A metal has a fcc lattice.The edge length of the unit cell is `404` pm ,the density of the metal is `2.72g cm^(-3)` . The molar mass of the metal is `(N_(A)`, Avorgadro's constant `=6.02xx10^(23)mol^(-1))`

Options

A. `20 g "mol"^(-1)`
B. `40 g "mol"^(-1)`
C. `30 g "mol"^(-1)`
D. `28 "g mol"^(-1)`

Answer

D

Explanation

`rho = (Z xx M)/(N_(A) xx a^(3))` for fcc, Z = 4
a = 404 pm = `404 xx 10^(-10) cm`
`2.72 = (4 xx M)/(6.02 xx 10^(23) xx (404 xx 10^(-10))^(3))`
`M = 27 "g mol"^(-1)`