Respuesta :
Explanation:
Suppose in 100 g of alloy contains 90% titanium 6% aluminum and 4% vanadium.
Mass of titanium = 90 g
Moles of titanium = [tex]\frac{90 g}{47.87 g/mol}=1.8800 mol[/tex]
Total number of atoms of titanium ,[tex]a_t=1.8800 mol\times N_A[/tex]
Mass of aluminum = 6 g
Moles of aluminium = [tex]\frac{6 g}{26.98 g/mol}=0.2223 mol[/tex]
Total number of atoms of aluminium,[tex]a_a=0.2223 mol\times N_A[/tex]
Mass of vanadium = 4 g
Moles of vanadium= [tex]\frac{4 g}{50.94 g/mol}=0.0785 mol[/tex]
Total number of atoms of vanadium[tex]a_v=0.0785 mol\times N_A[/tex]
Total number of atoms in an alloy = [tex]a_t+a_a+a_v[/tex]
Atomic percentage:
[tex]Atomic\%=\frac{\text{total atoms of element}}{\text{Total atoms in alloy}}\times 100[/tex]
Atomic percentage of titanium:
:[tex]\frac{a_t}{a_t+a_a+a_v}\times 100=\frac{1.8800 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=86.20\%[/tex]
Atomic percentage of Aluminium:
:[tex]\frac{a_a}{a_t+a_a+a_v}\times 100=\frac{0.2223 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=10.19\%[/tex]
Atomic percentage of vanadium
:[tex]\frac{a_v}{a_t+a_a+a_v}\times 100=\frac{0.0785 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=3.59\%[/tex]
