Cesium atoms are the largest neturally occurring atoms. The radius of `Cs` atom is `2.6 Å`. The number of moles of `Cs` atoms to be laid side by side to give a row of `Cs` atoms `2.5 cm` long is `x xx 10^(-17)`. Find the value of `x`.

Explanation

To solve the problem, we need to determine the number of moles of cesium (Cs) atoms that can be laid side by side to form a row of length 2.5 cm. Given that the radius of a cesium atom is 2.6 Å (angstroms), we will follow these steps:

### Step 1: Convert the radius of the cesium atom from angstroms to centimeters.
1 Å = \(1 \times 10^{-8}\) cm, so:
\[ \text{Radius of Cs atom} = 2.6 \, \text{Å} = 2.6 \times 10^{-8} \, \text{cm} \]

### Step 2: Calculate the diameter of a cesium atom.
The diameter \(d\) of a cesium atom is twice the radius:
\[ d = 2 \times \text{Radius} = 2 \times (2.6 \times 10^{-8} \, \text{cm}) = 5.2 \times 10^{-8} \, \text{cm} \]

### Step 3: Determine the number of cesium atoms that can fit into a row of length 2.5 cm.
To find the number of atoms \(N\) that can fit in a row of length \(L = 2.5 \, \text{cm}\):
\[ N = \frac{L}{d} = \frac{2.5 \, \text{cm}}{5.2 \times 10^{-8} \, \text{cm}} \approx 4.8077 \times 10^{7} \]

### Step 4: Calculate the number of moles of cesium atoms.
Using Avogadro's number \(N_A = 6.02 \times 10^{23}\) atoms/mol, we can find the number of moles \(n\):
\[ n = \frac{N}{N_A} = \frac{4.8077 \times 10^{7}}{6.02 \times 10^{23}} \approx 7.986 \times 10^{-17} \, \text{moles} \]

### Step 5: Express the result in the form \(x \times 10^{-17}\).
From our calculation, we have:
\[ n \approx 7.986 \times 10^{-17} \]
Thus, the value of \(x\) is approximately \(7.986\).

### Final Answer:
The value of \(x\) is \(7.986\).

---