1.

Calculate the density of silver which crystallizes in face- centred cubic form. The distance between nearest metal atoms is 287 pm (Molar mass of Ag = 107.87 g mol^(-1) , N_A = 6.022 xx 10^(23) mol^(-1)).

Answer»

Solution :Nearest neighbour distance in an FCC crystal = `1/2 xx "FACE diagonal" =a/sqrt2` or `287 xx 10^(-12) m = a/sqrt2`
`a=2.87 xz 1.414 xx 10^(-10)=4.058 xx 10^(-10)m`
= `405.8 xx 10^(-10)` cm
No. of atoms PER unit cell, Z=4
Atomic mass of silver, M=107.87 g `mol^(-1)`
Avogadro.s number, `N_A = 6.022 xx 10^(23) mol^(-1)`
Density of unit cell, `d = (Z xx M)/(N_A xx a^3 xx 10^(-30)) g cm^(-3)`
`:. d=(4 xx (107.87 g mol^(-1)))/(6.022 xx 10^(23) mol xx (405.8)^3 xx 10^(-30) cm^3)`
`= 10.72 g cm^(-3)`


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