Imagine that your initial absorbable reading for “low salt” fraction was too high, so you diluted your sample 4 times. When you measure the absorbable using 100 ul of the diluted sample , the absorbance is 0.6. What is the concentration of the original undiluted sample (mg/ml)

Answer 1

To determine the concentration of the original undiluted sample, we can use the dilution formula:

C1V1 = C2V2

Where:
C1 = concentration of the original undiluted sample
V1 = volume of the original undiluted sample
C2 = concentration of the diluted sample
V2 = volume of the diluted sample

We know that the diluted sample was diluted 4 times, so C2 = C1/4 and V2 = 100 ul.

Plugging in the values:
C1 * V1 = (C1/4) * 100 ul

Simplifying the equation:
V1 = (C1/4) * 100 ul / C1

Simplifying further:
V1 = 25 ul

Now, we need to determine the concentration (C1) of the original undiluted sample. We know that the absorbance of the diluted sample (A2) is 0.6.

The relationship between absorbance, concentration, and path length is given by Beer's Law:

A = ε * c * l

Where:
A = absorbance
ε = molar absorptivity (a constant for a given substance)
c = concentration
l = path length (typically 1 cm)

Assuming that the path length is 1 cm, we can rearrange the equation to solve for concentration:

c = A / (ε * l)

Let's assume ε and l are both constants.

Therefore, c2 = A2 / c1

Multiplying both sides by c1:
c1 * c2 = A2

Substituting the values:
c1 * 0.6 = A2

Now, we can solve for c1 (the concentration of the original undiluted sample):
c1 = A2 / 0.6

Substituting the given absorbance value:
c1 = 0.6 / 0.6 = 1

Therefore, the concentration of the original undiluted sample is 1 mg/ml.