Question

A certain element crystallizes in a face-centered cubic lattice. The density of the crystal is 22.67 g·cm–3, and the edge of the unit cell is 383.3 pm. Calculate the atomic mass of the element. (a) 192 g·mol–1 (b) 40 g·mol–1 (c) 183 g·mol–1 (d) 70 g·mol–1 (e) None of the above

Answer #1

Density = 22.67 g cm^{-3}

edge of unit call = 383.3 pm

volume of unit cell = a^{3} = 56314010.5 pm^{3}
= 56314010.5 * 10^{-30} cm^{3} = 5.631 *
10^{-23} cm^{3}

=> Mass of a unit cell = density * volume of unit cell =
22.67 * 5.631 * 10^{-23} g = 127.65 * 10^{-23}
g

In face centered cubie, 4 atoms are pesent.

Thus, mass of an atom = (127.65 * 10^{-23})/4 = 31.91 *
10^{-23} g/atom

Atomic mass = mass of one atom * Avagadro's number = 31.91 *
10^{-23} g/atom * 6.023 * 10^{23} atoms/mol =
192.21 g/mol

**The atomic mass of the element: (a) 192
g·mol ^{–1}**

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