An element with molar mas `2.7xx10^(-2)" kg mol"^(-1)` forms a cubic unit cell with edge length 405 pm. If its density is `2.7xx10^(-3)kgm^(-3)`, the radius of the element is approximately ______ `xx 10^(-12) m` (to the nearest integer).

Explanation

To find the radius of the element given its molar mass, density, and edge length of the cubic unit cell, we can follow these steps:

### Step 1: Understand the relationship between density, molar mass, and the unit cell.
The formula for density (d) is given by:
\[ d = \frac{Z \cdot m}{a^3 \cdot N_A} \]
where:
- \( Z \) = number of atoms per unit cell
- \( m \) = molar mass of the element
- \( a \) = edge length of the cubic unit cell
- \( N_A \) = Avogadro's number (approximately \( 6.022 \times 10^{23} \, \text{mol}^{-1} \))

### Step 2: Rearrange the formula to solve for \( Z \).
Rearranging the density formula gives:
\[ Z = \frac{d \cdot a^3 \cdot N_A}{m} \]

### Step 3: Substitute the known values into the equation.
Given:
- \( d = 2.7 \times 10^{-3} \, \text{kg/m}^3 \)
- \( a = 405 \, \text{pm} = 405 \times 10^{-12} \, \text{m} \)
- \( m = 2.7 \times 10^{-2} \, \text{kg/mol} \)
- \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)

Now substituting these values:
\[ Z = \frac{(2.7 \times 10^{-3}) \cdot (405 \times 10^{-12})^3 \cdot (6.022 \times 10^{23})}{2.7 \times 10^{-2}} \]

### Step 4: Calculate \( a^3 \).
Calculating \( a^3 \):
\[ a^3 = (405 \times 10^{-12})^3 = 6.634 \times 10^{-28} \, \text{m}^3 \]

### Step 5: Substitute \( a^3 \) back into the equation for \( Z \).
Now substituting \( a^3 \):
\[ Z = \frac{(2.7 \times 10^{-3}) \cdot (6.634 \times 10^{-28}) \cdot (6.022 \times 10^{23})}{2.7 \times 10^{-2}} \]

### Step 6: Simplify and calculate \( Z \).
Perform the calculations:
\[ Z = \frac{(2.7 \times 10^{-3}) \cdot (6.634 \times 10^{-28}) \cdot (6.022 \times 10^{23})}{2.7 \times 10^{-2}} \approx 4 \]

### Step 7: Determine the type of cubic unit cell.
Since \( Z = 4 \), the element likely belongs to a Face-Centered Cubic (FCC) structure.

### Step 8: Relate edge length \( a \) to atomic radius \( R \).
For an FCC structure, the relationship between edge length \( a \) and atomic radius \( R \) is given by:
\[ a = 2\sqrt{2}R \]
Thus, we can rearrange this to find \( R \):
\[ R = \frac{a}{2\sqrt{2}} \]

### Step 9: Substitute the value of \( a \) to find \( R \).
Substituting \( a = 405 \times 10^{-12} \, \text{m} \):
\[ R = \frac{405 \times 10^{-12}}{2\sqrt{2}} \approx 143.17 \times 10^{-12} \, \text{m} \]

### Step 10: Round the radius to the nearest integer.
The radius \( R \) is approximately:
\[ R \approx 143 \times 10^{-12} \, \text{m} \]

### Final Answer:
The radius of the element is approximately \( 143 \times 10^{-12} \, \text{m} \).

---