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Chromium crystallizes in a body centred cubic lattice, whose density is 7.20 g//cm^3. The length of the edge of unit cell is 288.4 pm. Calculate Avogadro's number. (Atomic mass of chromium=52)

Answer»


SOLUTION :d= 7.20 g `cm^(-3)`, a=288.4 PM, Z2 for bcc unit cell, M = 52 g `MOL^(-1)`, d = `(Z XX M)/(N_A xx a^3)` or `N_A=(Z xx M)/(d xx a^3)=(2xx52)/(7.20 xx (288.4)^3 xx 10^(-30)) =6.02 xx 10^(23) mol^(-1)`


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