Question

The atomic fraction (d) of tin in bronze (fcc) with a density of $7717$ kg m${}^{-3}$ and a lattice parameter of $3.903$ $\mathring{A}$ is :

(Atomic weight of Cu $=63.54$ amu and Sn $=118.7$ amu, where $1$ amu $=1.66\times {10}^{-27}$ kg)

• A. $0.01$
• B. $0.05$
• C. $0.10$
• D. $3.80$

Answer 1

The correct option is C $0.10$

The density of bronze is $7717$ kg m${}^{-3}$ or $7.717$ g cm${}^{-3}$.
The density is given by the expression: $\rho =\frac{Z\times M}{N_0\times a^3}$
$7.717=\frac{4\times M}{6.023\times {10}^{23}\times (3.903\times {10}^{-8}{)}^3}$
The molar mass of bronze is $M=69.19$ g/mol.
The atomic weight of Cu and Sn are $63.54$ g/mol and $118.7$ g/mol.
The atomic fraction of Sn is $d$ and the atomic fraction of Cu is $1-d$.
The molar mass of bronze is $69.19=63.54(1-d)+118.7d$
$69.19=63.54-63.54d+118.7d$
$d=\frac{5.65}{55.16}=0.10$
Hence, the atomic fraction of tin in bronze is $0.10$.