Question

A substitutional solid solution is formed between silver and gold. Calculate the weight percent of gold that should be added to silver to obtain an alloy with 5.5 x 1021 gold atoms per cm3 . Pure gold and silver have densities of 19.32 and 10.49 g/cm3 , respectively

Answer 1

no'of atoms in one mole=6.022×10²³

molar mass of gold and silver is = 196.96657 and 107.8682

This is how you solve it:

First off, look at a cm^3 of Silver. It contains 10.49 g/mol of silver. Take that number times it by Avogadro's constant and divide it by it's atomic weight.

[(10.49 g/mol)(6.023x10^23 atm/mol)] / (107.8682 g/mol) = 5.857x10^22 atm/cm^3 = 58.57x10^21 atm/cm^3

Now, let's looks at Gold. We need total mass of gold in the alloy. So, there is 5.5x10^21 atm/cm^3 of Gold. Take the atm/cm^3 times it's atomic weight and divide it by Avogadro's constant.

[(5.5x10^21 atm/cm^3)(196.96657 g/mol)] / (6.023x10^23 atm/mol) = 1.7986 g/cm^3

Then,
[(58.57x10^21) - (5.5x10^21)](107.8682 g/mol) / (6.023x10^23 atm/mol) = 9.505 g/cm^3 of Gold

Now we're trying to find the weight percent of Gold from the alloy.
From equation :

( 1.7986 g/cm^3) / [( 1.7986 g/cm^3) + (9.505 g/cm^3)] x100% = 15.91 wt%