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. `26"g mol"^(-1)`
D. `20"g mol"^(-1)`

Answer

C

Explanation

`M=(rhoxxa^(3)xxN_(0)xx10^(-30))/Z`
Edge length (a)`=404 "pm" =404`
No. of atoms per unit cell (z)=4
Density of the element `(rho)=2.72 g " cm"^(-3)`
Avogadro's No (No) `=6.022xx10^(23) " mol"^(-1)`
`M=((2.72 g " cm"^(-3))xx(404)^(3)xx(6.022xx10^(23) " mol"^(-1))xx10^(-30))/4`
`=27 g " mol"^(-1)`