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

Explanation

To find the molar mass of a metal with a face-centered cubic (FCC) lattice, given the edge length and density, we can follow these steps:

### Step 1: Understand the relationship between density, molar mass, and unit cell parameters.
The formula for density (d) in terms of the molar mass (M), the number of atoms per unit cell (Z), Avogadro's number (N_A), and the volume of the unit cell (V) is given by:
\[ d = \frac{Z \cdot M}{N_A \cdot V} \]

### Step 2: Calculate the volume of the unit cell.
For a cubic unit cell, the volume (V) can be calculated using the edge length (a):
\[ V = a^3 \]
Given that the edge length \( a = 404 \) pm, we need to convert this into centimeters:
\[ a = 404 \, \text{pm} = 404 \times 10^{-10} \, \text{cm} \]
Now, calculate the volume:
\[ V = (404 \times 10^{-10} \, \text{cm})^3 = 6.58 \times 10^{-23} \, \text{cm}^3 \]

### Step 3: Substitute the known values into the density formula.
We know:
- Density \( d = 2.72 \, \text{g/cm}^3 \)
- Number of atoms per unit cell for FCC \( Z = 4 \)
- Avogadro's number \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \)

Rearranging the density formula to solve for molar mass (M):
\[ M = \frac{d \cdot N_A \cdot V}{Z} \]

### Step 4: Plug in the values and calculate the molar mass.
Substituting the known values:
\[ M = \frac{2.72 \, \text{g/cm}^3 \cdot 6.02 \times 10^{23} \, \text{mol}^{-1} \cdot 6.58 \times 10^{-23} \, \text{cm}^3}{4} \]

Calculating the numerator:
\[ = 2.72 \cdot 6.02 \cdot 6.58 \]
\[ = 2.72 \cdot 39.03 \approx 106.25 \]

Now divide by 4:
\[ M = \frac{106.25}{4} \approx 26.56 \, \text{g/mol} \]

### Step 5: Round to the nearest whole number.
The molar mass is approximately \( 27 \, \text{g/mol} \).

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

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