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The density of a pure substance 'A' whose atoms pack in cubic close pack arrangement is 1g m/cc . If B atoms can occupy tetrahedral voids and all the tetrahedral voids are occupied by 'B' atoms what is the density of resulting solid in gm/cc . [Atomic mass (A) = 30 gm/mol and atomic mass (B) = 50gm/mol]


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Let the volume of f.c.c unit cell = V
$\rho_A = \large\frac{4\times M_A}{N_A. a^3}$
$\rho_B = \large\frac{8\times M_B}{N_A . a^3}$
$\large\frac{\rho_A}{\rho_B} = \large\frac{M_A}{2M_B} $
$\;\;\; = \large\frac{30}{2\times 50}$
$\;\;\; = \large\frac{30}{100}$
$\;\;\; = 0.3 $
$\rho_B = \large\frac{\rho_A }{\rho_B} = 3.33 g/cc$
Hence answer is (A)
answered Apr 7, 2014 by sharmaaparna1

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