The density of silver having atomic mass 107.8 gram `"mol"^(-1)` is 10.8 gram `cm^(-3)`. If the edge length of cubic unit cell is `4.05xx10^(-8)cm`, find the number of silver atoms in the unit cell. `(N_(A)=6.022xx10^(23),1A^(@)=10^(-8)cm)`

Explanation

To find the number of silver atoms in the unit cell, we can use the formula relating density, atomic mass, and the number of atoms in the unit cell. Here’s a step-by-step solution:

### Step 1: Understand the formula
The formula relating density (d), number of atoms in the unit cell (Z), atomic mass (M), edge length of the cubic unit cell (A), and Avogadro's number (N_A) is given by:

\[ d = \frac{Z \cdot M}{A^3 \cdot N_A} \]
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