JEE Main 2024 · 4 April · evening · Single correct · +4 / −1

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

  1. (a) S cm² mol^(-3 / 2) L
  2. (b) S cm ² mol^(-3 / 2) L^(-1 / 2)
  3. (c) S cm ² mol⁻¹ L^(1 / 2)
  4. (d) S cm² mol^(-3 / 2) L^(1 / 2)
Answer and solution

Answer (d) S cm² mol^(-3 / 2) L^(1 / 2)

Solution

Chemistry ma’am · 4 steps

Slope is just y-units divided by x-units, so divide the units of Λₘ by the units of c^(1/2).

  1. The straight line. For a strong electrolyte, the Debye–Hückel–Onsager (Kohlrausch) equation gives a straight line of Λₘ against √c. The slope is -A, and A has the units of Λₘ divided by the units of √c.

    Λₘ = Λₘ° - A√c

  2. Units of each axis. Λₘ is in S cm² mol⁻¹. Concentration is in mol L⁻¹, so √c is in mol^(1/2) L^(-1/2).
  3. Divide them. (S cm² mol⁻¹)/(mol^(1/2) L^(-1/2)) = S cm² mol^(-1-1/2) L^(+1/2). Dividing flips the signs of both exponents in the denominator.
  4. So. The slope is in S cm² mol^(-3/2) L^(1/2), which is option (d).

The idea to keep. For the unit of any slope, divide y-units by x-units. When a square root sits in the denominator, halve each exponent first, then flip its sign.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Electrochemistry

Asked about 170 times in JEE Main since 2020. This question is from 2024.