You made a 100.0 mL of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was
only 80.0 mL left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take 2.00 mL of the solution and add an excess of a concentration sodium chloride solution. You obtain a solid with a mass of 3.407 g. What was the concentration of the original lead(II) nitrate solution?
To solve this problem, we can simply calculate for the
dose by multiplying the volume of solution containing Selenium 75 and the
activity of the Selenium 75. That is: