A 200-gram alloy of silver and gold consists of 60 wt% Ag and 40 wt% Au. What are the percentages of Ag and Au atoms in this alloy?
The weight percentage of silver in the alloy is 60%, which means that there are 0.6 * 200 g = 120 g of silver in the alloy.
The weight percentage of gold in the alloy is 40%, which means that there are 0.4 * 200 g = 80 g of gold in the alloy.
The number of atoms of silver in the alloy is equal to the mass of silver divided by the atomic mass of silver. The atomic mass of silver is 107.868 g/mol. So the number of atoms of silver is 120 g / 107.868 g/mol = 1.116 moles of silver.
The number of atoms of gold in the alloy is equal to the mass of gold divided by the atomic mass of gold. The atomic mass of gold is 196.966 g/mol. So the number of atoms of gold is 80 g / 196.966 g/mol = 0.406 moles of gold.
The percentage of silver atoms in the alloy is equal to the number of silver atoms divided by the total number of atoms and multiplied by 100%. The total number of atoms is 1.116 moles + 0.406 moles = 1.522 moles. So the percentage of silver atoms is 1.116 moles / 1.522 moles * 100% = 73.2%.
The percentage of gold atoms in the alloy is equal to 100% – 73.2% = 26.8%.
Therefore, the percentages of Ag and Au atoms in the alloy are 73.2% and 26.8%, respectively.
I hope this helps! Let me know if you have any other questions.
