answer:Isopropyl alcohol marketed in drug stores is generally in solution with H20. Typical alcohol percentages range form 70 to 91%. Pure water is a very poor electrical conductor, but the addition of NaCl (salt) makes the water much more more conductive. 0.5% would be enough salt to have a measurable impact on electrical conductivity of water but would have no impact on the conductivity of pure C3H8O (isopropanol), since salt is imiscible in isoporpanol and pure isopropanol is non-conductive. Another curious fact is that the addition of a sufficient quantity of any metal salt including sodium chloride to an aqueous solution of isopropanol causes the alcohol to become insoluble in water, thus producing an effect called “layering out” as a separate layer if sufficient salt is added. So the fact that your lab result was as described in the OP tells me that you probably had 91% 2-proponol in aqueous solution and that your instruments were not sensitive enough to detect the conductance of a solution containing just 9% distilled water, but were able to detect conductance once NaCl was added. In water, salt breaks into free Na + and Cl – ions, which readily carry an electrical charge. Hope this helps explain the results.