Match List-I with List-IIList-IList IIUnitsTerms(A) mol/L(I) mole fra…

The correct answer is Option (3) → (A) - (IV), (B) - (I), (C) - (II), (D) - (III)

The unit of a quantity follows directly from its definition: write the defining ratio and put in the units of each part.

• (A) mol/L — This matches (IV) Molarity. Molarity is moles of solute per litre of solution, so its unit is mol L⁻¹.

      \( \displaystyle M=\frac{n_B}{V}\ \Rightarrow\ \frac{\mathrm{mol}}{\mathrm{L}} \)

• (B) no unit — This matches (I) mole fraction. It is a ratio of moles to moles, so the units cancel and it is a pure number.

      \( \displaystyle x_A=\frac{n_A}{n_A+n_B}\ \Rightarrow\ \frac{\mathrm{mol}}{\mathrm{mol}} \)

• (C) S m⁻¹ — This matches (II) Conductivity. Conductivity (κ) is the reciprocal of resistivity. From R = ρl/A, resistivity ρ = RA/l has the unit Ω m²/m = Ω m, so κ has the unit Ω⁻¹ m⁻¹ = S m⁻¹.

      \( \displaystyle \kappa=\frac{1}{\rho}\ \Rightarrow\ \frac{1}{\Omega\,\mathrm{m}}=\mathrm{S\,m^{-1}} \)

• (D) S m² mol⁻¹ — This matches (III) Molar conductivity. Molar conductivity is conductivity divided by the molar concentration (in mol m⁻³).

      \( \displaystyle \Lambda_m=\frac{\kappa}{c}\ \Rightarrow\ \frac{\mathrm{S\,m^{-1}}}{\mathrm{mol\,m^{-3}}}=\mathrm{S\,m^{2}\,mol^{-1}} \)

Shortcut: (C)–(II) appears only in Options (1) and (3), and (A)–(IV) then leaves only Option (3).