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An element of atomic mass 90 occurs in fce structure with cell edge of 500 pm. Calculate the Avogadro's number if the density is 4.2g cm^(-3)

Answer»


Solution :M=90, Z=4 for FCC structure, a = 500 pm, d=4.20g `cm^(-3)`
`N_A=(Z xx M)/(d xx a^3)=(4 xx 90)/(4.2 xx (500)^3 10^(-30))=6.86 xx 10^(23) mol^(-1)`


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