Following figure shows dependence of molar conductance of two electrolytes on concentration. Λ mᵒ is the limiting molar conductivity.
The number of incorrect statement(s) from the following is ___________
(A) Λ mᵒ for electrolyte A is obtained by extrapolation
(B) For electrolyte B, Λ m vs √c graph is a straight line with intercept equal to Λ mᵒ
(C) At infinite dilution, the value of degree of dissociation approaches zero for electrolyte B.
(D) Λ mᵒ for any electrolyte A and B can be calculated using λ° for individual ions

Solution

Chemistry ma’am · 2 steps

Read the curve shapes first: A shoots up near zero, so it is weak; B is a straight line, so it is strong.

1. Identify the electrolytes. Curve A rises very steeply as √c → 0, which is the typical shape for a weak electrolyte. Line B falls gently and in a straight line, which is the typical shape for a strong electrolyte.

   Λₘ = Λₘ° - A√c (strong electrolyte only)

   Statement	Verdict	Reason
   (A) Λₘ° of A by extrapolation	Incorrect	A is weak. Its curve shoots up near zero concentration, so it cannot be extended to meet the axis.
   (B) B gives a straight line with intercept Λₘ°	Correct	This is the equation above. Put c = 0 and the intercept is Λₘ°.
   (C) α → 0 at infinite dilution for B	Incorrect	At infinite dilution every electrolyte dissociates completely, so α → 1, not 0.
   (D) Λₘ° from ionic λ° for both	Correct	Kohlrausch's law, Λₘ° = ν₊λ₊° + ν₋λ₋°, works for strong and weak electrolytes alike.

2. So. Statements (A) and (C) are incorrect, so the number of incorrect statements is 2.

The common slip: 1. Taking extrapolation as valid for the weak electrolyte A leaves only (C) as incorrect, which gives 1 instead of 2.

The idea to keep. Extrapolating to zero concentration works only for strong electrolytes. For weak ones, find Λₘ° with Kohlrausch's law. At infinite dilution, α always goes to 1.

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