Question

You have been asked to measure the volume of a small lake. You dump in 5.0 L of a 2.0 M solution of a dye, which degrades with a half-life of 3.0 days. You wait exactly one week for the lake to become well mixed (during this time, assume no water is lost); you then take a 100 mL sample. The dye's concentration in this sample is 2.9*10-6 M. What is the lake's volume?

Answer 1

Solution :-

Lets first calculate the rate constant using the half life

K= 0.693 t1/2

K= 0.693 / 3.0 day

K= 0.231 day-1

1 week = 7 days

Now lets calculate the initial concentration before 7 days

ln([A]t/[A]0) = - K * t

ln(2.9*10^-6/ [A]o) = -0.231 day-1 * 7 days

ln(2.9*10^-6/ [A]o) = -1.617

(2.9*10^-6/ [A]o) = anti ln [-1.617]

(2.9*10^-6/ [A]o) = 0.1984933

[A]o = 2.9*10^-6 / 0.1984933

[A]o = 1.461*10^-5 M

Now lets calculate the volume of the lake

Initial volume of the dye = 5.0 L initial molarity = 2.0 M

final concentration after the dilution = 1.461*10^-5 M

now lets calculate the volume of the lake

M1V1 = M2V2

V2 = M1V1/M2

V2 = 2.0 M * 5.0 L / 1.461*10^-5 M

V2 = 6.84*10^5 L

So the volume of the lake water is 6.84*10^5 L