A metal has an fcc latticed.The edge length of the unit cell is `404` pm .The density of the metal is `2.72 g//cm^(-3)` .The molar mass of the metal is
`(N_(A)` Avogadro's constant `= 6.2 xx 10^(23) mol^(-1))`

Options

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

Answer

To solve the problem, we need to find the molar mass of a metal with a face-centered cubic (FCC) lattice, given the edge length and density. Let's go through the steps systematically.

### Step 1: Understand the FCC structure
In a face-centered cubic (FCC) lattice, there are atoms located at each corner of the cube and at the centers of each face. The effective number of atoms (Z) per unit cell in an FCC structure is calculated as follows:
- There are 8 corners, each contributing \( \frac{1}{8} \) of an atom.
- There are 6 faces, each contributing \( \frac{1}{2} \) of an atom.

Thus, the total number of atoms per unit cell (Z) is:
\[ Z = \left(8 \times \frac{1}{8}\right) + \left(6 \times \frac{1}{2}\right) = 1 + 3 = 4 \]

### Step 2: Convert the edge length to centimeters
The edge length of the unit cell is given as 404 picometers (pm). We need to convert this to centimeters (cm):
\[ \text{Edge length} = 404 \text{ pm} = 404 \times 10^{-12} \text{ m} = 404 \times 10^{-10} \text{ cm} \]

### Step 3: Calculate the volume of the unit cell
The volume (V) of the cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
where \( a \) is the edge length in cm. Thus,
\[ V = (404 \times 10^{-10} \text{ cm})^3 = 6.58 \times 10^{-29} \text{ cm}^3 \]

### Step 4: Use the density formula to find molar mass
The density (d) of the metal is given as 2.72 g/cm³. The relationship between density, molar mass (M), and the number of atoms per unit cell (Z) is given by:
\[ d = \frac{Z \cdot M}{V \cdot N_A} \]
Rearranging this formula to solve for molar mass (M):
\[ M = \frac{d \cdot V \cdot N_A}{Z} \]

### Step 5: Substitute the known values
Now, we can substitute the values into the equation:
- Density \( d = 2.72 \text{ g/cm}^3 \)
- Volume \( V = 6.58 \times 10^{-29} \text{ cm}^3 \)
- Avogadro's number \( N_A = 6.2 \times 10^{23} \text{ mol}^{-1} \)
- Number of atoms per unit cell \( Z = 4 \)

Substituting these values:
\[ M = \frac{2.72 \times 6.58 \times 10^{-29} \times 6.2 \times 10^{23}}{4} \]

### Step 6: Calculate the molar mass
Calculating the above expression:
\[ M = \frac{2.72 \times 6.58 \times 6.2}{4} \times 10^{-29 + 23} \]
\[ M = \frac{2.72 \times 6.58 \times 6.2}{4} \times 10^{-6} \]
Calculating the numerical part:
\[ M = \frac{106.73}{4} \times 10^{-6} \approx 26.68 \text{ g/mol} \]

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

### Final Answer
The correct option is **27 g/mol**.