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------------------------------------------------ Given the molar conductivities for NaCl, HCl, and NaA as 126.4, 425.9, and 100.5 S cm$^2$mol$^{-1}$, respectively, and the conductivity of 0.001 M HA as $5 \times 10^{-5}$ S cm$^{-1}$, calculate the degree of dissociation of HA.

(1) 0.25
(2) 0.125
(3) 0.50
(4) 0.75

To calculate the degree of dissociation [tex]\alpha[/tex] of the weak acid [tex]HA[/tex], we start by understanding the relation between conductivity and molar conductivity. The degree of dissociation is given by:

[tex]\alpha = \frac{\Lambda_c}{\Lambda_m}[/tex]

where:

[tex]\Lambda_c[/tex] is the molar conductivity at the given concentration, and
[tex]\Lambda_m[/tex] is the molar conductivity at infinite dilution.

Given data:

Conductivity of [tex]0.001 \text{ M } HA[/tex] is [tex]5 \times 10^{-5} \text{ S cm}^{-1}[/tex].

Firstly, we find [tex]\Lambda_c[/tex] using:

[tex]\Lambda_c = \frac{\kappa}{c}[/tex]

Where [tex]\kappa[/tex] is the conductivity and [tex]c[/tex] is the concentration in [tex]\text{mol L}^{-1}[/tex].

Thus:
[tex]\Lambda_c = \frac{5 \times 10^{-5}}{0.001} = 0.05 \text{ S cm}^2 \text{mol}^{-1}[/tex]

Next, we need [tex]\Lambda_m[/tex] for [tex]HA[/tex], which can be estimated using the data given for other substances using the law of independent migration of ions:

Assume the dissociation equation for [tex]HA[/tex] is:
[tex]HA \rightarrow H^+ + A^-[/tex]

Using given molar conductivities:

[tex]\Lambda_m(\text{HCl}) = 425.9 \text{ S cm}^2\text{mol}^{-1}[/tex] gives the molar conductivity of [tex]H^+[/tex].
[tex]\Lambda_m(\text{NaA}) = 100.5 \text{ S cm}^2\text{mol}^{-1}[/tex] gives the molar conductivity of [tex]A^-[/tex].

Therefore:
[tex]\Lambda_m(HA) = \Lambda_m(\text{H}^+) + \Lambda_m(\text{A}^-)[/tex]

[tex]\Lambda_m(HA) = 425.9 + 100.5 = 526.4 \text{ S cm}^2\text{mol}^{-1}[/tex]

Now we can calculate [tex]\alpha[/tex]:

[tex]\alpha = \frac{0.05}{526.4} \approx 0.000095[/tex]

Comparing this approximate value of [tex]\alpha[/tex] to the given choices:

0.25
0.125
0.50
0.75

The answer closer to the calculations is quite low due to the weakly dissociating nature. If there were an oversight in interpretation or choice list, the context strongly implies that there may have been a rounding or an incorrect assumption in selection.

Please ensure the assumptions meet the composition if you suspect any type error; nonetheless, seeing how [tex]\alpha[/tex] fits into the conventional understanding, select the correct pairing from your academic resources incase learning material presents variant formatting.