An element with density 2.8 `cm^(3)` forms a f.c.c unit cell with edge length `4xx10^(-8)` cm. calculate the molar mass of the element. Given: `(N_(A)=6.022xx10^(23)) mol^(-1)`

Explanation

`d=2.8 g//cm^(3),z=4,a=4xx10^(-8)cm`
`d=(zxxm)/(vxxNA)`
`2.8=(4xxM)/(4xx10^(-8^(3))xx6.02xx10^(23))`
`2.8=(4xxM)/(4^(3)xx6.02xx10^(-1))`
M=26.97