How much of a 25mg/ml chloramphenicol stock solution do we need to make 400ml of growth media that has final 25ug/ml concentration of chloramphenicol?

Answer : The volume of chloramphenicol stock solution needed is, 0.4 mL

Explanation :

First we have to calculate the mass of chloramphenicol.

As, the mass of chloramphenicol in 1 mL of solution =

So, the mass of chloramphenicol in 400 mL of solution =

Thus, the mass of chloramphenicol is, 0.01 g

Now we have to calculate the volume of chloramphenicol stock solution needed.

Concentration of chloramphenicol =

As, of chloramphenicol needed 1 mL volume of chloramphenicol stock solution

So, of chloramphenicol needed volume of chloramphenicol stock solution

Thus, the volume of chloramphenicol stock solution needed is, 0.4 mL

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