Question

to test a benzene content in a real soil sample, 10.0 g of sample was weighed into a 40 mL of vial and then 20.0 of methanol was added to it. The vial was then shaken for 30 minutes and the methanol extract was separated and analyzed on a GC-FID. The GC-FID was pre-calibrated with benzene standards and the calibration outcome data is shown in the following table. The sample extract was analized two times and the benzene peak areas are recorded in the table under calibration data. if the soil sample contains 20% of moisture

Benzene conc (mg/L)	o	5	10	20	50	75
Peak Area	0	1439	2889	6164	16156	23866
Peak area for sample extract= 5050, duplicate analysis = 5110

1. Plot the standard calibration curve (y-concentration in mg/L, x-peak area) and obtain the liner regression equation (y=ax+b). Attach your linear regression analysis results, give the equation and correlation coefficient R)

2. Report the mean concentration of benzene in soil sample in mg/kg (dry weight basis)

Answer 1

1) The calibration curve is obtained by directly plotting the data given in the question

The standard calibration curve with y axis as concentration in mg/L and x axis as peak area is shown below

You can see from the graph that the equation of the calibration curve is y=0.0031x + 0.503 and the correlation cefficient R for this linear regression analysis is 99.97%.

2)

From the data of the unknown samples we can see that peak areas obtained are 5050 and 5110

We can calculate the amount of benzene from these samples using the equation of the straight line obtained above.

y=0.0031x + 0.503 (substituting for x in the above equation from the values obtained)

y = (0.0031 x 5050) + 0.503

y = 16.158 mg/L

y = (0.0031 x 5110) + 0.503

y = 16.344 mg/L

for calculation we will use an average of these two values 16.252 mg/L, This is the mean of the two samples

This is the concentration of benzene in the sample obtained from

10.0 g of sample was weighed into a 40 mL of vial and then 20.0 of methanol was added to it. Since the sample as prepared here was directly inject from the procedure used above (as explained in the experimental; The vial was then shaken for 30 minutes and the methanol extract was separated and analyzed on a GC-FID.) the area obtained is in mg in the extract and not mg/L because the sample.

so 20 ml of the methanol extract has 16.252 mg of benzene which is the benzene in 10 g of the soil sample. SInce the soil sample has 20% moisture on a dry basis the soil sample is 10g - 20%of 10g = 10g - 2 g = 8g dry basis soil.

16.252 mg benzene is present in 8 g of the soil.

1 Kg soil will have 1000g/8g x 16.252 = 2031.37 mg/Kg mean concentration