The kinetic molecular theory attributes an average kinetic theory of `(3RT)/(2 N)` to each particle. What rms speed would a mist particle of mass `10^(-12)g` have at room temperature `27^(@)C` according to kinetic theory of gases? 

Explanation

To find the root mean square (rms) speed of a mist particle with a mass of \(10^{-12} \, \text{g}\) at room temperature \(27^\circ C\), we can follow these steps:

### Step 1: Convert the mass to kilograms
The mass of the particle is given as \(10^{-12} \, \text{g}\). To convert this to kilograms, we use the conversion factor \(1 \, \text{g} = 10^{-3} \, \text{kg}\).

\[ m = 10^{-12} \, \text{g} \times 10^{-3} \, \text{kg/g} = 10^{-15} \, \text{kg} \]
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