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

Step-by-step explanation:

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

Mass of titanium = 90 g

Moles of titanium =
`(90 g)/(47.87 g/mol)=1.8800 mol`

Total number of atoms of titanium ,
`a_t=1.8800 mol* N_A`

Mass of aluminum = 6 g

Moles of aluminium =
`(6 g)/(26.98 g/mol)=0.2223 mol`

Total number of atoms of aluminium,
`a_a=0.2223 mol* N_A`

Mass of vanadium = 4 g

Moles of vanadium=
`(4 g)/(50.94 g/mol)=0.0785 mol`

Total number of atoms of vanadium
`a_v=0.0785 mol* 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}}* 100`

Atomic percentage of titanium:

:
`(a_t)/(a_t+a_a+a_v)* 100=(1.8800 mol* N_A)/(1.8800 mol* N_A+0.2223 mol* N_A+0.0785 mol* N_A)* 100=86.20\%`

Atomic percentage of Aluminium:

:
`(a_a)/(a_t+a_a+a_v)* 100=(0.2223 mol* N_A)/(1.8800 mol* N_A+0.2223 mol* N_A+0.0785 mol* N_A)* 100=10.19\%`

Atomic percentage of vanadium

:
`(a_v)/(a_t+a_a+a_v)* 100=(0.0785 mol* N_A)/(1.8800 mol* N_A+0.2223 mol* N_A+0.0785 mol* N_A)* 100=3.59\%`