Question

Aluminum–lithium (Al–Li) alloys have been developed by the aircraft industry to reduce the weight and improve the performance of its aircraft. A commercial aircraft skin material having a density of 2.14 g/cm3 is desired. Compute the concentration of Li (in wt%) that is required. The densities of aluminum and lithium are 2.71 and 0.534 g/cm3, respectively.

Answer 1

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  $2.14 g/cm^3$

Volume of an alloy = V

$V=\frac{100 g}{2.14 g/cm^3}=46.73 cm^3$

Volume of aluminum = $v_1$

Mass of aluminum = x

Density of aluminum = $d_1=2.71 g/cm^3$

$v_1=\frac{x}{d_1}$

Volume of lithium = $v_2$

Mass of lithium = y

Density of lithium= $d_2=0.534 g/cm^3$

$v_2=\frac{y}{d_2}$

$x+y=100$ ..[1]

$v_1+v_2=V$

$\frac{x}{d_1}+\frac{y}{d_2}=46.73 cm^3$

$\frac{x}{2.71}+\frac{y}{0.534}=46.73$..[2]

Solving [1] and [2] :

we get :

x = 93.46 g

y = 6.54 g

Concentration of Li (in wt%) that is required:

$Li\%=\frac{6.54 g}{100 g}\times 100=6.54\%$