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.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

thats a little if u want the answer just look it up on google thats what i do haha : )