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Question: The atomic fraction (d) of tin in bronze (fcc) with a density of \(7717 kg{m}^{-3}\) and a lattice p...

The atomic fraction (d) of tin in bronze (fcc) with a density of 7717kgm37717 kg{m}^{-3} and a lattice parameter of 3.903A03.903 \overset {0}{A} is:
(Atomic weight of Cu = 63.54 amu and Sn = 118.7 amu, where 1amu=1.66×1027kg1 amu = 1.66 \times {10}^{-27}kg)
A. 0.01
B. 0.05
C. 0.10
D. 3.80

Explanation

Solution

The atomic fraction can be defined as the ratio of the atoms of one kind to the atoms of the other kind. It can be used to find the composition of a certain element in a mixture or an alloy with respect to the other element.

Complete step by step answer: To solve this question, we first need to see how we can find out the density of the fcc lattice. The density is given by the formula:
ρ=Z×MN0×a3\rho = \cfrac { Z \times M }{ { N }_{ 0 }\times { a }^{ 3 } } ------(1)
where ρ\rho is the density of the lattice, Z is the number of atoms per unit cell, N0{N}_{0} is the Avogadro's number and a is the lattice parameter. The equation (1) can also be written as:
M=ρ×N0×a3ZM = \cfrac { { \rho \times N }_{ 0 } \times { a }^{ 3 } }{ Z } ------(2)
Now, it is given that ρ=7717kg/m3=7.717g/cm3\rho = 7717 kg/{m}^{3} = 7.717 g/{cm}^{3}, Z=4, N0=6.023×1023{N}_{0} = 6.023 \times {10}^{23} and a=3.903×108ma = 3.903 \times {10}^{-8}m. Substituting the values in eq. (2), we get
M=7.717×6.023×1023×(3.903×108)34M = \cfrac { { 7.717 \times }6.023 \times { 10 }^{ 23 } \times { (3.903 \times { 10 }^{ -8 } })^{ 3 } }{ 4 }
    M=69.19g/mol\implies M = 69.19g/mol
Therefore, the molar mass of Bronze, M = 69.19 g/mol.
Let us assume that the atomic fraction of Sn is dd, then the atomic fraction of Cu becomes 1d1-d.
Now, the molar mass of bronze is given as the summation of the product of atomic mass of each component and the atomic fraction of the respective component.
So, M=(AtomicmassofSn×AtomicfractionofSn)+(AtomicmassofCu×AtomicfractionofCu)M=(Atomic mass of Sn \times Atomic fraction of Sn) + (Atomic mass of Cu \times Atomic fraction of Cu)------(3)
It is given that the atomic mass of Sn = 118.7 amu and the atomic mass of Cu = 63.54 amu. Substituting these values if eq. (3), we get
69.19=(118.7×d)+(63.54×(1d))69.19 = (118.7 \times d) + (63.54 \times (1-d))
    69.19=63.5463.54d+118.7d\implies 69.19 = 63.54 - 63.54d + 118.7d
    d=5.6555.16=0.10\implies d = \cfrac {5.65}{55.16} = 0.10

Therefore, the atomic fraction is 0.10. Hence, the correct answer is option (C).

Note: There is a difference between the atomic fraction and the mole fraction. Atomic fraction is the ratio of the atoms of one kind to the atoms of the other kind. Whereas, the mole fraction is the ratio of the number of moles of a particular compound to the total number of moles.