## Step-by-Step Solution ### **Given Data** - **Diameter of aerosol particles, \( d \) = 2.5 μm = \( 2.5 \times 10^{-6} \) m** - **Density of particles, \( \rho \) = 105 kg/m³** - **Aerosol concentration = 70 μg/m³ = \( 70 \times 10^{-6} \) g/m³ = \( 70 \times 10^{-9} \) kg/m³** - **Breathing rate = 30 m³/day** - **Duration = 8 hours** --- ### **Step 1: Volume of Air Breathed in 8 Hours** - There are 24 hours in a day. - Volume breathed in 8 hours: \[ \text{Volume in 8 hours} = \frac{8}{24} \times 30~\text{m}^3 = 10~\text{m}^3 \] --- ### **Step 2: Total Mass of Aerosol Inhaled** - Aerosol concentration = \( 70 \times 10^{-9} \) kg/m³ - Total mass inhaled: \[ \text{Total mass} = 70 \times 10^{-9}~\text{kg/m}^3 \times 10~\text{m}^3 = 700 \times 10^{-9}~\text{kg} = 7 \times 10^{-7}~\text{kg} \] --- ### **Step 3: Mass of a Single Particle** #### **a) Volume of One Particle (sphere):** \[ V = \frac{4}{3}\pi r^3 \] - \( r = \frac{d}{2} = \frac{2.5 \times 10^{-6}}{2} = 1.25 \times 10^{-6}~\text{m} \) \[ V = \frac{4}{3}\pi (1.25 \times 10^{-6})^3 \] \[ V = \frac{4}{3}\pi (1.953125 \times 10^{-18}) \] \[ V \approx 8.181 \times 10^{-18}~\text{m}^3 \] #### **b) Mass of One Particle:** \[ \text{Mass} = \text{Volume} \times \text{Density} \] \[ \text{Mass} = 8.181 \times 10^{-18}~\text{m}^3 \times 105~\text{kg/m}^3 \] \[ \text{Mass} \approx 8.590 \times 10^{-15}~\text{kg} \] --- ### **Step 4: Number of Particles Inhaled** \[ \text{Number of particles} = \frac{\text{Total mass inhaled}}{\text{Mass of one particle}} \] \[ \text{Number of particles} = \frac{7 \times 10^{-7}}{8.590 \times 10^{-15}} \] \[ = 8.15 \times 10^7 \] --- ## **Final Answer** \[ \boxed{8.2 \times 10^7} \] A person ingests approximately **\( 8.2 \times 10^7 \)** aerosol particles over the course of 8 hours.