JEE Chemistry 2024 Question with Solution

Given: thickness of silver coating = $0.05 \, \text{cm}$, area of plate = $0.05 \, \text{m}^2$, density of silver = $7.9 \, \text{g/cm}^3$, atomic mass of silver = $108 \, \text{g/mol}$.

Find: the value of the blank in _____ \times 10^{23} for the number of silver atoms deposited.

First convert area into $\text{cm}^2$.

$$A = 0.05 \times 10000 = 500 \, \text{cm}^2$$

Now calculate the volume of silver coating.

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

Mass of deposited silver:

$$\text{Mass} = 25 \times 7.9 = 197.5 \, \text{g}$$

Moles of silver:

$$\text{Moles of silver} = \frac{197.5}{108}$$

Number of atoms:

$$\text{Number of atoms} = \frac{197.5}{108} \times 6.023 \times 10^{23} \approx 11.01 \times 10^{23}$$

Therefore, the value of the blank is 11.

Given: a silver layer of thickness $0.05 \, \text{cm}$ is deposited over area $0.05 \, \text{m}^2$. Also, $d = 7.9 \, \text{g/cm}^3$ and molar mass of Ag is $108 \, \text{g/mol}$.

Find: the number multiplying $10^{23}$ in the total atom count.

Use the sequence: area conversion $\rightarrow$ volume $\rightarrow$ mass $\rightarrow$ moles $\rightarrow$ atoms.

Convert area:

$$1 \, \text{m}^2 = 10000 \, \text{cm}^2$$ $$0.05 \, \text{m}^2 = 0.05 \times 10000 = 500 \, \text{cm}^2$$

Then volume of coating is:

$$V = A \times t = 500 \times 0.05 = 25 \, \text{cm}^3$$

Mass of silver deposited:

$$m = Vd = 25 \times 7.9 = 197.5 \, \text{g}$$

Moles of silver:

$$n = \frac{m}{M} = \frac{197.5}{108} \approx 1.8287$$

Atoms of silver:

$$N = nN_A = 1.8287 \times 6.022 \times 10^{23} \approx 11.011 \times 10^{23}$$

Hence, the required numerical value is 11.