Answer
$N_{Mo}= 1.73 \times 10^{22} atoms/cm^{3}$
Work Step by Step
Given:
$c_{1} = c_{Mo} = 16.4 wt%$
$A_{1} = A_{Mo} = 95.94g/mol$
$p_{1} = p_{Mo} = 10.22 g/cm^{3}$
$p_{2} = p_{W} = 19.30 g/cm^{3}$
Required:
number of Nb atoms per cubic centimeter for niobium-vanadium alloy
Solution:
Using Equation 4.18:
$N_{Mo} = \frac{N_{A}c_{Mo}}{(\frac{c_{Mo}A_{Mo}}{p_{Mo}})+(\frac{A_{Mo}}{p_{W}}(100-c_{Mo})} = \frac{(6.022 \times 10^{23} atoms/mol)(16.4 wt)}{\frac{(16.4 wt)(95.94 g/mol)}{10.22 g/cm^{3}} + \frac{95.94 g/mol}{19.30 g/cm^{3}}(100- 16.4 wt)}= 1.73 \times 10^{22} atoms/cm^{3}$
