Question

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

Answer #1

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

Volume of unit cell = a^{3} (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/ N_{A} (Here M = molar mass and
N_{A} = Avogadro's number)

Density = z.M / a^{3}. N_{A}

For ccp structure, z = 4

Given Density = 19.27 g/cm^{3} and edge length, a = 408
pm

Volume of unit cell = a^{3} = (408pm)^{3} = (408
x 10^{-12}m) = (408 x 10^{-10} cm)^{3} =
6.79 x 10^{-23} cm^{3}

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

M = d.N_{A}.a^{3} / z = 19.27 x 6.022 x
10^{23} 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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