Amixture is 20.00 mole% methyl alcohol, 60.0 mole% methyl acetate, and 20.0 mole% acetic acid.
what is the mass of a sample containing 45.0 kmol of methyl acetate?

4714.950 kilograms  is the mass of a sample containing 45.0 kmol of methyl acetate.

Explanation:

Moles of methyl acetate ==45.0 kmol= 45000 mol

Mole percentage of methyl acetate = 60.0%

Total moles in the sample = n

Mole percentage of methyl alcohol = 20.0%

Moles of methyl alcohol = n_2

Mass of methyl alcohol =  

=

Mole percentage of acetic acid  = 20.0%

Moles of acetic acid = n_3

Mass of acetic acid=

Mass of methyl methyl acetate=

Mass of sample: 480,600 g + 3,333,600 g + 900,750 g = 4714950 g

4714950 g = 4714.950 kg

(1 kg = 1000 g)

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

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