A metal has a fcc lattice.The edge length of the unit cell is `404` pm ,the density of the metal is `2.72g cm^(-3)` . The molar mass of the metal is `(N_(A)`, Avorgadro's constant `=6.02xx10^(23)mol^(-1))`

Options

A. `30 g mol^(-1)`
B. `27 g mol^(-1)`
C. `20 g mol^(-1)`
D. `40 g mol^(-1)`

Explanation

To find the molar mass of the metal with a face-centered cubic (FCC) lattice, we can use the relationship between density, molar mass, and the unit cell volume. Here are the steps to solve the problem:

### Step 1: Determine the number of atoms in the FCC unit cell (z)
In a face-centered cubic (FCC) lattice:
- There are 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- There are 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom to the unit cell.

Calculating the total number of atoms (z):
\[ z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 1 + 3 = 4 \]

### Step 2: Calculate the volume of the unit cell (V)
The edge length (a) of the unit cell is given as \( 404 \) pm. First, we need to convert this to centimeters:
\[ a = 404 \, \text{pm} = 404 \times 10^{-10} \, \text{cm} \]

Now, we calculate the volume of the unit cell:
\[ V = a^3 = (404 \times 10^{-10})^3 \, \text{cm}^3 \]

Calculating \( V \):
\[ V = 404^3 \times (10^{-10})^3 = 66.4 \times 10^{-30} \, \text{cm}^3 = 6.64 \times 10^{-29} \, \text{cm}^3 \]

### Step 3: Use the density formula to find the molar mass (M)
The density (d) of the metal is given as \( 2.72 \, \text{g/cm}^3 \). The formula relating density, molar mass, and unit cell volume is:
\[ d = \frac{z \cdot M}{V \cdot N_A} \]
Rearranging this to solve for M:
\[ M = \frac{d \cdot V \cdot N_A}{z} \]

Substituting the known values:
- \( d = 2.72 \, \text{g/cm}^3 \)
- \( V = 6.64 \times 10^{-29} \, \text{cm}^3 \)
- \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \)
- \( z = 4 \)

Calculating M:
\[ M = \frac{2.72 \cdot 6.64 \times 10^{-29} \cdot 6.02 \times 10^{23}}{4} \]

Calculating the numerator:
\[ 2.72 \cdot 6.64 \times 10^{-29} \cdot 6.02 \times 10^{23} = 1.08 \times 10^{-5} \, \text{g} \]

Now, dividing by 4:
\[ M = \frac{1.08 \times 10^{-5}}{4} = 2.7 \times 10^{-6} \, \text{g/mol} \]

### Final Calculation
To convert to grams per mole:
\[ M \approx 27 \, \text{g/mol} \]

### Conclusion
The molar mass of the metal is approximately \( 27 \, \text{g/mol} \).

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