The length of the unit cell edge of a lattice metal is 350 pm .Thus volume of atoms in one mole of the metal is

Options

A. `9.21 cm^(3)`
B. `7.25 cm^(3)`
C. `1.21 cm^(3)`
D. `4.00 cm^(3)`

Answer

A

Explanation

In a bcc struture , the atoms lough each other along the body diagonaql.

Radius of the atom `= r`
`AB^(2) = BC^(2) + AC^(2)`
`BC^(2) = BD^(2) + CD^(2) = a^(2) + a^(2) = 2a^(2)`
`AB^(2) = 2a^(2) = 2a^(2) + a^(2) = 3a^(2)`
`AB = sqrt(3a)`

But `AB = 4r`
`4r = sqrt(2a) rArr r = sqrt(2a) //4`
`r = sqrt(3) xx (350)/(4)` pm
`= sqrt(3) xx (350)/(4) xx 10^(-12) m`
`= sqrt(3) xx (350)/(4) xx 10^(-19) cm`
`= 154.14 xx 10^(-19) cm`
Volume of one atoms `(4)/(2) pi r^(3) = (4)/(3) pi (154.14 xx 10^(-19))^(3) cm^(3)`
`= 1.53 xx 10^(-23) cm^(3)`
volume of `N_(0)` atoms is one mole of the metal
`= 6.02 xx 10^(23) xx 153 xx 10^(-23) cm^(3)`
`= 9.21 cm^(3)`