$\text{If the molar conductivity (Λₘ) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its degree of dissociation will be}$

$\text{[Assume Λ°₊ = 349.6 S cm² mol⁻¹ and Λ°₋ = 50.4 S cm² mol⁻¹]}$$\text{Degree of dissociation (α) is given as}$

To find the degree of dissociation of the monobasic weak acid, we can use the relationship between its molar conductivity at a given concentration and its molar conductivity at infinite dilution.

Step 1: Identify the given values
- Concentration of the weak acid solution, $C=0.050{\text{mol L}}^{-1}$
- Molar conductivity of the solution, ${\Lambda }_m=90{\text{S cm}}^2{\text{mol}}^{-1}$
- Limiting molar conductivity of the cation, ${\Lambda }_{+}^{\circ }=349.6{\text{S cm}}^2{\text{mol}}^{-1}$
- Limiting molar conductivity of the anion, ${\Lambda }_{-}^{\circ }=50.4{\text{S cm}}^2{\text{mol}}^{-1}$

Step 2: Calculate the limiting molar conductivity of the weak acid (${\Lambda }_m^{\circ }$)
According to Kohlrausch's Law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting ionic conductivities of its individual ions:

${\Lambda }_m^{\circ }={\Lambda }_{+}^{\circ }+{\Lambda }_{-}^{\circ }$

Substitute the given values into the equation:

${\Lambda }_m^{\circ }=349.6+50.4=400.0{\text{S cm}}^2{\text{mol}}^{-1}$

Step 3: Calculate the degree of dissociation ($\alpha$)
The degree of dissociation ($\alpha$) is defined as the ratio of molar conductivity at a specific concentration to the molar conductivity at infinite dilution:

$\alpha =\frac{{\Lambda }_m}{{\Lambda }_m^{\circ }}$

Substitute the values of ${\Lambda }_m$ and ${\Lambda }_m^{\circ }$ into the equation:

$\alpha =\frac{90}{400}=0.225$

Therefore, the degree of dissociation of the weak acid is 0.225.