel known as a conductivity cell.

(N/A) conductivity cell is an instrument used to measure the ionic conductivity or resistivity of an electrolytic solution.
It is available in several designs,and two simple ones are shown in the diagram.
Construction: Basically,it consists of two platinum electrodes coated with platinum black (finely divided metallic $Pt$ is deposited on the electrodes electrochemically).
These electrodes have an area of cross-section equal to $A$ and are separated by a distance $l$.
Therefore,the solution confined between these electrodes acts as a column of length $l$ and area of cross-section $A$.
Uses: By using a conductivity cell,the conductivity and resistivity of an unknown electrolytic solution can be measured.

(N/A) The quantity $\left(\frac{l}{A}\right)$ is called the cell constant,denoted by the symbol $G^*$. Here,$l$ is the distance between the two electrodes of the cell,and $A$ is the area of cross-section of the electrodes. Direct measurement of $l$ and $A$ is often inconvenient and unreliable.
$(A)$ Measurement of cell resistance: The cell constant is usually determined by measuring the resistance of the cell containing a solution whose conductivity is already known. For this purpose,we generally use $KCl$ solutions,whose conductivity is known accurately at various concentrations and temperatures.
$(B)$ Procedure: The conductivity cell is filled with a $KCl$ solution of known concentration and is then connected to a Wheatstone bridge using an $AC$ current source to measure its resistance.
$(C)$ Calculation of cell constant $G^*$: $G^* = \left(\frac{l}{A}\right) = R \cdot \kappa$
Where,$R$ is the measured resistance of the $KCl$ solution,and $\kappa$ is the conductivity of the $KCl$ solution obtained from standard tables.
The table below provides reference values for $KCl$ solutions at $298.15 \ K$:

Molarity (mol $L^{-1}$)	Molarity (mol $m^{-3}$)	Conductivity $(S \ cm^{-1})$	Conductivity $(S \ m^{-1})$	Molar Conductivity $(S \ cm^{2} \ mol^{-1})$	Molar Conductivity $(S \ m^{2} \ mol^{-1})$
$1.000$	$1000$	$0.1113$	$11.13$	$111.3$	$111.3 \times 10^{-4}$
$0.100$	$100.0$	$0.0129$	$1.29$	$129.0$	$129.0 \times 10^{-4}$
$0.010$	$10.00$	$0.00141$	$0.141$	$141.0$	$141.0 \times 10^{-4}$

(N/A) To measure the resistivity of an electrolytic solution,we first determine the conductivity cell constant $G^*$. By using this cell,we measure the resistance and ionic conductivity of the solution.
$(A)$ Measurement of resistance of solution: The setup for the measurement of the resistance is shown in the figure. It consists of two known resistances $R_3$ and $R_4$,a variable resistance $R_1$,and the conductivity cell having the unknown resistance $R_2$. The Wheatstone bridge is fed by an oscillator $O$ (a source of $a.c.$ power in the audio frequency range $550$ to $5000$ cycles per second). $P$ is a suitable detector (a headphone or other electronic device),and the bridge is balanced when no current passes through the detector.
$(B)$ Calculation for resistance of solution: In the balanced condition,the unknown resistance $R_2$ of the solution is obtained by the formula:
$R_2 = \frac{R_1 R_4}{R_3} = R$
These days,inexpensive conductivity meters are available. Electric resistance $R$ is measured in ohm $(\Omega)$. The relationship between resistance,resistivity,and cell constant is:
$R = \rho \left( \frac{l}{A} \right) = \frac{1}{\kappa} \left( \frac{l}{A} \right) = \frac{G^*}{\kappa}$
Where,
$R = \text{Resistance}$
$G^* = \text{Cell constant} = \frac{l}{A}$
$\rho = \text{Resistivity}$
$\kappa = \text{Conductivity of solution}$
Thus,resistivity $\rho$ can be calculated as $\rho = R \left( \frac{A}{l} \right) = \frac{R}{G^*}$.

(A) $(i)$ First,determine the cell constant $(G^*)$ using a standard solution of known conductivity.
$(ii)$ Determine the resistance $(R)$ of the unknown solution using a conductivity cell.
$(iii)$ Calculate the conductivity $(\kappa)$ of the solution using the equation: $\kappa = \frac{G^*}{R}$.
The $SI$ unit of conductivity is $S \ m^{-1}$,and it is also commonly expressed as $S \ cm^{-1}$.

(N/A) The conductivity of solutions of different electrolytes in the same solvent and at a given temperature differs due to the charge and size of the ions in which they dissociate,the concentration of ions,or the ease with which the ions move under a potential gradient. It,therefore,becomes necessary to define a physically more meaningful quantity called molar conductivity. It is denoted by the symbol $\Lambda_{m}$ (Greek,lambda).
It is related to the conductivity of the solution by the equation:
$\Lambda_{m} = \frac{k}{c}$
Where,
$k =$ conductivity of the solution (unit: $S \ m^{-1}$)
$c =$ concentration of the solution (unit: $mol \ m^{-3}$)
$\Lambda_{m} =$ molar conductivity of the solution
Therefore,the unit of $\Lambda_{m}$ is $S \ m^{2} \ mol^{-1}$.
In practical units,if $k$ is in $S \ cm^{-1}$ and concentration $(c)$ is in $mol \ L^{-1}$ (molarity),the formula is:
$\Lambda_{m} = \frac{k \times 1000}{Molarity}$
The unit becomes $S \ cm^{2} \ mol^{-1}$ or $\Omega^{-1} \ cm^{2} \ mol^{-1}$.
Conversion factor:
$1 \ S \ m^{2} \ mol^{-1} = 10^{4} \ S \ cm^{2} \ mol^{-1}$

(N/A) Both conductivity $(k)$ and molar conductivity $(\Lambda_{m})$ change with the concentration of the electrolyte.
Conductivity $(k)$ always decreases with a decrease in concentration (dilution),for both weak and strong electrolytes.
This is because the number of ions per unit volume that carry the current in the solution decreases upon dilution.
Conductivity is defined as the conductance of one unit volume of solution kept between two platinum electrodes with unit area of cross-section and unit length. If $G$ is the conductance,$A$ is the cross-sectional area,and $l$ is the distance between electrodes,then $G = k \left( \frac{A}{l} \right)$. For $A = 1$ and $l = 1$,$G = k$.
Molar conductivity $(\Lambda_{m})$ is the conductance of the volume $V$ of solution containing one mole of electrolyte kept between two electrodes. It is related to conductivity by the equation $\Lambda_{m} = k V$. As concentration decreases,$V$ increases,which leads to an increase in molar conductivity.

(N/A) Strong electrolyte: Electrolytes that undergo complete or maximum ionization in their aqueous solution are known as strong electrolytes. Their solutions exhibit high conductivity. Examples include $KCl, NaCl, KNO_{3}, NaNO_{3}, MgCl_{2}, CaCl_{2}, MgSO_{4}$,etc. Salts of strong acids and strong bases are typically strong electrolytes.
$(b)$ Concentration of solution and value of $\Lambda_{m}$: For strong electrolytes,$\Lambda_{m}$ increases slowly with dilution and can be represented by the Kohlrausch equation:
$\Lambda_{m} = \Lambda_{m}^{\circ} - A c^{1/2}$
Where:
$\Lambda_{m} =$ molar conductivity of the strong electrolyte.
$\Lambda_{m}^{\circ} =$ limiting molar conductivity of the strong electrolyte.
$c =$ concentration of the solution in $mol \ L^{-1}$.
$A =$ constant,which represents the negative slope of the graph.

Strong electrolyte	$NaCl, KCl, CaCl_{2}, MgSO_{4}$
Valency or Type	$1-1, 1-1, 2-1, 2-2$

The value of $A$ depends on: $(i)$ the type of electrolyte (charge on ions) like $1-1, 2-1, 2-2$ etc.,$(ii)$ temperature,and $(iii)$ pressure.

(N/A) The Kohlrausch equation is given by: $\Lambda_m = \Lambda_m^o - A \sqrt{c}$.
To verify this,we calculate $c^{1/2}$ for each concentration:

$c / mol \ L^{-1}$	$c^{1/2} / (mol \ L^{-1})^{1/2}$	$\Lambda_m / S \ cm^2 \ mol^{-1}$
$0.000198$	$0.01407$	$148.61$
$0.000309$	$0.01758$	$148.29$
$0.000521$	$0.02283$	$147.81$
$0.000989$	$0.03145$	$147.09$

Plotting $\Lambda_m$ on the y-axis and $c^{1/2}$ on the x-axis yields a straight line.
From the slope of the line,$A = -\frac{\Delta \Lambda_m}{\Delta c^{1/2}} = -\frac{147.09 - 148.61}{0.03145 - 0.01407} \approx -87.46 \ S \ cm^2 \ mol^{-1} (mol \ L^{-1})^{-1/2}$.
Extrapolating the line to $c^{1/2} = 0$ (y-intercept),we get $\Lambda_m^o \approx 150.0 \ S \ cm^2 \ mol^{-1}$.

(N/A) Kohlrausch examined $\Lambda_{m}^{o}$ values for several strong electrolytes and observed certain regularities. He noted that the difference in $\Lambda_{m}^{o}$ of the electrolytes $NaX$ and $KX$ for any $X$ is nearly constant. For example,at $298 \ K$ :
$[\Lambda_{m(KCl)}^{o} - \Lambda_{m(NaCl)}^{o}] = [\Lambda_{m(KBr)}^{o} - \Lambda_{m(NaBr)}^{o}] = [\Lambda_{m(KI)}^{o} - \Lambda_{m(NaI)}^{o}] = 23.4 \ S \ cm^{2} \ mol^{-1}$
Similarly,$[\Lambda_{m(NaBr)}^{o} - \Lambda_{m(NaCl)}^{o}] = [\Lambda_{m(KBr)}^{o} - \Lambda_{m(KCl)}^{o}] = 1.8 \ S \ cm^{2} \ mol^{-1}$
Kohlrausch Law: Based on the above observations,he enunciated the Kohlrausch law of independent migration of ions.
Law: The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
If $\lambda_{Na^{+}}^{o}$ is the limiting molar conductivity of sodium ions and $\lambda_{Cl^{-}}^{o}$ is the limiting molar conductivity of chloride ions,then:
$\Lambda_{m(NaCl)}^{o} = \lambda_{Na^{+}}^{o} + \lambda_{Cl^{-}}^{o}$
In general,if an electrolyte on dissociation gives $\nu_{+}$ cations and $\nu_{-}$ anions,then its limiting molar conductivity is given by:
$\Lambda_{m}^{o} = \nu_{+} \lambda_{+}^{o} + \nu_{-} \lambda_{-}^{o}$
Here,$\lambda_{+}^{o}$ and $\lambda_{-}^{o}$ are the limiting molar conductivities of the cation and anion,respectively.

(N/A) At infinite dilution (i.e.,concentration $c \rightarrow 0$),an electrolyte dissociates completely $(\alpha = 1)$. However,at such low concentrations,the conductivity of the solution is so low that it cannot be measured accurately.
Therefore,the value of $\Lambda_{m}^{\circ}$ cannot be determined by extrapolating the graph of $\Lambda_{m}$ versus $c^{1/2}$.
Instead,$\Lambda_{m}^{\circ}$ for weak electrolytes is obtained using Kohlrausch's law of independent migration of ions.
According to this law,$\Lambda_{m}^{\circ} = \nu_{+} \lambda_{+}^{\circ} + \nu_{-} \lambda_{-}^{\circ}$,where $\lambda^{\circ}$ represents the limiting molar conductivity of individual ions.
Furthermore,the dissociation constant $(K_{a})$ of a weak electrolyte like acetic acid can be calculated using the relation $K_{a} = \frac{c \alpha^{2}}{1 - \alpha}$,where $\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}}$.

(N/A) Law: The law states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
For example,if the limiting molar conductivity of positive and negative ions is $\lambda_{m^{+}}^{\circ}$ and $\lambda_{m^{-}}^{\circ}$,respectively,then the limiting molar conductivity of the solution $(\Lambda_{m}^{\circ})$ will be as follows:
$\Lambda_{m}^{\circ} = v_{+} \lambda_{m^{+}}^{\circ} + v_{-} \lambda_{m^{-}}^{\circ}$
Explanation:
If $\lambda_{m}^{\circ}$ of $K^{+} = 73.5$ and $\lambda_{m}^{\circ}$ of $Br^{-} = 78.1 \ S \ cm^{2} \ mol^{-1}$,then the limiting molar conductivity of $KBr$ solution at infinite dilution is as follows:
$\Lambda_{m}^{\circ}(KBr) = \lambda_{m}^{\circ}(K^{+}) + \lambda_{m}^{\circ}(Br^{-})$
$= 73.5 + 78.1$
$= 151.6 \ S \ cm^{2} \ mol^{-1}$
Limiting molar conductivity of some ions at $298 \ K$ temperature:

Ion | $\lambda^{\circ} / (S \ cm^{2} \ mol^{-1})$	Ion | $\lambda^{\circ} / (S \ cm^{2} \ mol^{-1})$
$H^{+} \ | \ 349.6$	$OH^{-} \ | \ 199.1$
$Na^{+} \ | \ 50.1$	$Cl^{-} \ | \ 76.3$
$K^{+} \ | \ 73.5$	$Br^{-} \ | \ 78.1$
$Ca^{2+} \ | \ 119.0$	$CH_{3}COO^{-} \ | \ 40.9$
$Mg^{2+} \ | \ 106.0$	$SO_{4}^{2-} \ | \ 160.0$

(N/A) Kohlrausch's law of independent migration of ions states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte. Mathematically,$\Lambda_{m}^{\circ} = \nu_{+} \lambda_{+}^{\circ} + \nu_{-} \lambda_{-}^{\circ}$,where $\nu_{+}$ and $\nu_{-}$ are the number of cations and anions per formula unit of electrolyte,and $\lambda_{+}^{\circ}$ and $\lambda_{-}^{\circ}$ are the limiting molar conductivities of the individual ions.
Applications:
$1$. Calculation of limiting molar conductivity $(\Lambda_{m}^{\circ})$ for weak electrolytes: The law allows the determination of $\Lambda_{m}^{\circ}$ for weak electrolytes by using the limiting molar conductivities of strong electrolytes.
$2$. Calculation of degree of dissociation $(\alpha)$: For a weak electrolyte at a given concentration $c$,the degree of dissociation is given by $\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}}$,where $\Lambda_{m}$ is the molar conductivity at concentration $c$.
$3$. Calculation of dissociation constant $(K_{a})$: The dissociation constant is calculated using the formula $K_{a} = \frac{c \alpha^{2}}{1 - \alpha} = \frac{c \Lambda_{m}^{2}}{\Lambda_{m}^{\circ}(\Lambda_{m}^{\circ} - \Lambda_{m})}$.

(N/A) The graph of $\Lambda_{m}$ versus $c^{1/2}$ for aqueous solutions of strong and weak electrolytes is shown below:

Feature	Strong Electrolyte (e.g.,$KCl$)	Weak Electrolyte (e.g.,$CH_{3}COOH$)
Graph Nature	Linear graph.	Non-linear (curved) graph.
Kohlrausch Equation	Follows $\Lambda_{m} = \Lambda_{m}^{o} - A c^{1/2}$.	Does not follow this linear relation.
Determination of $\Lambda_{m}^{o}$	Can be obtained by extrapolating the graph to zero concentration (intercept).	Cannot be obtained by extrapolation as the curve becomes parallel to the y-axis at low concentrations.

Uses:
$1$. For strong electrolytes,the graph allows the determination of molar conductivity at infinite dilution $(\Lambda_{m}^{o})$ using the intercept.
$2$. For weak electrolytes,$\Lambda_{m}^{o}$ cannot be determined graphically; it is calculated using Kohlrausch's Law of independent migration of ions.

(B) $\Lambda_{m}$ is the molar conductivity of the solution.
$k$ is the specific conductivity (conductivity) of the solution.
$\rho$ is the resistivity of the solution $= 10 \ \Omega \ cm$.
$k = \frac{1}{\rho} = \frac{1}{10} = 0.1 \ \Omega^{-1} \ cm^{-1} = 0.1 \ S \ cm^{-1}$.
Concentration of the solution $c = 0.5 \ M = 0.5 \ mol \ L^{-1}$.
The formula for molar conductivity is $\Lambda_{m} = \frac{1000 \times k}{c}$.
Substituting the values: $\Lambda_{m} = \frac{1000 \times 0.1}{0.5} = \frac{100}{0.5} = 200 \ S \ cm^{2} \ mol^{-1}$.

(N/A) Cell constant $(G^{*})$ is given by:
$G^{*} = \frac{l}{A} = \frac{1.5 \ cm}{3.8 \ cm^{2}} = 0.3947 \ cm^{-1}$
Conductivity $(k)$ is calculated as:
$k = \frac{G^{*}}{R} = \frac{0.3947 \ cm^{-1}}{30.0 \ \Omega} = 0.013157 \ S \ cm^{-1}$
Molar conductivity $(\Lambda_{m})$ is calculated using the formula:
$\Lambda_{m} = \frac{1000 \times k}{c}$
$\Lambda_{m} = \frac{1000 \times 0.013157 \ S \ cm^{-1}}{0.05 \ mol \ L^{-1}} = 263.14 \ S \ cm^{2} \ mol^{-1}$

(N/A) The conductivity $k$ is the reciprocal of resistivity $\rho$:
$k = \frac{1}{\rho} = \frac{1}{5 \times 10^{-3} \ \Omega \ cm} = 200 \ S \ cm^{-1}$
The formula for molar conductivity $\Lambda_{m}$ is:
$\Lambda_{m} = \frac{1000 \times k}{c}$
Substituting the values:
$\Lambda_{m} = \frac{1000 \times 200}{0.08} \ S \ cm^{2} \ mol^{-1}$
$\Lambda_{m} = \frac{200000}{0.08} \ S \ cm^{2} \ mol^{-1}$
$\Lambda_{m} = 2.5 \times 10^{6} \ S \ cm^{2} \ mol^{-1}$

According to Kohlrausch's law of independent migration of ions:
$NH_4OH \rightarrow NH_{4(aq)}^{+} + OH_{(aq)}^{-}$
$\Lambda_{m}^{\circ}(NH_4OH) = \lambda_{m}^{\circ}(NH_{4}^{+}) + \lambda_{m}^{\circ}(OH^{-})$
Given data:
$(I) \Lambda_{m}^{\circ}(NH_4Cl) = \lambda_{m}^{\circ}(NH_{4}^{+}) + \lambda_{m}^{\circ}(Cl^{-}) = 129.8 \ S \ cm^2 \ mol^{-1}$
$(II) \Lambda_{m}^{\circ}(KOH) = \lambda_{m}^{\circ}(K^{+}) + \lambda_{m}^{\circ}(OH^{-}) = 248.0 \ S \ cm^2 \ mol^{-1}$
$(III) \Lambda_{m}^{\circ}(KCl) = \lambda_{m}^{\circ}(K^{+}) + \lambda_{m}^{\circ}(Cl^{-}) = 126.0 \ S \ cm^2 \ mol^{-1}$
To obtain $\Lambda_{m}^{\circ}(NH_4OH)$,we perform the operation $(I) + (II) - (III)$:
$\Lambda_{m}^{\circ}(NH_4OH) = \Lambda_{m}^{\circ}(NH_4Cl) + \Lambda_{m}^{\circ}(KOH) - \Lambda_{m}^{\circ}(KCl)$
$\Lambda_{m}^{\circ}(NH_4OH) = 129.8 + 248.0 - 126.0$
$\Lambda_{m}^{\circ}(NH_4OH) = 251.8 \ S \ cm^2 \ mol^{-1}$

Step $1$: Calculate the cell constant $(G^*)$.
$G^* = \kappa \times R = (2.768 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}) \times (164 \ \Omega) = 0.4539 \ cm^{-1}$.
Step $2$: Calculate the conductivity $(\kappa)$ of $0.05 \ M \ AgNO_3$.
$\kappa = \frac{G^*}{R} = \frac{0.4539 \ cm^{-1}}{75.8 \ \Omega} = 5.988 \times 10^{-3} \ \Omega^{-1} \ cm^{-1} \approx 5.99 \times 10^{-3} \ \Omega^{-1} \ cm^{-1}$.
Step $3$: Calculate the molar conductivity $(\Lambda_m)$ of $AgNO_3$.
$\Lambda_m = \frac{1000 \times \kappa}{M} = \frac{1000 \times 5.988 \times 10^{-3}}{0.05} = 119.76 \ \Omega^{-1} \ cm^2 \ mol^{-1}$.

(N/A) The formula for molar conductivity is $\Lambda_m = \frac{\kappa \times 1000}{C}$.
Given conductivity $\kappa = 2.06 \times 10^{-3} \, S \, cm^{-1}$ and concentration $C = 0.02 \, M$.
Substituting the values: $\Lambda_m = \frac{2.06 \times 10^{-3} \times 1000}{0.02} = \frac{2.06}{0.02} = 103 \, S \, cm^2 \, mol^{-1}$.

(A) Given: Distance $(l) = 2 \, cm$,Area $(A) = 4.0 \, cm^2$,Resistance $(R) = 25 \, \Omega$,Molarity $(C) = 0.5 \, M$.
Cell constant $(G^*) = \frac{l}{A} = \frac{2}{4} = 0.5 \, cm^{-1}$.
Conductivity $(\kappa) = \frac{G^*}{R} = \frac{0.5}{25} = 0.02 \, S \, cm^{-1}$.
Molar conductivity $(\Lambda_m) = \frac{\kappa \times 1000}{C} = \frac{0.02 \times 1000}{0.5} = \frac{20}{0.5} = 40 \, S \, cm^2 \, mol^{-1}$.

(N/A) Given: For $0.0100 \ M \ KCl$,$\kappa = 0.00141 \ S \ cm^{-1}$ and $R = 161.8 \ \Omega$.
$(i)$ Cell constant $(G^*)$ = $\kappa \times R = 0.00141 \ S \ cm^{-1} \times 161.8 \ \Omega = 0.2281 \ cm^{-1}$.
$(ii)$ For $0.005 \ M \ NaOH$,$R = 190 \ \Omega$. Specific conductivity $(\kappa)$ = $G^* / R = 0.2281 \ cm^{-1} / 190 \ \Omega = 1.2 \times 10^{-3} \ S \ cm^{-1}$.
$(iii)$ Molar conductivity $(\Lambda_m)$ = $(\kappa \times 1000) / M = (1.2 \times 10^{-3} \ S \ cm^{-1} \times 1000) / 0.005 \ M = 240 \ S \ cm^2 \ mol^{-1}$.

(D) Metallic conductivity depends on the nature of the material,the number of valence electrons per atom,and temperature. As temperature increases,metallic conductivity decreases due to the vibration of metal ions.
Ionic (electrolytic) conductivity depends on the nature of the electrolyte,the size of the ions produced and their solvation,the nature of the solvent and its viscosity,the concentration of the electrolyte,and temperature. As temperature increases,ionic conductivity increases due to decreased viscosity and increased kinetic energy.