Given:
Diffusion constant = 1.71 x 10⁻⁶ m²/s.
Let's find the average distance a perfume molecule moves in air.
Apply the formula:
![x_(rms)=√(2Dt)](https://img.qammunity.org/2023/formulas/physics/college/7srpxmbl38o97fsk9elgduxa9n32h3pq5f.png)
Where:
D is the diffusion constant
t is the time = 1 second.
Thus, we have:
![\begin{gathered} x_(rms)=\sqrt{2*1.71*10^(-6)*1.00} \\ \\ x_(rms)=\sqrt{3.42*10^(-6)} \\ \\ x_(rms)=1.85*10^{-3\text{ }}m \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/pa0pbs1ue1k13l35p2p6xnf9q5hi48ve0m.png)
Therefore, the average distance a perfume molecule moves in one second in air is 1.85 x 10⁻³ m
ANSWER:
1.85 x 10⁻³ m