It has been found that when sodium benzoate and ascorbic acid are present in a b

Question

It has been found that when sodium benzoate and ascorbic acid are present in a beverage a reaction can occur that produces benzene which is a reported carcinogen. WHO has set a limit for benzene in drinking water of 10.0ng/ml (ppb). Several beverages were monitored for benzene by a GC-MS method. The sample preparation involved passing 100.00ml of the beverage through a C-18SEP. The benzene was eluted from the SPE using 2ml of dichloromethane and analyzed. The following data was obtained: Calculate the concentration of benzene in the beverages. Do any exceed the allowed level. Assume a density of 1 for all samples.

Explanation / Answer

to get this, plot first a standard curve, and then, calculate concentrations of the beverage. With the standard plot a curve of C vs A. The data I obtained is the following:

y = 5.4082 + 17.9704x
r2 = 0.999969

Now, with this calibration curve, let's calculate concentration of the beverage:
beverage A = 7948 - 5.4082/17.9704 = 441.98 ng/mL
beverage B = 2001 - 5.4082/17.9704 = 111.05 ng/mL
beverage C = 1973 - 5.4082/17.9704 = 109.49 ng/mL
beverage D = 5329 - 5.4082/17.9704 = 296.24 ng/mL

Now to calculate the concentration of benzene, remember that we need to calculate a new concentration cause this is the preparation that were diluted so:
C = Co * (2mL/100mL)

Ca = 441.98 * 2/100 = 8.839 ng/mL
Cb = 111.05 * 2/100 = 4.442 ng/mL
Cc = 109.45 * 2/100 = 2.189 ng/mL
Cd = 296.24 * 2/100 = 5.925 ng/mL

Hope this helps

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