Answer
$1.6\times10^{-15}kg$
Work Step by Step
Here we use equation 14.6 $\frac{1}{2}mv_{rms}^{2}=\frac{3}{2}kT$ to find the mass of the smoke particle.
$$\frac{1}{2}mv_{rms}^{2}=\frac{3}{2}kT=>m=\frac{3kT}{v_{rms}^{2}}$$
Let's plug known values into this equation.
$m=\frac{3(1.38\times10^{-23}J/K)(301\space K)}{(2.8\times10^{-3}m/s)^{2}}=1.6\times10^{-15}kg$
Mass of the smoke particle = $1.6\times10^{-15}kg$
