Silver has a cubic unit cell with a cell edge of 408 pm. Its density is 10.6 g `cm^(-3)`. How many atoms of silver are there in the unit cell? What is the structure of silver?

Explanation

To solve the problem, we need to determine the number of atoms of silver in its cubic unit cell and identify its structure. Here’s a step-by-step solution:

### Step 1: Understand the given data
- **Edge length (a)** = 408 pm = 408 × 10^(-10) cm (since 1 pm = 10^(-12) m, we convert it to cm)
- **Density (d)** = 10.6 g/cm³
- **Atomic mass of silver (m)** = 108 g/mol
- **Avogadro's number (Na)** = 6.022 × 10²³ atoms/mol

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