In this exam question, it comes to electrical conductivity of a weak electrolyte. Formic acid is a weak acid with a ks value of 1.7 times 10 to the negative 4th mole per liter and dissociates in water into protons and formate ions incompletely. One mole of formic acid is diluted to a water to a volume of 1 liter, determine the degree of dissociation alpha and the electric conductivity. According to Oswald's dilution law, alpha squared O, O, 1 minus alpha times C equals ks. Substituting the numerical values and solvents that equi-tretic equation for alpha, we obtain a degree of dissociation of 0.013, that is 1.3%. By adding the limiting conductivity of protons and formate ions, we obtain the limiting equivalent conductivity lambda of formic acid to 404.4 Z-mance centimeter squared per mole. Because of the incompleteness of formic acid, the equivalent conductivity is reduced to only 1.3% of this value, that is 5.34, Z-mance centimeter squared per mole. From this value, which is also the molar equivalent conductivity, the conductivity caper is obtained by simply multiplying by concentration, concentration in moles per milliliters unit. We finally get 5.24 millis-e-mans per centimeter as a specific conductivity. Thank you for watching.