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). Y=12.235x
To find the concentration of the original undiluted sample, we can use the dilution factor formula.
Dilution factor = initial volume/final volume
Since the sample was diluted 4 times, the dilution factor is 4.
Now, we can calculate the concentration of the original undiluted sample using the formula provided:
Y = 12.235x
Where Y is the absorbance and x is the concentration (in mg/ml).
Given that the absorbance (Y) is 0.6, we can substitute this value into the equation:
0.6 = 12.235x
Now, we can solve for x (the concentration):
x = 0.6/12.235
x ≈ 0.049 mg/ml
However, this concentration is for the diluted sample. To find the concentration of the original undiluted sample, we need to multiply this value by the dilution factor:
Concentration of original undiluted sample = 0.049 mg/ml * 4
Concentration of original undiluted sample ≈ 0.196 mg/ml
Therefore, the concentration of the original undiluted sample is approximately 0.196 mg/ml.
To determine the concentration of the original undiluted sample (mg/ml), we can use the dilution factor and the absorbance measurement.
First, let's calculate the dilution factor. In this case, you diluted the sample four times. Each dilution reduced the concentration of the sample by a factor of 4.
Next, let's calculate the corrected absorbance of the undiluted sample. Since the absorbance was measured using the diluted sample, we need to correct it to account for the dilution.
The corrected absorbance (A_corrected) can be calculated using the formula:
A_corrected = A_measured × dilution factor
In this case, the measured absorbance (A_measured) is 0.6, and the dilution factor is 4.
A_corrected = 0.6 × 4 = 2.4
Now, let's use the equation Y = 12.235x to calculate the original concentration (x).
To solve for x, divide both sides of the equation by 12.235:
x = A_corrected / Y
Substituting the values:
x = 2.4 / 12.235 ≈ 0.196 mg/ml
Therefore, the concentration of the original undiluted sample is approximately 0.196 mg/ml.
To find the concentration of the original undiluted sample, we can use the dilution equation:
C1V1 = C2V2
Where:
C1 = concentration of the original undiluted sample (mg/ml)
V1 = volume of the original undiluted sample (ml)
C2 = concentration of the diluted sample (mg/ml)
V2 = volume of the diluted sample (ml)
In this case, we have:
C2 = (0.6 mg/ml) [Since it is given that the absorbance of the diluted sample is 0.6]
V2 = 100 ul = 0.1 ml [Since it is given that you measure the absorbance using 100 ul of the diluted sample]
We also know that the diluted sample was prepared by diluting the original sample 4 times. Therefore, the dilution factor is 4.
This means that:
V1 = 4V2 = 4(0.1 ml) = 0.4 ml
Let's substitute the values into the dilution equation and solve for C1:
C1(0.4 ml) = (0.6 mg/ml)(0.1 ml)
C1 = (0.6 mg/ml)(0.1 ml) / (0.4 ml)
C1 = 0.06 mg/ml
Therefore, the concentration of the original undiluted sample is 0.06 mg/ml.