Silver crystalilises in a fcc lattice. The edge length of its unit is `4.077xx10^(-8)xxcm` and its density is `10.5g cm^(-3)`. Claculate on this basis the atomic mass of silver `(N_(A)=6.02xx10^(23) "mol"^(-1))`

Explanation

To calculate the atomic mass of silver based on the given data, we can use the formula for density in relation to the unit cell of a crystal lattice. The steps are as follows:

### Step 1: Understand the formula for density
The density (d) of a crystal can be expressed using the formula:
\[ d = \frac{Z \cdot M}{A^3 \cdot N_A} \]
Where:
- \( d \) = density of the substance (g/cm³)
- \( Z \) = number of atoms per unit cell (for FCC, \( Z = 4 \))
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