The decreasing order of electrical conductivity of the following aqueous solutions is:
(A) 0.1 M Formic acid,
(B) 0.1 M Acetic acid,
(C) 0.1 M Benzoic acid.

We recall that the electrical conductivity $$\kappa$$ of an aqueous electrolyte is proportional to the total concentration of the ions actually present in solution. For a weak monoprotic acid HA of analytical concentration $$C$$ we have

$$\text{HA}\rightleftharpoons \text{H}^{+}+\text{A}^{-}$$

Its acid-dissociation constant is defined by the formula

$$K_a=\dfrac{[\text{H}^{+}][\text{A}^{-}]}{[\text{HA}]}\,.$$

If the degree of ionisation is $$\alpha$$, then at equilibrium

$$[\text{H}^{+}]=\alpha C,\qquad [\text{A}^{-}]=\alpha C,\qquad [\text{HA}]=C(1-\alpha).$$

Substituting these expressions in the definition of $$K_a$$ gives

$$K_a=\dfrac{(\alpha C)(\alpha C)}{C(1-\alpha)}=\dfrac{\alpha^{\,2}C}{1-\alpha}\,.$$

For weak acids $$\alpha\lt\!\lt1$$, so $$1-\alpha\approx1$$ and we obtain the very useful relation

$$\alpha\approx\sqrt{\dfrac{K_a}{C}}\;.$$

Because each acid furnishes the same types of ions (mainly the very mobile $$\text{H}^{+}$$ ion), the molar conductivity $$\Lambda_m$$ is roughly proportional to $$\alpha$$, and the specific conductivity $$\kappa=C\Lambda_m$$ is therefore also proportional to $$\alpha$$ when the concentration $$C$$ is fixed. So, for the three 0.1 M acids given, the larger the $$K_a$$, the larger the $$\alpha$$ and hence the greater the conductivity.

Now we list the literature values of $$K_a$$ for the three acids at 298 K:

$$K_a(\text{Formic acid}) = 1.8\times10^{-4}$$

$$K_a(\text{Benzoic acid}) = 6.3\times10^{-5}$$

$$K_a(\text{Acetic acid}) = 1.75\times10^{-5}$$

Clearly,

$$K_a(\text{Formic}) \gt K_a(\text{Benzoic}) \gt K_a(\text{Acetic}).$$

Using $$\alpha\approx\sqrt{\dfrac{K_a}{C}}$$ with the common concentration $$C=0.1\ \text{M}$$, we obtain numerically

$$\alpha_{\text{Formic}} \approx \sqrt{\dfrac{1.8\times10^{-4}}{0.1}} = \sqrt{1.8\times10^{-3}} \approx 4.2\times10^{-2},$$

$$\alpha_{\text{Benzoic}} \approx \sqrt{\dfrac{6.3\times10^{-5}}{0.1}} = \sqrt{6.3\times10^{-4}} \approx 2.5\times10^{-2},$$

$$\alpha_{\text{Acetic}} \approx \sqrt{\dfrac{1.75\times10^{-5}}{0.1}} = \sqrt{1.75\times10^{-4}} \approx 1.3\times10^{-2}.$$

Thus we obtain the sequence

$$\alpha_{\text{Formic}} \gt \alpha_{\text{Benzoic}} \gt \alpha_{\text{Acetic}},$$

and, because conductivity tracks $$\alpha$$, we directly conclude

$$\kappa(0.1\ \text{M Formic}) \gt \kappa(0.1\ \text{M Benzoic}) \gt \kappa(0.1\ \text{M Acetic}).$$

Writing this in the order requested, the decreasing order of electrical conductivity is

(A) 0.1 M Formic acid > (C) 0.1 M Benzoic acid > (B) 0.1 M Acetic acid.

Hence, the correct answer is Option B.