Q.Resistance of a conductivity cell filled with $0.1\ mol\ L^{-1}$ KCl solution is $100\ \Omega$. If the resistance of the same cell when filled with $0.02\ mol\ L^{-1}$ KCl solution is $520\ \Omega$, calculate the conductivity and molar conductivity of $0.02\ mol\ L^{-1}$ KCl solution. The conductivity of $0.1\ mol\ L^{-1}$ KCl solution is $1.29\ S\ m^{-1}$.

Step 1 — Recall the defining relation

Conductivity ($\kappa$) of a solution relates to the measured resistance ($R$) and the cell constant ($G^*$, also called $l/A$) as:

Step 2 — Substitute the given values

Step 3 — Conclusion

The calculated cell constant is 1.26 cm$^{-1}$, matching option (B).

The first thing to notice is the units of the rate constant: $3.4 \times 10^{-5}\,\text{s}^{-1}$. A rate constant with units of $\text{s}^{-1}$ is the hallmark of a first-order reaction. For a first-order reaction, the rate depends linearly on the concentration of exactly one reactant — here, that reactant is $\text{PCl}_5$.

Why does this matter? Because the rate law is not something you guess; it is determined experimentally. But the problem gives you the rate constant’s units, and those units tell you the order. If $k$ had units like $\text{L mol}^{-1}\text{s}^{-1}$, it would be second order. Here, it’s clearly first order.

So the rate law is:

Now we simply plug in the numbers.

1. Write the first-order rate law For the decomposition $\text{PCl}_5 \rightarrow \text{PCl}_3 + \text{Cl}_2$, the rate of disappearance of $\text{PCl}_5$ is:

2. Substitute the given values Rate = $1.02 \times 10^{-4}\ \text{mol L}^{-1}\text{s}^{-1}$ $k = 3.4 \times 10^{-5}\ \text{s}^{-1}$ So:

3. Solve for $[\text{PCl}_5]$ Divide both sides by $3.4 \times 10^{-5}$:

Concept and Intuition

Conductivity measures how easily ions move through a solution under an electric field. Anything that slows ions down — like a thick (viscous) solvent, large ion size, or a heavy solvation shell — reduces conductivity. Temperature increases kinetic energy, helping ions overcome these drag forces, so electrolytic conductance increases with temperature. The trick is to spot which statement contradicts this physical picture.

Step-by-step reasoning

1. Statement (A): “Electrolytic conductance increases with increase in temperature.”

   • As temperature rises, ions move faster (higher kinetic energy), the solvent becomes less viscous, and ion dissociation often increases. All these factors increase conductance.
   • This is a well-known fact — true for electrolytes. So (A) is correct.

2. Statement (B): “Conductivity does not depend on the viscosity of the solution.”

   • Viscosity is a measure of a fluid’s resistance to flow. Ions moving through a viscous solution experience greater frictional drag, reducing their mobility and thus the conductivity.
   • For example, a concentrated sugar solution (high viscosity) conducts electricity much worse than pure water, even if the same salt is dissolved.
   • Therefore, conductivity does depend on viscosity. This statement is false.

3. Statement (C): “Conductivity depends on the size of the ions.”

   • Larger ions move more slowly through the solvent (Stokes’ law: drag force ∝ radius). Smaller ions (like $\text{K}^+$) are more mobile than larger ones (like $\text{Cs}^+$), all else equal.
   • So ion size directly affects conductivity. This statement is true.

4. Statement (D): “Conductivity depends on the solvation of the ions.”

   • Solvation (hydration in water) wraps ions in a shell of solvent molecules, increasing the effective size of the moving particle. Heavily solvated ions (e.g., $\text{Li}^+$) move slower than lightly solvated ones (e.g., $\text{K}^+$).
   • Thus solvation influences mobility and conductivity. This statement is true.

Concept: molar conductivity Lambda_m = kappa x 1000 / C , and conductance G = kappa x (A/l) = 1/R.

Step 1 - conductivity:

kappa = Lambda_m x C / 1000 = 194.5 x 0.05 / 1000 = 9.725 x 10^-3 S cm^-1

Step 2 - cell constant:

l/A = 1.0 cm / 3.0 cm^2 = 0.3333 cm^-1

Step 3 - resistance:

R = (l/A) / kappa = 0.3333 / 9.725 x 10^-3 = 34.27 ohm

ANSWER: D

Concept: properties of conductance and cell quantities.

A - kappa (conductivity) = conductance of unit volume; it depends on the number of ions per unit VOLUME, so it DECREASES on dilution (i.e. with decrease in concentration) for BOTH strong and weak electrolytes -> R

B - E0_cell (standard cell potential) does not depend on the amount of substance -> an INTENSIVE property -> P

C - Molar conductivity Lambda_m = kappa x 1000/C: it INCREASES with dilution (more complete dissociation / greater ionic mobility) -> S

D - dG_cell = -nFE_cell: it is proportional to n, the amount of substance -> an EXTENSIVE property -> Q

So A=R, B=P, C=S, D=Q.

ANSWER: B

The cell constant of a conductivity cell is a fixed geometric factor that relates the measured resistance of a solution to its conductivity. It depends only on the distance between the electrodes and their area, not on the solution being measured.

The fundamental relation is:

where $\kappa$ is the conductivity (in S cm$^{-1}$), $R$ is the resistance (in $\Omega$), and $G^*$ is the cell constant (in cm$^{-1}$).

This means: if you know the conductivity of a solution and you measure its resistance in a particular cell, you can directly calculate the cell constant.

1. Identify the given data

   • Resistance, $R = 1500\ \Omega$
   • Conductivity, $\kappa = 0.146 \times 10^{-3}\ \text{S cm}^{-1}$
   • Temperature: 298 K (consistent for both values)

2. Apply the formula

Substitute the values:

3. Calculate step by step

Molar conductance relates to specific conductance (conductivity) by:

with $\Lambda_m = 1.06\times10^{2}\ \text{ohm}^{-1}\text{cm}^2\text{mol}^{-1}$ and $C = 0.1\ \text{mol L}^{-1}$:

Silver nitrate precipitates only chloride ions that are free in solution — those outside the coordination sphere of the metal. Ligands bound inside the sphere are not available to $\text{Ag}^+$. So the number of moles of $\text{AgCl}$ formed per mole of complex tells us exactly how many chlorides are ionisable.

1. Interpret the precipitation data. $\text{AgNO}_3$ reacts only with free $\text{Cl}^-$:

1 mole of complex yields 2 moles of $\text{AgCl}$, so there are 2 moles of free $\text{Cl}^-$ per mole of complex. Both chloride ions are therefore outside the coordination sphere.

2. Deduce the coordination sphere. The dot formula $\text{PdCl}_2 \cdot 4\text{NH}_3$ hides the true structure. With both chlorides free, the coordination sphere holds only the 4 ammonia ligands around palladium:

3. Determine the ions in solution. On dissolving, it dissociates completely:

Each formula unit gives 3 ions: one divalent cation and two monovalent anions.

4. Classify the electrolyte. One cation to two anions is a 1:2 electrolyte.

The key here is to understand what molar conductance and specific conductance actually represent, and how they are related.

Specific conductance ($\kappa$) is the conductance of a solution held between two electrodes of unit area placed one unit apart. It depends only on the concentration and nature of ions present in that volume. Molar conductance ($\Lambda_m$), on the other hand, is the conductance of all ions produced from one mole of electrolyte dissolved in a given volume. The two are linked by the concentration of the solution.

The defining relation is:

But here the question asks for the ratio $\frac{\Lambda_m}{\kappa}$, not $\Lambda_m$ itself. Let’s work it out step by step.

1. Write the ratio in terms of concentration. From the formula above:

So the ratio is simply the reciprocal of the concentration expressed in $\mathrm{mol \, cm^{-3}}$.

2. Convert the given concentration to the right units. The concentration is $0.01$ M, which means $0.01$ moles per litre. Since $1 \, \mathrm{L} = 1000 \, \mathrm{cm}^3$, we have:

3. Take the reciprocal to get the ratio.

Notice the units: $\mathrm{cm^3 \, mol^{-1}}$ — this is a volume per mole, which makes perfect sense. It tells you the volume of solution that contains exactly one mole of NaCl at this concentration.

4. Match with the options. The result $10^5 \ \mathrm{cm^3 \, mol^{-1}}$ corresponds exactly to option (A). Option (D) has $\mathrm{cm^2}$, which is area, not volume — so it’s dimensionally wrong for this ratio.

The question tests a fundamental distinction: what makes a substance conduct electricity? Conductivity depends on the presence of mobile charge carriers — electrons or ions. Metals like copper have delocalised electrons that drift easily under a voltage. Ionic compounds, however, conduct only when their ions are free to move, which happens in the molten state or in solution. In the solid state, the ions are fixed at lattice points and can only vibrate, not migrate.

Let’s examine each option:

1. Solid NaCl — Sodium chloride in its crystalline form has Na⁺ and Cl⁻ ions arranged in a regular, repeating lattice. These ions are held tightly by strong electrostatic forces. They cannot move from their positions, so there are no mobile charge carriers. Solid NaCl is an electrical insulator.

2. Cu (copper metal) — Copper is a metallic conductor. Its atoms release their outermost electrons into a “sea” of delocalised electrons that flow freely through the metal lattice. When a potential difference is applied, these electrons drift, producing a current. Copper is an excellent conductor.

3. Fused NaCl — “Fused” means molten. When solid NaCl is heated above its melting point (801 °C), the lattice breaks down. The Na⁺ and Cl⁻ ions become free to move through the liquid. In the molten state, NaCl conducts electricity because the ions themselves act as charge carriers.

4. Brine solution — Brine is a concentrated solution of NaCl in water. In water, the ionic lattice dissolves completely, releasing Na⁺ and Cl⁻ ions into the solution. These hydrated ions are mobile and can migrate toward electrodes, making brine a good conductor of electricity.