1.

Sodium crystallizes in the cubic lattice and the edge of the unit cell is 430 pm. Calculate the number of atoms in a unit cell. (Atomic mass of Na = 23.0 density = 0.9623g cm^(-3), N_A = 6.023 xx 10^(23) mol^(-1)).

Answer»

Solution :Length of edge = 430 pm = `430 xx 10^(-12) m=430 xx 10^(-10)` cm
Volume of unit CELL = `a^3 =(430 xx 10^(-10))^3 cm^3 = 79.5 xx 10^(-24) cm^3`
Mass of one atom= `("Atomic mass")/("Avogadro.s No.") = (23)/(6.023xx10^(23))=3.82 xx 10^(-23)` g
Mass of unit cell = `"Volume" xx "density" = 79.5 xx 10^(-24) xx 0.9623 = 76.5 xx 10^(-24)` g
No. of atoms in unit cell = `("Mass of unit cell")/("Mass of an atom") =(76.5 xx 10^(-24))/(3.82 xx 10^(-23)) = 2`


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