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

`V=a^(3)`
`V=(404xx10^(-10))^(3)`
`=6.6xx10^(-23)`
`p=(ZxxM_(w))/(N_(A)xxV)`
`2.72=(4xxM_(w))/(6.02xx10^(23)xx6.6xx10^(-23))`
`M_(w)=(2.72xx6.023xx6.6)/(4)`
`=27g//mol.`