Let's solve the problem step by step. --- ## **Step 1. Understanding the Problem** We have a **two-cell setup** for measuring diffusivity. The **cell constant is unknown**, so we must first determine it using a calibration experiment with a system of known diffusivity (formic acid-water). **General setup:** - Two cells separated by a barrier. - Initially, one cell has pure solvent (cell 1), the other has solution (cell 2). - Concentrations in both cells change over time as diffusion occurs. We assume the system is described by the **"two-cell method"** (also called the "interdiffusion cell" or "Stirred-cell" method). --- ## **Step 2. Formula for the Two-Cell Experiment** The **two-cell experiment** gives the following formula for diffusivity: \[ D_{AB} = \frac{K}{t} \ln \left( \frac{(C_{A2,0} - C_{A1,0})}{(C_{A2,t} - C_{A1,t})} \right) \] Where: - \( D_{AB} \) = Diffusivity (m²/s) - \( K \) = cell constant (unit: m³) - \( t \) = time (s) - \( C_{A2,0} \), \( C_{A1,0} \) = initial concentrations in cell 2 and 1 (k-mol/m³) - \( C_{A2,t} \), \( C_{A1,t} \) = concentrations after time t --- ## **Step 3. Calibration Experiment (Find Cell Constant \( K \))** ### **Given:** - Known \( D_{AB} \) (formic acid in water): \( 1.52 \times 10^{-9} \) m²/s - Initial concentrations: - Cell 1: \( C_{A1,0} = 0 \) - Cell 2: \( C_{A2,0} = 0.12 \) k-mol/m³ - After \( t = 55 \) hrs \( 28 \) min: - Cell 1: \( C_{A1,t} = 0.045 \) k-mol/m³ - Cell 2: \( C_{A2,t} = 0.065 \) k-mol/m³ #### **Convert time to seconds:** \[ 55 \text{ hr } 28 \text{ min } = 55 \times 60 + 28 = 3328 \text{ min } = 3328 \times 60 = 199,680 \text{ s} \] #### **Plug into the formula and solve for \( K \):** \[ 1.52 \times 10^{-9} = \frac{K}{199,680} \ln \left( \frac{0.12 - 0}{0.065 - 0.045} \right) \] \[ 0.065 - 0.045 = 0.02 \] \[ \frac{0.12}{0.02} = 6 \] \[ \ln(6) = 1.7918 \] \[ 1.52 \times 10^{-9} = \frac{K}{199,680} \times 1.7918 \] \[ K = \frac{1.52 \times 10^{-9} \times 199,680}{1.7918} \] Calculate numerator: \[ 1.52 \times 10^{-9} \times 199,680 = 3.034 \times 10^{-4} \] Divide by 1.7918: \[ K = \frac{3.034 \times 10^{-4}}{1.7918} = 1.693 \times 10^{-4} \text{ m}^3 \] --- ## **Step 4. Use \( K \) to Find Methanol Diffusivity** ### **Given:** - Initial concentrations: - Cell 1: \( 0 \) - Cell 2: \( 0.085 \) k-mol/m³ - After \( t = 75 \) hr \( 36 \) min: - Cell 1: \( 0.0325 \) k-mol/m³ - Cell 2: \( 0.050 \) k-mol/m³ #### **Convert time to seconds:** \[ 75 \text{ hr } 36 \text{ min } = 75 \times 60 + 36 = 4536 \text{ min } = 4536 \times 60 = 272,160 \text{ s} \] #### **Plug into the formula:** \[ D_{AB} = \frac{K}{t} \ln \left( \frac{0.085 - 0}{0.050 - 0.0325} \right) \] \[ 0.050 - 0.0325 = 0.0175 \] \[ \frac{0.085}{0.0175} = 4.8571 \] \[ \ln(4.8571) = 1.580 \] \[ D_{AB} = \frac{1.693 \times 10^{-4}}{272,160} \times 1.580 \] Calculate: \[ \frac{1.693 \times 10^{-4}}{272,160} = 6.225 \times 10^{-10} \] \[ D_{AB} = 6.225 \times 10^{-10} \times 1.580 = 9.832 \times 10^{-10} \text{ m}^2/\text{s} \] --- ## **Step 5. Final Answer** \[ \boxed{ D_{AB} \text{ (Methanol in water at 25°C)} = 9.83 \times 10^{-10} \text{ m}^2/\text{s} } \] --- ### **Summary Table** | Step | Calculation | Value | |-----------------|--------------------------------------------------|-----------------------| | Calibration K | \( K = \frac{1.52 \times 10^{-9} \times 199,680}{1.7918} \) | \( 1.693 \times 10^{-4} \) m³ | | Methanol D | \( D_{AB} = \frac{K}{t} \ln \left( \frac{C_{A2,0} - C_{A1,0}}{C_{A2,t} - C_{A1,t}} \right) \) | \( 9.83 \times 10^{-10} \) m²/s | --- ### **Boxed Final Answer** \[ \boxed{ \text{The molecular diffusivity of methanol in water at 25°C is: } \mathbf{9.83 \times 10^{-10}~\text{m}^2/\text{s}} } \]