The atomic fraction `(d)` of tin in bronze (fcc) with a density of `7717 kg m^(-3)` and a lattice parameter of `3.903 Å` is `(Aw Cu = 63.54, Sn = 118.7, 1 amu = 1.66 xx 10^(-27 kg))`

Options

A. `0.01`
B. `0.05`
C. `0.10`
D. `3.8`

Explanation

To find the atomic fraction of tin (Sn) in bronze (an alloy of copper (Cu) and tin) given the density, lattice parameter, and atomic weights, we can follow these steps:

### Step 1: Understand the relationship between density, mass, and volume.
The density (ρ) of a crystal can be expressed as:
\[ \rho = \frac{\text{Mass of atoms in unit cell}}{\text{Volume of unit cell}} \]

### Step 2: Calculate the volume of the unit cell.
Given the lattice parameter \( a = 3.903 \, \text{Å} = 3.903 \times 10^{-10} \, \text{m} \), the volume \( V \) of the cubic unit cell is:
\[ V = a^3 = (3.903 \times 10^{-10})^3 \, \text{m}^3 \]

Calculating this gives:
\[ V \approx 5.95 \times 10^{-29} \, \text{m}^3 \]

### Step 3: Set up the equation for mass of atoms in the unit cell.
Let \( n_{\text{Cu}} \) be the number of copper atoms and \( n_{\text{Sn}} \) be the number of tin atoms in the unit cell. The mass of the atoms can be expressed as:
\[ \text{Mass} = n_{\text{Cu}} \times A_{\text{Cu}} \times \text{amu to kg conversion} + n_{\text{Sn}} \times A_{\text{Sn}} \times \text{amu to kg conversion} \]
Where:
- \( A_{\text{Cu}} = 63.54 \, \text{g/mol} = 63.54 \times 10^{-3} \, \text{kg/mol} \)
- \( A_{\text{Sn}} = 118.7 \, \text{g/mol} = 118.7 \times 10^{-3} \, \text{kg/mol} \)
- \( 1 \, \text{mol} = 6.022 \times 10^{23} \, \text{atoms} \)

The mass of one atom in kg is:
\[ \text{Mass of Cu atom} = \frac{63.54 \times 10^{-3}}{6.022 \times 10^{23}} \approx 1.055 \times 10^{-25} \, \text{kg} \]
\[ \text{Mass of Sn atom} = \frac{118.7 \times 10^{-3}}{6.022 \times 10^{23}} \approx 1.973 \times 10^{-25} \, \text{kg} \]

### Step 4: Write the density equation.
Substituting the mass into the density equation:
\[ \rho = \frac{n_{\text{Cu}} \cdot 1.055 \times 10^{-25} + n_{\text{Sn}} \cdot 1.973 \times 10^{-25}}{V} \]
Given that \( \rho = 7717 \, \text{kg/m}^3 \), we have:
\[ 7717 = \frac{n_{\text{Cu}} \cdot 1.055 \times 10^{-25} + n_{\text{Sn}} \cdot 1.973 \times 10^{-25}}{5.95 \times 10^{-29}} \]

### Step 5: Solve for \( n_{\text{Cu}} \) and \( n_{\text{Sn}} \).
Rearranging gives:
\[ n_{\text{Cu}} \cdot 1.055 \times 10^{-25} + n_{\text{Sn}} \cdot 1.973 \times 10^{-25} = 7717 \cdot 5.95 \times 10^{-29} \]
Calculating the right side:
\[ 7717 \cdot 5.95 \times 10^{-29} \approx 4.58 \times 10^{-25} \]

Now, we can express:
\[ 1.055 n_{\text{Cu}} + 1.973 n_{\text{Sn}} = 4.58 \]

### Step 6: Assume a value for \( n_{\text{Cu}} \).
Assuming \( n_{\text{Cu}} = 4 \) (as a common assumption for bronze), we can substitute:
\[ 1.055 \cdot 4 + 1.973 n_{\text{Sn}} = 4.58 \]
Calculating gives:
\[ 4.22 + 1.973 n_{\text{Sn}} = 4.58 \]
\[ 1.973 n_{\text{Sn}} = 0.36 \]
\[ n_{\text{Sn}} \approx 0.182 \]

### Step 7: Calculate the atomic fraction of tin.
The atomic fraction \( d \) of tin in bronze is given by:
\[ d = \frac{n_{\text{Sn}}}{n_{\text{Cu}} + n_{\text{Sn}}} \]
Substituting the values:
\[ d = \frac{0.182}{4 + 0.182} \approx \frac{0.182}{4.182} \approx 0.0436 \approx 0.05 \]

### Final Answer
The atomic fraction of tin in bronze is approximately \( 0.05 \).

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