Redox Reactions and Electrochemistry NEET 2026: Conductivity

What is the answer to the NEET 2026 limiting conductivity question?

Option A, 80.0 siemens square centimetres per mole, is correct. This supplied NEET 2026 Chemistry question from Redox Reactions and Electrochemistry tests molar conductivity of strong electrolytes. First recover the salt’s limiting conductivity, then subtract the cation’s contribution.

The strong electrolyte XY supplies one cation and one anion per formula unit:

X Y \to X^{+} + Y^{-}

All measurements are at 298 K. The supplied straight-line plot gives:

Slope of Λ_{m} versus \sqrt{c} = - 90.0 S c m^{2} m o l^{- 3 / 2} L^{1 / 2}

c = 0.01 m o l L^{- 1}, Λ_{m} = 145.0 S c m^{2} m o l^{- 1}

λ_{X^{+}}^{0} = 74.0 S c m^{2} m o l^{- 1}

Find the anion’s limiting molar conductivity: $λ_{Y^{-}}^{0}$

Options, all in siemens square centimetres per mole:

The question-bank difficulty is medium. The question bank’s expected solving time is 90 seconds, not an official exam time allowance.

How do you recover the salt’s limiting conductivity from the negative slope?

Λ_{m} = Λ_{m}^{0} - A \sqrt{c}

The official strong-electrolyte relation gives an intercept of 154.0 siemens square centimetres per mole. The negative slope places the measured conductivity below that intercept. Recovering the intercept therefore requires adding the concentration correction, not subtracting it again.

Compare the relation with the straight-line equation: $y = b + m x$

y = Λ_{m}, x = \sqrt{c}, b = Λ_{m}^{0}, m = - A

The slope is negative, while the coefficient is positive:

- A = - 90.0 \Rightarrow A = 90.0

The coefficient named A is unrelated to option A. With concentration entered in moles per litre:

c = 0.01, \sqrt{c} = \sqrt{0.01} = 0.1

Substitute into the official relation:

Λ_{m} = Λ_{m}^{0} - 90.0 \times 0.1

145.0 = Λ_{m}^{0} - 9.0

Move the negative correction to the other side:

Λ_{m}^{0} = 145.0 + 9.0 = 154.0 S c m^{2} m o l^{- 1}

The slope coefficient alone is not a conductivity. Multiplying it by the square root of concentration produces the conductivity correction:

(90.0 S c m^{2} m o l^{- 3 / 2} L^{1 / 2}) (0.1 m o l^{1 / 2} L^{- 1 / 2}) = 9.0 S c m^{2} m o l^{- 1}

As concentration approaches zero, the negative-slope line rises towards its zero-concentration intercept. The limiting value must therefore exceed the measured finite-concentration value.

How do you separate the ionic contributions and obtain option A?

Subtract the cation’s limiting contribution from the salt’s limiting conductivity. This gives the anion value of 80.0 siemens square centimetres per mole. This second operation uses independent ionic contributions at infinite dilution, not the concentration correction from the first step.

For the salt XY:

Λ_{m}^{0} (X Y) = λ_{X^{+}}^{0} + λ_{Y^{-}}^{0}

Both coefficients are one because each formula unit supplies one cation and one anion. Substitute the recovered salt intercept and the supplied cation value:

154.0 = 74.0 + λ_{Y^{-}}^{0}

λ_{Y^{-}}^{0} = 154.0 - 74.0 = 80.0 S c m^{2} m o l^{- 1}

The final answer is option A. Verify both stages, starting with the ionic sum: $74.0 + 80.0 = 154.0$

Then check the measured point:

154.0 - 90.0 \times 0.1 = 145.0

Subtract the cation contribution only after recovering the salt’s intercept. Subtracting it directly from the measured conductivity mixes a finite-concentration salt value with an infinite-dilution ionic value.

Why does treating the slope as conductivity give the wrong option C?

Taking the slope’s magnitude as the anion conductivity confuses two different quantities. The slope describes how the salt’s molar conductivity changes with the square root of concentration. It does not give an individual ion’s limiting contribution. One possible erroneous method is:

Slope = - 90.0 \Rightarrow incorrectly report λ_{Y^{-}}^{0} = 90.0

That selects option C, but the units expose the error:

Slope units:

S c m^{2} m o l^{- 3 / 2} L^{1 / 2}

Requested ionic conductivity units:

S c m^{2} m o l^{- 1}

Removing a minus sign does not change units. This shortcut also never uses either the measured salt value, 145.0, or the cation contribution, 74.0. It cannot separate the two ions.

Repair the method in this order:

Use the slope to calculate the correction at the stated concentration.
Add that correction to recover the salt’s intercept.
Subtract the cation’s limiting contribution.

Which three related questions check whether you understand the method?

These related practice questions check slope recovery, solution conductivity and whether the straight-line method applies to weak electrolytes. They are based on the supplied data and concepts from the same chapter, not additional verified NEET PYQs.

What is the slope if the limiting value is 154.0 and the measured value is 145.0 at 0.01 M?

The slope is negative 90.0 in the supplied slope units. Use the finite-concentration point and the zero-concentration intercept. The horizontal coordinate is the square root of concentration, not concentration itself:

m = \frac{145.0 - 154.0}{0.1 - 0} = - 90.0 S c m^{2} m o l^{- 3 / 2} L^{1 / 2}

The negative sign agrees with the measured value being below the intercept. Dividing by 0.01 instead of 0.1 would use the wrong horizontal axis.

What is the conductivity of the original 0.01 M solution?

The solution conductivity is 0.00145 siemens per centimetre. Use the measured molar conductivity because the question asks about the original solution, not infinite dilution:

Λ_{m} = \frac{1000 κ}{c}

Here, conductivity is entered in siemens per centimetre and concentration in moles per litre:

κ = \frac{Λ_{m} c}{1000} = \frac{145.0 \times 0.01}{1000} = 0.00145 S c m^{- 1}

The factor 1000 converts litres to cubic centimetres. It is a unit conversion, not new experimental data.

Can you use the same straight-line extrapolation directly for a weak electrolyte?

No. A weak electrolyte’s degree of ionisation changes with concentration, so its molar conductivity does not follow this simple strong-electrolyte straight-line treatment.

Instead, obtain its limiting molar conductivity using appropriate limiting ionic contributions, with the correct stoichiometric coefficients. Before choosing a formula, identify the electrolyte type and whether the requested quantity belongs to the solution, the salt or one ion.

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Frequently asked questions

What is the answer to the NEET 2026 limiting conductivity question?

For the supplied XY question, option A, 80.0 S cm² mol⁻¹, is correct. First recover the salt’s limiting molar conductivity: 145.0 + 90.0 × √0.01 = 154.0 S cm² mol⁻¹. Then subtract the cation’s limiting contribution: 154.0 − 74.0 = 80.0 S cm² mol⁻¹.

How do I calculate limiting molar conductivity from a negative slope?

Use Λm = Λm⁰ − A√c, where the graph’s slope is −A. A slope of −90.0 gives A = 90.0, so at c = 0.01 mol L⁻¹ the correction is 9.0 S cm² mol⁻¹. Add this correction to the measured value of 145.0 to obtain the intercept, 154.0 S cm² mol⁻¹.

Why is 90.0 not the anion’s limiting molar conductivity?

The value 90.0 is the slope’s magnitude, not an ionic conductivity. The slope has units of S cm² mol⁻³ᐟ² L¹ᐟ², which differ from the requested S cm² mol⁻¹. Use the slope to recover the salt’s limiting conductivity, then subtract the cation’s limiting contribution.

Can I use straight-line extrapolation for a weak electrolyte?

No, a weak electrolyte does not follow this simple strong-electrolyte straight-line treatment because its degree of ionisation changes with concentration. Obtain its limiting molar conductivity from appropriate limiting ionic contributions, using the correct stoichiometric coefficients.

Redox Reactions and Electrochemistry NEET 2026: Ionic Conductivity

What is the answer to the NEET 2026 limiting conductivity question?

How do you recover the salt’s limiting conductivity from the negative slope?

How do you separate the ionic contributions and obtain option A?

Why does treating the slope as conductivity give the wrong option C?

Which three related questions check whether you understand the method?

What is the slope if the limiting value is 154.0 and the measured value is 145.0 at 0.01 M?

What is the conductivity of the original 0.01 M solution?

Can you use the same straight-line extrapolation directly for a weak electrolyte?

Frequently asked questions

What is the answer to the NEET 2026 limiting conductivity question?

How do I calculate limiting molar conductivity from a negative slope?

Why is 90.0 not the anion’s limiting molar conductivity?

Can I use straight-line extrapolation for a weak electrolyte?

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