This question involves calculating the number of silver atoms on a given surface area. To solve this, we need to relate the macroscopic properties (density, atomic weight) to the microscopic property (number of atoms). The key is to find the number of atoms per unit area.

Step 1: Understand the Structure
Silver has a face-centered cubic (FCC) structure. In an FCC lattice, atoms are located at the corners and the centers of all the faces of the cube.

Step 2: Relate Density to Lattice Parameter
The density (ρ) of a crystal is given by: $\rho =\frac{z\times M}{N_A\times a^3}$ where:
z = number of atoms per unit cell (for FCC, z = 4)
M = molar mass = 108 g/mol
N_A = Avogadro's number = 6.022 × 10²³ mol^–1
a = lattice parameter (edge length of the unit cell, in cm)
ρ = density = 10.5 g/cm³

Step 3: Calculate the Lattice Parameter (a)
Rearranging the density formula to solve for a³: $a^3=\frac{z\times M}{N_A\times \rho }=\frac{4\times 108}{6.022\times {10}^{23}\times 10.5}{\text{cm}}^3$ We will calculate the numerical value later if needed.

Step 4: Find the Number of Atoms per Unit Area on a Surface
The problem asks for the number of atoms on a surface area of 10^–12 m². For an FCC crystal, the most densely packed plane is the (111) plane. However, a simpler and common approach for such problems is to consider the number of atoms per unit area on the surface of the crystal.

The surface area of one face of the unit cell is a². For an FCC structure, the number of atoms on a particular face depends on the plane. For the (100) plane, the number of atoms per face is: $\frac{\text{Atoms}}{\text{face}}=2$ (The 4 corner atoms contribute 4 × (1/4) = 1 atom, and the center atom contributes 1 atom, total 2 atoms).

Therefore, the number of atoms per unit area on the (100) surface is: $\frac{\text{Number of atoms}}{\text{Area}}=\frac{2}{a^2}$ Let's denote this as n (atoms/cm²).

Step 5: Relate a² to Known Quantities
From Step 3, we have a³. We need a². Notice that: $a^2=\frac{a^3}{a}$ This seems messy. A more efficient way is to find n directly from the density and molar mass.

The number of atoms per unit volume (N_v) is: $N_v=\frac{\rho \times N_A}{M}$ For a rough estimate of surface atoms, the number of atoms per unit area (n) can be approximated by (N_v)^2/3. This is because the number of atoms along one edge is proportional to (N_v)^1/3, and the number on a face is the square of that, (N_v)^2/3.

Therefore: $n\approx {\left(\frac{\rho \times N_A}{M}\right)}^{2/3}$ This is a standard approach for such problems.

Step 6: Calculate n and Then the Number for the Given Area
First, calculate N_v: $N_v=\frac{10.5\text{g/cm³}\times 6.022\times {10}^{23}{\text{mol}}^{-1}}{108\text{g/mol}}=\frac{6.3231\times {10}^{24}}{108}{\text{cm}}^{-3}\approx 5.8547\times {10}^{22}{\text{cm}}^{-3}$

Now, calculate n ≈ (N_v)^2/3: $n\approx {(5.8547\times {10}^{22})}^{2/3}$ First, find the cube root of N_v: ${\left(N_v\right)}^{1/3}\approx {(5.8547\times {10}^{22})}^{1/3}={\left(5.8547\right)}^{1/3}\times {\left({10}^{22}\right)}^{1/3}\approx 1.80\times {10}^{22/3}\approx 1.80\times {10}^{7.333}$ Since 10^7.333 ≈ 10⁷ × 10^0.333 ≈ 10⁷ × 2.15, the cube root is approximately 1.80 × 2.15 × 10⁷ ≈ 3.87 × 10⁷ cm^–1 (atoms along an edge per cm).

Now, n (atoms/cm²) is the square of this: $n\approx {(3.87\times {10}^7)}^2=14.9769\times {10}^{14}\approx 1.50\times {10}^{15}{\text{cm}}^{-2}$

Step 7: Convert Area and Find Total Atoms
Given area = 10^–12 m². We need to convert this to cm². 1 m² = 10⁴ cm², so: $\text{Area}={10}^{-12}{\text{m}}^2\times {10}^4{\text{cm}}^2/{\text{m}}^2={10}^{-8}{\text{cm}}^2$

Number of atoms on this surface = n × Area = $(1.50\times {10}^{15}{\text{cm}}^{-2})\times \left({10}^{-8}{\text{cm}}^2\right)=1.50\times {10}^7$

This is expressed as y × 10^x, where y = 1.50 and x = 7.

Final Answer: The value of x is $7$.

Related Topics & Formulae

Solid State Chemistry: This problem is rooted in understanding the structure of solids, particularly metallic crystals like silver which have an FCC lattice. Key concepts include the unit cell, coordination number, and packing efficiency.

Density of a Crystal: $\rho =\frac{z\times M}{N_A\times a^3}$ This formula connects the macroscopic property of density to the microscopic structure (atoms per cell, cell size).

Atoms on a Surface: The number of atoms per unit area on a crystal surface can be estimated from the number of atoms per unit volume using the relation n ≈ (N_v)^2/3, where N_v = (ρ × N_A)/M. This is a crucial concept in surface science and nanotechnology.