1.

An element ‘X’ (At mass = 40 g mol–1) having f.c.c. structore, has unit celledge length of 400 pm.Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’ (N_A = 6.022 × 10^23 mol^–1)

Answer»

SOLUTION :`d=(zM)/(a^3N_A)`
=`(4 times 40)/((4 times 10^-8) times 6.022 times 10^-23)`
`=4.15g//cm^3`
No of UNIT cells=total no of atoms 4
=`[4/40 times 6.022 times 10^23]//4`
`=1.5 times 10^22`


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