Question

0.05 cm thick coating of silver is deposited on a plate of 0.05 m² area. The number of silver atoms deposited on plate are _____ × 10²³. (At mass Ag = 108,d = 7.9 g/cm³)

Answer 1

Correct Answer: 11

Step 1. Calculate the volume of silver coating:
\(Volume  = 0.05 \times 0.05 \times 10000 = 25 \, \text{cm}^3\)

Step 2. Calculate the mass of silver deposited:
\(Mass\ of silver = 25 \times 7.9 \, \text{g}\)

Step 3. Calculate moles of silver atoms:
\(\text{Moles of silver } = \frac{25 \times 7.9}{108}\)

Step 4. Calculate the number of atoms:
\(\text{Number of  atoms} = \frac{25 \times 7.9}{108} \times 6.023 \times 10^{23} = 11.01 \times 10^{23}\)
 

Answer 2

The problem asks for the total number of silver atoms deposited in a thin coating on a plate. We are given the thickness of the coating, the area of the plate, the atomic mass of silver, and the density of silver.

Concept Used:

To find the total number of atoms, we need to follow a sequence of calculations:

1. Calculate the Volume of the coating: The volume of the thin coating is the product of its area and thickness.
2. Calculate the Mass of the coating: The mass is the product of the volume and the given density.
3. Calculate the Number of Moles: The number of moles is the mass divided by the molar mass (atomic mass in g/mol).
4. Calculate the Number of Atoms: The number of atoms is the number of moles multiplied by Avogadro's number (\(N_A \approx 6.022 \times 10^{23}\) atoms/mol).

Step-by-Step Solution:

Step 1: List the given values and convert them to a consistent unit system.

• Thickness, \(t = 0.05 \, \text{cm}\)
• Area, \(A = 0.05 \, \text{m}^2\). We convert this to cm². Since \(1 \, \text{m} = 100 \, \text{cm}\), \(1 \, \text{m}^2 = (100 \, \text{cm})^2 = 10000 \, \text{cm}^2\). \[ A = 0.05 \times 10000 \, \text{cm}^2 = 500 \, \text{cm}^2 \]
• Density, \(d = 7.9 \, \text{g/cm}^3\)
• Atomic mass of Silver (Ag), \(M = 108 \, \text{g/mol}\)

Step 2: Calculate the volume of the silver coating.

\[ \text{Volume} = A \times t = 500 \, \text{cm}^2 \times 0.05 \, \text{cm} = 25 \, \text{cm}^3 \]

Step 3: Calculate the mass of the deposited silver.

\[ \text{Mass} = \text{Volume} \times d = 25 \, \text{cm}^3 \times 7.9 \, \text{g/cm}^3 = 197.5 \, \text{g} \]

Step 4: Calculate the number of moles of silver.

\[ \text{Moles} = \frac{\text{Mass}}{M} = \frac{197.5 \, \text{g}}{108 \, \text{g/mol}} \approx 1.8287 \, \text{mol} \]

Final Computation & Result:

Step 5: Calculate the total number of silver atoms.

\[ \text{Number of Atoms} = \text{Moles} \times N_A \] \[ \text{Number of Atoms} = 1.8287 \times 6.022 \times 10^{23} \] \[ \text{Number of Atoms} \approx 11.011 \times 10^{23} \]

The problem asks for the answer in the format \( \_\_\_\_ \times 10^{23} \). The value for the blank is approximately 11.011.

The number of silver atoms deposited on the plate is approximately 11.01 \( \times 10^{23} \).