Niobium forms a substitutional solid solution with vanadium. Compute the weight

Question

Niobium forms a substitutional solid solution with vanadium. Compute the weight percent of niobium that must be added to vanadium to yield an alloy that contains 1.55 times 10%22 Nb atoms per cubic centimeter. The densities of pure Nb and V are 8.57 and 6.10 g/cm^3, respectively. Gold forms a substitutional solid solution with silver. Compute the number of gold atoms per cubic centimeter for a silver-gold alloy that contains 10 wt% Au and 90 wt% Ag. The densities of pure gold and silver are 19.32 and 10.49 g/cm^3, respectively. Nitrogen from a gaseous phase is to be diffused into pure iron at 700 degree C. If the surface concentration is maintained at 0.1 wt% N, what will be the concentration 1 mm from the-surface after 10 h? The diffusion coefficient for nitrogen in iron at 700 degree C 2.5 times 10^11 m^2/s.

Explanation / Answer

1. Nb = 1.55 x 10^22 atoms/cm^3

atomic mass Nb = 92.906 g/mol

Using densities,

the weight percent Nb to be added to vanadium would be be (C),

C = 100/[1+ (6.023 x 10^23 x 6.10/92.906 x 1.55 x 10^22) - (6.10/8.57)]

    = 35.23%

2. For gold

atomic mass Au = 196.97 g/mol

we get,

number of gold atoms per cm^3 (N)

N = 6.023 x 10^23 x 10/[(10 x 196.97/19.32) + (196.97 x 90/10.49)]

    = 3.36 x 10^21 atoms/cm^3

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