To find the dissociation constant of acetic acid, we will use the relationship between molar conductivity,$\\$degree of dissociation, and the dissociation constant.$\\$Given:$\\ \Lambda_m = 20 \, \mathrm{S \, cm^2 \, mol^{-1}} \\ c = 0.007 \, \mathrm{M} \\ \Lambda_{{H}^+}^0 = 350 \, \mathrm{S \, cm^2 \, mol^{-1}} \\ \Lambda_{{CH}_3{COO}^-}^0 = 50 \, \mathrm{S \, cm^2 \, mol^{-1}} \\$First, calculate the limiting molar conductivity $\Lambda^0: \\ \Lambda^0 = \Lambda_{{H}^+}^0 + \Lambda_{{CH}_3{COO}^-}^0 \\ \Lambda^0 = 350 + 50 \\ \Lambda^0 = 400 \, \mathrm{S \, cm^2 \, mol^{-1}} \\$The degree of dissociation $\alpha$ is given by:$\\ \alpha = \frac{\Lambda_m}{\Lambda^0} \\ \alpha = \frac{20}{400} \\ \alpha = 0.05 \\$The dissociation constant $K_a$ is given by:$\\ K_a = c \alpha^2 \\ K_a = 0.007 \times (0.05)^2 \\ K_a = 0.007 \times 0.0025 \\ K_a = 0.0000175 \, \mathrm{mol \, L^{-1}} \\$Therefore, the dissociation constant of acetic acid is $1.75 \times 10^{-5} \, \mathrm{mol \, L^{-1}}. \\$This corresponds to Option 3.