Germanium forms a substitutional solid solution with silicon. Compute the weight

Question

Germanium forms a substitutional solid solution with silicon. Compute the weight percent of germanium that must be added to silicon to yield an alloy that contains 4.43 1021 Ge atoms per cubic centimeter. The densities of pure Ge and Si are 5.32 and 2.33 g/cm3, respectively. Also, the atomic weights for germanium and silicon are 72.64 and 28.09 g/mol, respectively.

wt% Germanium forms a substitutional solid solution with silicon. Compute the weight percent of germanium that must be added to silicon to yield an alloy that contains 4.43 times 1021 Ge atoms per cubic centimeter. The densities of pure Ge and Si are 5.32 and 2.33 g/cm3, respectively. Also, the atomic weights for germanium and silicon are 72.64 and 28.09 g/mol, respectively.

Explanation / Answer

First off, look at a cm^3 of Silicon. It contains 2.33 g/mol of silver. Take that number times it by Avogadro's constant and divide it by it's atomic weight.

[(2.33 g/mol)(6.203x10^23 atm/mol)] / (28.09 g/mol) = 5.0x10^22 atm/cm^3 = 50.0x10^21 atm/cm^3

Now, let's looks at Germanium. We need total mass of Ge in the alloy. So, there is 3.43x10^21 atm/cm^3 of Ge. Take the atm/cm^3 times it's atomic weight and divide it by Avogadro's constant.

[(3.43x10^21 atm/cm^3)(72.64 g/mol)] / (6.203x10^23 atm/mol) = 0.4137 g/cm^3

Then,
[(50.0x10^21) - (3.43x10^21)](28.09 g/mol) / (6.203x10^23 atm/mol) = 2.17 g/cm^3 of Ge

Now we're trying to find the weight percent of Ge from the alloy.
From equation 4.3:

(0.4137 g/cm^3) / [(0.4137 g/cm^3) + (2.17 g/cm^3)] x100% = 16.01 wt%


The solution showed equation 4.19:

C = 100 / [1 + [(Na)(Density a)/(Nnb)(Anb)] - (Density v /Density nv)]
You can try doing this way.

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