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, we will use the given data and apply the relevant formulas step by step.

### Step 1: Understand the given data
- Molar mass (M) = \(2.7 \times 10^{-2} \, \text{kg/mol}\)
- Edge length of the cubic unit cell (a) = \(405 \, \text{pm} = 405 \times 10^{-12} \, \text{m}\)
- Density (d) = \(2.7 \times 10^{-3} \, \text{kg/m}^3\)
- Avogadro's number (N_A) = \(6.022 \times 10^{23} \, \text{mol}^{-1}\)

### Step 2: Use the density formula
The density (d) of a substance in a cubic unit cell can be expressed as:
\[ d = \frac{z \cdot M}{N_A \cdot a^3} \]
where:
- \(z\) = number of atoms per unit cell
- \(M\) = molar mass
- \(N_A\) = Avogadro's number
- \(a\) = edge length of the unit cell

### Step 3: Rearranging the formula to find z
Rearranging the formula gives us:
\[ z = \frac{d \cdot N_A \cdot a^3}{M} \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ z = \frac{(2.7 \times 10^{-3}) \cdot (6.022 \times 10^{23}) \cdot (405 \times 10^{-12})^3}{2.7 \times 10^{-2}} \]

### Step 5: Calculate \(a^3\)
Calculating \(a^3\):
\[ a^3 = (405 \times 10^{-12})^3 = 6.64 \times 10^{-31} \, \text{m}^3 \]

### Step 6: Substitute \(a^3\) back into the equation
Now substituting \(a^3\) back into the equation for \(z\):
\[ z = \frac{(2.7 \times 10^{-3}) \cdot (6.022 \times 10^{23}) \cdot (6.64 \times 10^{-31})}{2.7 \times 10^{-2}} \]

### Step 7: Simplify the equation
Calculating the numerator:
\[ (2.7 \times 10^{-3}) \cdot (6.022 \times 10^{23}) \cdot (6.64 \times 10^{-31}) = 1.07 \times 10^{-7} \]
Now substituting this back into the equation for \(z\):
\[ z = \frac{1.07 \times 10^{-7}}{2.7 \times 10^{-2}} \approx 3.96 \approx 4 \]

### Step 8: Determine the relationship between radius (R) and edge length (a)
For a face-centered cubic (FCC) structure, the relationship between the radius (R) and the edge length (a) is given by:
\[ 4R = \sqrt{2}a \]
Thus:
\[ R = \frac{\sqrt{2}}{4} a \]

### Step 9: Substitute the edge length into the equation for R
Substituting \(a = 405 \times 10^{-12} \, \text{m}\):
\[ R = \frac{\sqrt{2}}{4} (405 \times 10^{-12}) \approx \frac{1.414}{4} \cdot 405 \times 10^{-12} \]
Calculating this gives:
\[ R \approx 0.1768 \times 10^{-9} \, \text{m} = 176.8 \times 10^{-12} \, \text{m} \]

### Step 10: Round to the nearest integer
Rounding \(176.8\) to the nearest integer gives:
\[ R \approx 177 \times 10^{-12} \, \text{m} \]

### Final Answer
The radius of the element is approximately \(177 \times 10^{-12} \, \text{m}\).