Step 1: Principle of conductometric titration

Conductance depends on:

1. Number of ions present
2. Charge on ions
3. Ionic mobility

Here, we compare slopes using limiting ionic conductivity values $\Lambda_0$.

At equivalence point, the nature of ions present changes, so the graph shows a break.

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Step 2: Analyse each titration

(P) Titrate: $KCl$, Titrant: $AgNO_3$

Reaction: $Ag^+ + Cl^- \rightarrow AgCl(s)$

Initially, solution contains $K^+$ and $Cl^-$. As $AgNO_3$ is added, $Cl^-$ is removed as precipitate and is replaced by $NO_3^-$.

Compare ionic conductivities: $\lambda^0(Cl^-) = 7.6, \qquad \lambda^0(NO_3^-) = 7.2$

So before equivalence, conductance slightly decreases. After equivalence, excess $AgNO_3$ adds free $Ag^+$ and $NO_3^-$, so conductance increases sharply.

Thus graph: small decrease up to equivalence, then increase. So $P \to 3$.

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(Q) Titrate: $AgNO_3$, Titrant: $KCl$

Reaction: $Ag^+ + Cl^- \rightarrow AgCl(s)$

Initially, solution contains $Ag^+$ and $NO_3^-$. As $KCl$ is added, $Ag^+$ is removed and replaced by $K^+$.

Compare ionic conductivities: $\lambda^0(Ag^+) = 6.2, \qquad \lambda^0(K^+) = 7.4$

So before equivalence, conductance increases slightly. After equivalence, excess $KCl$ adds free $K^+$ and $Cl^-$, so conductance increases more steeply.

Thus graph: increase before equivalence, steeper increase after equivalence. So $Q \to 4$.

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(R) Titrate: $NaOH$, Titrant: $HCl$

Reaction: $NaOH + HCl \rightarrow NaCl + H_2O$

Initially, solution contains $Na^+$ and $OH^-$. Since $\lambda^0(OH^-) = 19.9$ is high, replacing $OH^-$ by $Cl^-$ $\lambda^0(Cl^-) = 7.6$ causes conductance to decrease sharply up to equivalence.

After equivalence, excess $HCl$ adds highly mobile $H^+$: $\lambda^0(H^+) = 35.0$ so conductance increases sharply.

Thus graph: V-shaped, sharp decrease then sharp increase. So $R \to 2$.

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(S) Titrate: $NaOH$, Titrant: $CH_3COOH$

Reaction: $NaOH + CH_3COOH \rightarrow CH_3COONa + H_2O$

Initially, solution contains $Na^+$ and $OH^-$. As acetic acid is added, $OH^-$ is replaced by $CH_3COO^-$.

Compare: $\lambda^0(OH^-) = 19.9, \qquad \lambda^0(CH_3COO^-) = 4.1$

So conductance decreases strongly up to equivalence. After equivalence, because $CH_3COOH$ is a weak acid, it ionizes only slightly, so conductance increases only slightly after equivalence.

Thus graph: decrease up to equivalence, then gentle increase. So $S \to 5$.

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Step 3: Final matching

We get: $P-3, \quad Q-4, \quad R-2, \quad S-5$

This corresponds to Option C.

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Step 4: Comparison with stored answer

Stored correct answer = C

Our derived answer = C

So the answer agrees.