An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is `vec(v), m` is its mass and `k_(s)` is Bolzmann constant, then its temperature will be `(mv^(2))/(nk_(B))` , where n is __________ .

Explanation

To solve the problem, we need to derive the expression for the temperature of an HCl molecule in terms of its root mean square (RMS) velocity, mass, and Boltzmann's constant.

### Step-by-Step Solution:

1. **Understanding RMS Velocity**:
The root mean square (RMS) velocity \( V_{\text{rms}} \) of gas molecules can be expressed using the formula:
\[ V_{\text{rms}} = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the absolute temperature, and \( M \) is the molar mass of the gas.

2. **Using Boltzmann's Constant**:
We can also express the RMS velocity in terms of Boltzmann's constant \( k_B \):
\[ V_{\text{rms}} = \sqrt{\frac{3k_B T}{m}} \]
where \( m \) is the mass of a single molecule of the gas.

3. **Squaring the RMS Velocity**:
Squaring both sides of the equation gives us:
\[ V_{\text{rms}}^2 = \frac{3k_B T}{m} \]

4. **Rearranging for Temperature**:
Rearranging the equation to solve for temperature \( T \) yields:
\[ T = \frac{m V_{\text{rms}}^2}{3 k_B} \]

5. **Identifying the Value of \( n \)**:
The problem states that the temperature can also be expressed as:
\[ T = \frac{m V^2}{n k_B} \]
By comparing the two expressions for temperature, we can see that:
\[ n = 3 \]

### Final Answer:
The value of \( n \) is \( 3 \).