Question

The common titanium alloy known as T-64 has a composition of 90 weight% titanium 6 wt% aluminum and 4 wt% vanadium. Calculate the concentrations as atomic percents.

Answer 1

Explanation:

Suppose in 100 g of alloy contains 90% titanium 6% aluminum and 4% vanadium.

Mass of titanium = 90 g

Moles of titanium = $\frac{90 g}{47.87 g/mol}=1.8800 mol$

Total number of atoms of titanium ,$a_t=1.8800 mol\times N_A$

Mass of aluminum = 6 g

Moles of aluminium = $\frac{6 g}{26.98 g/mol}=0.2223 mol$

Total number of atoms of aluminium,$a_a=0.2223 mol\times N_A$

Mass of vanadium  = 4 g

Moles of vanadium= $\frac{4 g}{50.94 g/mol}=0.0785 mol$

Total number of atoms of vanadium$a_v=0.0785 mol\times N_A$

Total number of atoms in an alloy = $a_t+a_a+a_v$

Atomic percentage:

$Atomic\%=\frac{\text{total atoms of element}}{\text{Total atoms in alloy}}\times 100$

Atomic percentage of titanium:

:$\frac{a_t}{a_t+a_a+a_v}\times 100=\frac{1.8800 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=86.20\%$

Atomic percentage of Aluminium:

:$\frac{a_a}{a_t+a_a+a_v}\times 100=\frac{0.2223 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=10.19\%$

Atomic percentage of vanadium

:$\frac{a_v}{a_t+a_a+a_v}\times 100=\frac{0.0785 mol\times N_A}{1.8800 mol\times N_A+0.2223 mol\times N_A+0.0785 mol\times N_A}\times 100=3.59\%$