1.

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))

Answer»

30 g `mol^(-1)`
27 g `mol^(-1)`
20 g `mol^(-1)`
40 g `mol^(-1)`

Solution :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)`


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