Answer:
6.54% is the required concentration of lithium in an alloy (Al–Li).
Explanation:
Suppose 100 grams of an alloy of aluminium and lithium.
Density of an alloy = d [tex]2.14 g/cm^3[/tex]
Volume of an alloy = V
[tex]V=\frac{100 g}{2.14 g/cm^3}=46.73 cm^3[/tex]
Volume of aluminum = [tex]v_1[/tex]
Mass of aluminum = x
Density of aluminum = [tex]d_1=2.71 g/cm^3[/tex]
[tex]v_1=\frac{x}{d_1}[/tex]
Volume of lithium = [tex]v_2[/tex]
Mass of lithium = y
Density of lithium= [tex]d_2=0.534 g/cm^3[/tex]
[tex]v_2=\frac{y}{d_2}[/tex]
[tex]x+y=100[/tex] ..[1]
[tex]v_1+v_2=V[/tex]
[tex]\frac{x}{d_1}+\frac{y}{d_2}=46.73 cm^3[/tex]
[tex]\frac{x}{2.71}+\frac{y}{0.534}=46.73[/tex]..[2]
Solving [1] and [2] :
we get :
x = 93.46 g
y = 6.54 g
Concentration of Li (in wt%) that is required:
[tex]Li\%=\frac{6.54 g}{100 g}\times 100=6.54\%[/tex]
