A prospector wants to know how much gold is in the ore that he has been mining. A sample weighing 11.2456 grams is dissolved in Aqua Regia (a mixture of HCl and HNO3). That sample is then transferred to a 200.00 mL volumetric flask and diluted to the mark with water. This is called solution 1. A volumetric pipet is used to transfer 10.00 mL of solution 1 to a second 500.00 mL volumetric flask. The second flask is filled to the mark resulting in solution 2. 5.00 mL of solution 2 is analyzed and found to contain 13.45 ppm Au. Answer each of the questions below. 1. What is the concentration of Au in solution 1. The units should be ppm 2. How many micrograms of Au are in the 5.00 mL sample that was measured? 3. How many micrograms of gold were in the 500.00 mL flask containing solution 2. 4. How many grams Au were in the flask containing solution 1 5. What is the %Au in the original sample?
States of Matter
The substance that constitutes everything in the universe is known as matter. Matter comprises atoms which in turn are composed of electrons, protons, and neutrons. Different atoms combine together to give rise to molecules that act as a foundation for all kinds of substances. There are five states of matter based on their energies of attraction, namely solid, liquid, gases, plasma, and BEC (Bose-Einstein condensates).
Chemical Reactions and Equations
When a chemical species is transformed into another chemical species it is said to have undergone a chemical reaction. It consists of breaking existing bonds and forming new bonds by changing the position of electrons. These reactions are best explained using a chemical equation.
A prospector wants to know how much gold is in the ore that he has been mining. A sample weighing 11.2456 grams is dissolved in Aqua Regia (a mixture of HCl and HNO3). That sample is then transferred to a 200.00 mL volumetric flask and diluted to the mark with water. This is called solution 1. A volumetric pipet is used to transfer 10.00 mL of solution 1 to a second 500.00 mL volumetric flask. The second flask is filled to the mark resulting in solution 2. 5.00 mL of solution 2 is analyzed and found to contain 13.45 ppm Au. Answer each of the questions below.
1. What is the concentration of Au in solution 1. The units should be ppm
2. How many micrograms of Au are in the 5.00 mL sample that was measured?
3. How many micrograms of gold were in the 500.00 mL flask containing solution 2.
4. How many grams Au were in the flask containing solution 1
5. What is the %Au in the original sample?
Mass of the gold sample taken for analysis = 11.2456 g
11.2456 g is diluted to 200.00 mL to form solution-1.
10 mL of solution-1 is diluted to 500.00 mL to form solution-2
Required conversion factors:
1 ppm = 1 mg/L
1 mg = 1000 g
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