Question

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.

Answer 1

# gold atoms per cm3

assume Volumes are additive so

Davg = Dgold*Xgold + Dsilver*XSilver

Davg = (19.32*0.1) + (10.49*0.9) = 11.373 g/cm3

in 1 cm3; then

11.373 grams of alloy are present

MW of alloy = xsilver*MWsilver + xgold*MWgold = 0.9*107.8682 +0.1*196.96657 = 116.778037 g /mol

find moles of alloy

mol = mass/MW = 11.373 /116.778037 = 0.09738 mol of alloy

atoms of alloy in 1cm3 = 0.09738*6.022*10^23 = 5.8642236*10^22 atoms of alloy

since only 10% are gold then

0.1*(5.8642236*10^22) = 5.86422*10^21 atoms of gold in those cm3