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

Options

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

Answer

B

Explanation

Density is given by
`d=(zxxM)/(N_(A)a^(3))`, where z=number of formula units
present in unit cell, which is 4 for fcc
a=edge length of unit cell. M=molecular mass.
`2.72=(4xxM)/(6.02xx10^(23)xx(404xx10^(-10))^(3))`
`(because 1"pm"=10^(-10)cm)`
`M=(2.72**6.02**(404)^(3))/(4**10^(7))=26.99`
=27g `mol^(-1)`