If the molar conductivity ${\Lambda }_m$ 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 ${\Lambda }_{+}^{°}$ = $349.6$ S cm² mol^-1 and ${\Lambda }_{-}^{°}$ = $50.4$ S cm² mol^-1.]

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:

  ${\Lambda }_{+}^{°}$ = $349.6$ S cm² mol^‑1

  ${\Lambda }_{‑}^{°}$ = $50.4$ 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:

$\Lambda °\; =\; \Lambda °(H⁺)\; +\; \Lambda °(A⁻)$

Insert the given values:

$\Lambda °\; =\; 349.6\; +\; 50.4\; =\; 400.0$ S cm² mol^‑1

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

$\alpha \; =\; \backslash frac\{\Lambda \_m\}\{\Lambda °\}$

Now substitute the measured conductivity and the calculated Λ°:

$\alpha \; =\; \backslash frac\{90\}\{400.0\}$

Carry out the division:

$\alpha \; =\; 0.225$

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