Vanadium forms a substitutional solid solution with iron. Compute the number of vanadium atoms per cubic centimeter for a vanadium-iron alloy that contains 10 wt% V and 90 wt% Fe. The density of pure vanadium and iron are 6.10 and 7.30 \( g/cm^3 \), respectively.
wt% V = 10 wt% \( \quad \rho _{\nu} \) = 6.10 \( g/cm^3 \)
wt% Fe = 90 wt% \( \quad \rho _{\nu} \) = 7.30 \( g/cm^3 \)
\( atoms / cm^3 \)
\(A_{V} = \) 50.942 \( g/mol \)
\(A_{Fe} = \) 55.845 \( g/mol \)
\(N_{A} = 6.022x10^{23} atom/mol \)
total mass = 100g
atoms V = \( {N_{A} atoms} \over {1 mol} \) \( {1 mol} \over A_{V} \) 10g = 10g \(6.022 x 10^{23} atoms \over 1 mol \) \(1 mol \over 50.942 g \)
volume of alloy = \( 10g \over 6.10 g/cm^3 \) + \( 90g \over 7.30 g/cm^3 \)
8.45 x \(10^{21} \) \( atoms \over {cm^3} \)