NEET (UG)-2025 Chemistry Code-47 Question Paper with Solutions

# Q1 of 45

If the molar conductivity Λm of a 0.050 mol L-1 solution of a monobasic weak acid is 90 S cm2 mol-1, its extent (degree) of dissociation will be
[Assume Λ + ° = 349.6 S cm2 mol-1 and Λ - ° = 50.4 S cm2 mol-1.]

Options
A.

0.215

B.

0.115

C.

0.125


D.

0.225


Show Answer
Correct Answer
D

0.225


Solution

We are given:

• Concentration of the weak acid solution: 0.050 mol L‑1

• Measured molar conductivity of this solution: 90 S cm2 mol‑1

• Molar conductivity of the fully dissociated ions at infinite dilution:

  Λ+° = 349.6 S cm2 mol‑1

  Λ° = 50.4 S cm2 mol‑1

For a monobasic weak acid HA, the limiting molar conductivity (when the acid is completely dissociated) is the sum of the ionic conductivities:

Λ° = Λ°(H⁺) + Λ°(A⁻)

Insert the given values:

Λ° = 349.6 + 50.4 = 400.0 S cm2 mol‑1

The degree of dissociation (α) relates the observed molar conductivity (Λm) to the limiting value (Λ°) by:

α = \frac{Λ_m}{Λ°}

Now substitute the measured conductivity and the calculated Λ°:

α = \frac{90}{400.0}

Carry out the division:

α = 0.225

Therefore, the extent (degree) of dissociation of the weak acid solution is 0.225.

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