Question

consider the oil and vinegar lab with hexane and h2o forming 2 layers in the test tube. which of the following is miscible (soluble) in h2o? a) ch3ch2ch2ch2ch2ch2ch2ch3 b) ccl4 c) ch3oh d) ch3ch2ch2och2ch2ch3

Explanation:

• For option a: \(CH_{3}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}CH_{2}CH_{3}\) (octane) is a non - polar hydrocarbon. Water (\(H_{2}O\)) is polar. According to the "like dissolves like" principle, non - polar substances are not soluble in polar water.
• For option b: \(CCl_{4}\) is a non - polar molecule (symmetrical tetrahedral structure). It will not dissolve in polar water.
• For option c: \(CH_{3}OH\) (methanol) has a polar \(-OH\) group. The \(-OH\) group can form hydrogen bonds with water molecules. Hydrogen bonding is a strong intermolecular force that allows methanol to be miscible (soluble) in water.
• For option d: \(CH_{3}CH_{2}CH_{2}OCH_{2}CH_{2}CH_{3}\) (dipropyl ether) has a relatively non - polar hydrocarbon part and a less polar ether functional group compared to the \(-OH\) group in methanol. The non - polar part dominates, so it is not miscible in water.

Answer:

C. \(CH_{3}OH\)