Question

# An element crystallizes in a cubic close pack structure. The edge of a unit cell is...

An element crystallizes in a cubic close pack structure. The edge of a unit cell is 408 pm and the density of the element is 19.27 g/cm^3. From the atomic mass of the element, the element is :

(a) Ag (b) Au (c) W (d) P

Density of the unit cell = mass of unit cell / volume of unit cell

Volume of unit cell = a3 (where a = edge length)

Mass of unit cell = z x m (z = no. of atom in unit cell and m = mass of one atom)

Also, m = M/ NA   (Here M = molar mass and NA = Avogadro's number)

Density = z.M / a3. NA

For ccp structure, z = 4

Given Density = 19.27 g/cm3 and edge length, a = 408 pm

Volume of unit cell = a3 = (408pm)3 = (408 x 10-12m) = (408 x 10-10 cm)3 = 6.79 x 10-23 cm3

Put all the values in above formula and calculate unknown value of M.

M = d.NA.a3 / z = 19.27 x 6.022 x 1023 x 6.79 x 10-23 / 4 = 196.26 g /mol

Atomic mass of Au = 196.9u

Therefore the element is Au

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