Question

Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold alloy that contains 10 wt% Au and 90 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm3, respectively. Weight of Au is 196.97 g/mol.

Answer 1

See answer below

Step-by-step explanation:

In order to do this, we need to use the following expression:

N'Au = C'Au * NAu (1)

Where:

N'Au: number of gold atoms per cubic centimeter.

C'Au = Atomic percentage of the Gold in the gold silver alloy.

NAu = Number of gold atoms

However, we do not have C'Au and neither NAu. We need to calculate those first.

To get the percentage of the gold we use the following expression:

C'Au = (CAu * MMAg) / (CAu * MMAg + CAg * MMAu) (2)

Where:

CAu and CAg: Weight percentage of the gold and the silve

MMAu and MMag: Atomic weights of the gold and the silver

We already have that data, so, replacing all the given data into (2) we have:

C'Au = (0.10 * 107.87) / (0.10 * 107.87 + 0.90 * 196.97)

C'Au = 10.787 / 188.06

C'Au = 0.0574 (or 5.74%)

We have C'Au, now we need to find the number of atoms of gold in the gold silver alloy. In order to do that, we use the following expression:

NAu = N * ρ / MMAu (3)

N is the avogadro's number which is 6.02x10²³ atoms. So replacing this value, the density and molecular weight of gold, the atoms are:

NAu = 6.02x10²³ * 19.32 / 196.97

NAu = 5.9x10²² atoms/cm³

Now that we have all the required data from (1), we just need to replace the obtained values:

N'Au = 5.9x10²² * 0.0574

N'Au = 3.39x10²¹ atoms/cm³

Hope this helps