In an experiment, two unknown compounds (one an alcohol and the other an ether) of equal molecular mass were dissolved in water. the result of the experiment is shown in the table.

solubility comparison
unknown compound solubility (g/100 ml water)
a 7.7
b 6.8

which of the following correctly explains the identity of compound b and its solubility?
it is an alcohol because the lack of hydrogen bonding between the oh− groups make it less soluble.
it is an ether because it is unable to form a hydrogen bond, so it is less soluble than the alcohol.
it is an ether because in the absence of a highly electronegative atom like o, f, or n, it lacks hydrogen bonding and is thus less soluble.
it is an alcohol because dispersion forces between the oh− group and organic chain makes it less soluble.

The right answer is B. (B is ether)

The solubility in water of alcohols depends on the same two factors as previously, but which are here antagonistic:

* The carbon chain, hydrophobic, tends to make the molecule insoluble.

* The hydrophilic hydroxyl group (thanks to its hydrogen bonds) tends to render the molecule soluble.

Thus, the alcohols are all the more soluble in water that the number of alcohol functions is high. For example, butanediols are soluble in all proportions while butan-1-ol has a solubility of 77 g L-1.

idk

explanation:

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when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.