3. Suppose that you do an AA spectroscopic analysis for Ca 2+ in yogurt. 1.5 g o

Question

3. Suppose that you do an AA spectroscopic analysis for Ca2+ in yogurt. 1.5 g of yogurt is digested with 20.0 mL of HNO3 for ~ 10 minutes. The mixture is filtered into a 100 mL volumetric flask and diluted to the line with water to make Solution A. Then, two successive dilution are made:

10 mL of Solution A is added to another 100 mL volumetric flask and diluted with water to the mark to make Solution B.

10 mL of Solution B is added to another 100 mL volumetric flask and diluted with water to the mark to make Solution C.

When solution C is analyzed using AA spectroscopy, it is found to have a concentration of 0.215 ppm Ca2+.

a. What is the dilution factor?

b. What is the concentration of Ca2+ (ppm) in Solution A?

c. How many g of Ca are in the original 1.5 g of yogurt?

d. What is the %Ca (w/w) in the yogurt?

Explanation / Answer

Initially you had a solution which is 20mL.

20mL----> 100mL Dilution factor = 1/5

then 10mL ---> 100mL dilution factor = 10/100 = 1/10

and again 10mL ----> 100mL    dilution = 1/10

overall dilution = 1/5 *1/10 *1/10 = 1/500th

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concentration of Ca in solution B = 0.215ppm *100/ 10 = 2.15 ppm

concentration of Ca in solution A =2.15ppm *100mL/20mL = 10.75 ppm

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grams of Ca

10.75ppm = 10.75mg/L

20mL solution conatin = 10.75 *20/1000mL =2.15 mg = 0.00215g Ca

1.5g yogurt contains 0.00215g Ca

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%Ca = 0.00215 *100/1.5 =0.14 %

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