Chemistry: The density of lead, which has the FCC structure, is 11.36 . The atomic weight of lead is 207.19 . Use Avogadro s number: 6.02210. Calculatethe lattice parameter(Enter your answer to three significant figures.) = 2.75*10^21 the atomic radius of lead(Enter your answer to three significant figures.) =

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The density of lead, which has the FCC structure,

860jay

01.02.2022

$a_o = 4.95 *10^{-8} \ cm$

$r = 1.7500 * 10^{-8} \ cm$

Explanation:

From the question we are told that

The density of lead is $\rho_l = 11.36 \ g/cm^3$

The atomic weight is $M_w = 207.19 \ g/mol$

The Avogadro's number is $N_a = 6.022 * 10^{23} \ \frac{atom}{mol}$

Generally for FCC the number of atom is n = 4

Generally the volume of a unit FCC cell is mathematically represented

$V = \frac{n * M_w }{N_a * \rho_l }$

$V = \frac{4 * 207.19 }{ 6.022 * 10^{23} * 11.36 }$

$V = 1.211 5 *10^{-22} \ cm^3$

Generally the lattice parameter is mathematically represented as

$a_o = \sqrt[3]{V }$

$a_o = \sqrt[3]{1.211 5 *10^{-22} }$

=> $a_o = 4.95 *10^{-8} \ cm$

Generally the radius of the lead is mathematically represented as

$r = \frac{a_o * \sqrt{2} }{4}$

=> $r = \frac{4.95 *10^{-8} * \sqrt{2} }{4}$

=> $r = 1.7500 * 10^{-8} \ cm$

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