A sample of crushed ore (0.97142 g) is acid digested and diluted to a volume of

Question

A sample of crushed ore (0.97142 g) is acid digested and diluted to a volume of 50.00 mL. A 2.00 mL aliquot of this first solution is placed in a 100.00 mL volumetric flask and diluted to the mark with buffer containing a complexing agent selective for thorium. Then 100 microliters of this second solution are placed in a 12-mL plastic test tube along with 9.00 mL of ultrapure water and 0.90 mL ultrapure electronic grade concentrated HNO3. The final solution in the test tube is introduced into an electrochemical analyzer and the thorium concentration was 1.33 x 107 M. What is the percent thorium in the original ore sample?

Explanation / Answer

We have 1.33 x 10^-7 M thorium in 10 mL solution

moles of Thorium = molarity x volume = 1.33 x 10^-7 x 0.01 = 1.33 x 10^-9 moles

So thorium concentration in 100 microlitre = 1.33 x 10^-9/0.0001 = 1.33 x 10^-5 M

moles of Thorium in 100 mL = 1.33 x 10^-5/0.100 = 1.33 x 10^-4 moles

So, 0.0665 M thorium is present in 50 mL solution

moles of Thorium = 0.0665 x 0.05 = 3.325 x 10^-3 moles

mass of thorium = 3.325 x 10^-3 x 232.04 = 0.7715 g

percentage of thorium in the sample ore = 0.7715/0.97142 x 100 = 79.42%

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