Answer
$N_{Nb} = 9.30 \times 10^{21} atoms/cm^{3}$
Work Step by Step
Given:
$c_{1} = c_{Nb} = 24 wt%$
$A_{1} = A_{Nb} = 92.91 g/mol$
$p_{1} = p_{Nb} = 8.57 g/cm^{3}$
$p_{2} = p_{V} = 6.10 g/cm^{3}$
Required:
number of Nb atoms per cubic centimeter for niobium-vanadium alloy.
Solution:
Using Equation 4.18:
$N_{Nb} = \frac{N_{A}c_{Nb}}{(\frac{c_{Nb}A_{Nb}}{p_{Nb}})+(\frac{A_{Nb}}{p_{V}}(100-c_{Nb})} = \frac{(6.022 \times 10^{23} atoms/mol)(24 wt)}{\frac{(24 wt)(92.91 g/mol)}{8.57 g/cm^{3}} + \frac{92.91 g/mol}{6.10 g/cm^{3}}(100- 15 wt)} \\= 9.30 \times 10^{21} atoms/cm^{3}$
