Question

.- Niobium forms a substitutional solid solution with vanadium. Compute the number of niobium atoms per cubic centimeter for a niobium-vanadium alloy that contains 24 wt% Nb and 76 wt% V. The densities of pure niobium and vanadium are 8.57 and 6.10 g/cm3, respectively. slader

Answer 1

Answer:

Number of atoms per per cubic centimeter for a niobium-vanadium alloy:

$1.109\times 10^{22} atoms/cm^3$

Explanation:

Suppose in 100 grams of solid sample has 24% of Nb and 76% of V.

Mass of Nb = 24 g

Volume of the Nb = v

Density of Nb = $d=8.57 g/cm^3$

$v=\frac{24 g}{8.57 g/cm^3}=2.80 cm^3$

Mass of V = 76 g

Volume of the V = v'

Density of V = $d'=6.10g/cm^3$

$v'=\frac{76 g}{6.10 g/cm^3}=12.46 cm^3$

Volume of the solid alloy$V_a$ = v + v' = $2.80 cm^3+12.46cm^3=15.26 cm^3$

Moles of Nb = $\frac{24 g}{93 g/mol}=0.2580 mol$

Number of atoms of Nb in 0.2580 mol = N

$N=n\times N_A$

$N=0.2580 mol\times 6.022\times 10^{23}=1.556\times 10^{23} atoms$

Number of atoms per per cubic centimeter for a niobium-vanadium alloy:

$=\frac{N}{V_a}=\frac{1.556\times 10^{23} atoms}{15.26 cm^3}$

$=1.109\times 10^{22} atoms/cm^3$

Answer 2

The number of Nb atoms per cubic cm of alloy has been $\rm \bold{1.109\;\times\;10^2^2}$.

The alloy has been assumed to be of 100 grams. The mass of vanadium and niobium has been:

The mass of niobium = 24 g

The mass of vanadium = 76 g

The density has been defined as mass per unit volume. The volume of vanadium and niobium has been given as:

$D=\dfrac{m}{v}$

The volume of niobium has been:

$8.57=\dfrac{24}{v_N_b} \\v_N_b=\dfrac{24}{8.57}\\v_N_b=2.80\;\rm cm^3$

The volume of niobium in the alloy has been $\rm \bold{2.80\;cm^3}$.

The volume of vanadium has been:

$6.10=\dfrac{76}{v_V} \\v_V=\dfrac{76}{6.10}\\v_V=12.46\;\rm cm^3$

The volume of vanadium in the alloy has been $\rm \bold{12.46\;cm^3}$.

The total volume of the alloy has been the sum of the volume of alloy constituents.

$V=2.80\;+\;12.46\;\rm cm^3\\\textit V=15.26\;cm^3$

The total volume of alloy has been $\rm \bold{15.26\;cm^3}$.

The moles of Nb in 24 grams has been given by:

$\rm Moles=\dfrac{mass}{molar\;mass}$

Substituting the values for Nb:

$\rm Moles_N_b=\dfrac{24}{93}\\Moles_N_b=0.2580\;mol$

The alloy has been constituent of 0.2580 mol Nb.

According to the Avogadro law, the number of atoms in a mole of sample has been equivalent to the Avogadro number.

The number of atoms of Nb has been:

$\rm 1\;mol=6.023\;\times\;10^2^3\;atoms\\0.258\;mol=6.023\;\times\;10^2^3\;\times\;0.258\;atoms\\0.258\;mol=1.556\;\times\;10^2^3\;atoms$

The number of atoms of Nb in alloy has been $\rm \bold{1.556\;\times\;10^2^3}$ atoms.

The number of atoms of Nb per cubic cm of alloy has been:

$\rm atoms/cm^3=\dfrac{1.556\;\times\;10^2^3}{15.26} \\atoms/cm^3=1.109\;\times\;10^2^2$

The number of Nb atoms per cubic cm of alloy has been $\rm \bold{1.109\;\times\;10^2^2}$.

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