A metal has a foc lattice. The edge length of the unit cell is 404 pm. The density of the metal is `2.72 g cm^(-3)` . The molar mass of the metal is `(N_A " Avogadro's constant " = 6.02 xx 10^(23) mol^(-1))`

Options

A. `27 g mol^(-1)`
B. `20 g mol^(-1)`
C. `40 g mol^(-1)`
D. `30 g mol^(-1)`

Answer

A

Explanation

`d= (ZM)/(N_Aa^3) " " ( Z = 4` for fcc)
`M=(dxxN_A a^3)/Z = (2.72xx6.02 xx 10^(23) xx (404 xx 10^(-10))^3)/4`
`M= 26.99 ~~ 27 g mol^(-1)`