The value of $$\alpha$$ is ___.

The limiting (infinite dilution) molar conductivity of the weak monobasic acid at 298 K is given as $$\Lambda_\infty = 4 \times 10^{2}\;S\,cm^{2}\,mol^{-1}$$.

For any weak electrolyte, the molar conductivity at concentration $$c$$ is related to its degree of dissociation $$\alpha$$ by the formula
$$\Lambda_m = \alpha\,\Lambda_\infty \quad -(1)$$

For the original solution the molar conductivity is stated to be $$y \times 10^{2}\;S\,cm^{2}\,mol^{-1}$$. Using $$(1)$$: $$y \times 10^{2} = \alpha \,(4 \times 10^{2})$$ Dividing both sides by $$10^{2}$$ yields $$y = 4\alpha \quad -(2)$$

After 20-fold dilution the molar conductivity becomes $$3y \times 10^{2}\;S\,cm^{2}\,mol^{-1}$$. Let the new degree of dissociation be $$\alpha'$$. Again using $$(1)$$: $$3y \times 10^{2} = \alpha'\,(4 \times 10^{2})$$ $$3y = 4\alpha' \quad -(3)$$

From $$(2)$$, substitute $$y = 4\alpha$$ into $$(3)$$: $$3(4\alpha) = 4\alpha' \;\Rightarrow\; \alpha' = 3\alpha \quad -(4)$$

For a weak monobasic acid the Ostwald dilution law gives the acid-dissociation constant
$$K_a = \dfrac{c\alpha^{2}}{1-\alpha} \quad -(5)$$ where $$c$$ is the initial concentration.

After dilution, the concentration becomes $$c/20$$, so $$K_a = \dfrac{(c/20)\,\alpha'^{2}}{1-\alpha'} \quad -(6)$$

Equating $$(5)$$ and $$(6)$$ because $$K_a$$ is a constant: $$\dfrac{c\alpha^{2}}{1-\alpha} = \dfrac{(c/20)\,\alpha'^{2}}{1-\alpha'}$$ The concentration $$c$$ cancels, giving $$\frac{\alpha^{2}}{1-\alpha} = \frac{\alpha'^{2}}{20\,(1-\alpha')} \quad -(7)$$

Insert $$\alpha' = 3\alpha$$ from $$(4)$$ into $$(7)$$: $$\frac{\alpha^{2}}{1-\alpha} = \frac{(3\alpha)^{2}}{20\,[\,1-3\alpha\,]}$$

Cross-multiplying:
$$20\alpha^{2}(1-3\alpha) = 9\alpha^{2}(1-\alpha)$$

Expand both sides:
$$20\alpha^{2} - 60\alpha^{3} = 9\alpha^{2} - 9\alpha^{3}$$

Bring all terms to one side and collect like powers of $$\alpha$$:
$$(20-9)\alpha^{2} + (-60 + 9)\alpha^{3} = 0$$ $$11\alpha^{2} - 51\alpha^{3} = 0$$ Factor out $$\alpha^{2}$$ (discarding the trivial root $$\alpha = 0$$): $$11 - 51\alpha = 0$$

Solve for $$\alpha$$:
$$\alpha = \frac{11}{51} \approx 0.2157$$

Thus the degree of dissociation of the original solution is approximately $$0.216$$, which lies in the range $$0.20 - 0.22$$.

Final answer: 0.20 - 0.22