JEE Main · 2024 · Shift-IIeasyEC-095

For a strong electrolyte, a plot of molar conductivity against (concentration)1/2 is a straight line, with a negative…

Electrochemistry · Class 12 · JEE Main Previous Year Question

Question

For a strong electrolyte, a plot of molar conductivity against (concentration)1/2^{1/2} is a straight line, with a negative slope, the correct unit for the slope is

Options
  1. a

    S cm2 mol−3/2 L−1/2\mathrm{S\ cm^2\ mol^{-3/2}\ L^{-1/2}}

  2. b

    S cm2 mol−3/2 L1/2\mathrm{S\ cm^2\ mol^{-3/2}\ L^{1/2}}

    ✓
  3. c

    S cm2 mol−3/2 L\mathrm{S\ cm^2\ mol^{-3/2}\ L}

  4. d

    S cm2 mol−1 L1/2\mathrm{S\ cm^2\ mol^{-1}\ L^{1/2}}

Correct Answerb

S cm2 mol−3/2 L1/2\mathrm{S\ cm^2\ mol^{-3/2}\ L^{1/2}}

Detailed Solution

Strategy: Use the Debye-Hückel-Onsager equation to derive the units of the slope (AA). The equation is linear in C\sqrt{C}: Λm=Λm0−AC\Lambda_m = \Lambda_m^0 - A\sqrt{C}.

Dimensional Analysis:

  • Λm\Lambda_m (Molar conductivity) units =Scm2mol−1= \mathrm{S cm^2 mol^{-1}}
  • CC (Concentration) units =molL−1= \mathrm{mol L^{-1}}
  • C\sqrt{C} units =mol1/2L−1/2= \mathrm{mol^{1/2} L^{-1/2}}

Units of Slope (AA): [A]=[Λm][C]=Scm2mol−1mol1/2L−1/2[A] = \frac{[\Lambda_m]}{[\sqrt{C}]} = \frac{\mathrm{S cm^2 mol^{-1}}}{\mathrm{mol^{1/2} L^{-1/2}}} [A]=Scm2mol−1⋅mol−1/2⋅L1/2[A] = \mathrm{S cm^2 mol^{-1} \cdot mol^{-1/2} \cdot L^{1/2}} [A]=Scm2mol−3/2L1/2[A] = \mathrm{S cm^2 mol^{-3/2} L^{1/2}}

Answer: (b) \ceS cm2 mol−3/2 L1/2\boxed{\text{Answer: (b) } \ce{S\ cm^2\ mol^{-3/2}\ L^{1/2}}}

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