For an alloy that consists of 352 wt Pb and 648 wt Sn what i

For an alloy that consists of 35.2 wt% Pb and 64.8 wt% Sn, what is the composition (a) of Pb (in at%), and (b) of Sn (in at%)? The atomic weights for Pb and Sn are 207.2 and 118.7 g/mol, respectively.

Solution

let the mass of alloy be m
therefore amount of Pb = m1 = 0.352 m
and amount of Sn =m2 = 0.648 m

number of moles of Pb (n1)= m1/207.2 =0.352/207.2= 1.698*10^(-3)*m moles
number of atoms of Pb (N1)= N_A * n1 = N_A * 1.698*10^(-3) atoms

number of moles of Sn (n2)= m2/118.7=0.648/118.7 = 5.45*10^(-3)*m moles
number of atoms of Sn (N2)= N_A * n2 = N_A * 5.45*10^(-3) atoms

where N_A = avagadro number

a) atom percent of Pb = N1*100/(N1+N2) = 23.75 %

b) atom percent of Sn = N2*100/(N1+N2) = 76.24 %