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

Answer

C

Explanation

Density , `rho=(ZxxM)/(a^3xxN_Axx10^(-30))`
`therefore 2.72=(4xxM)/((404)^3xx(6.02xx10^23)xx10^(-30))` or `M=(2.72xx(404)^3xx6.02xx10^23xx10^(-30))/4`
`=27 "g mol"^(-1)`