Sample preparation for the AA/ICP experiment involves many steps. The procedure

Question

Sample preparation for the AA/ICP experiment involves many steps. The procedure is similar to the one described below: A 2.7630 g tablet was ground into a uniform powder. 0.5235 g of the powder was then digested in 15 mL of nitric acid and then diluted to 50.0 mL with water to make the stock unknown solution. A 5.00 mL aliquot of the stock was then diluted to 50mL, and this dilute solution was analyzed. Analysis indicated the dilute solution contained 41.8 ppm of calcium. What was the mass of calcium in the tablet (in mg)?

Explanation / Answer

Ca = 41.8 ppm = 41.8 mg/L

this is present in the 50 mL used:

C1*V1 = C2*V2

C1*5 = 41.8*50

C1 = 418 ppm

those 5 mL where from 50 mL which were diluted

therefore, we have:

50 mL of sample, C = 418 mg/L

mass of Ca = C*V = 50*10^-3*418 = 20.900 mg = 20.9 mg

those were present in

0.5235 g of sample

so:

0.5235 g = 20.9 mg

2.7630 g = x ?

x = 2.763/0.5265*20.9

x = 109.68 mg of Ca in the sample

so..

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