Act like a helpful tutor and exlain step-by-step solution with explanation and final answer:
( give me answer without using bullet-points)1. A topical drug delivery patch contains a medicated gel lay

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Act like a helpful tutor and exlain step-by-step solution with explanation and final answer:
( give me answer without using bullet-points)1. A topical drug delivery patch contains a medicated gel layer that releases an active
drug into the skin while leaving behind an inactive carrier solvent. The diffusion
barrier consists of a stagnant stratum corneum layer that is 1.5 mm thick. At the
inner surface (point 1), in contact with the medicated gel, the drug concentration is
12.5 wt%, and the density of the solution at this point is 950 kg/m?. At the outer
surface (point 2), which interfaces with deeper skin layers, the drug concentration
decreases to 4.5 wt%, and the density at this point is 980 kg/m. The diffusivity of the
drug in the skin tissue is 0.600 x 10-2 m?/s. Using these conditions, determine the
steady-state diffusive flux of the drug across the skin barrier. How much of the drug
is delivered through a 5 cm square (5°x5”) patch in 1 day? Recall pa=x.*p. Do not
consider this a dilute solution.

VIPSolutions · Accepted Answer

### Step-by-Step Solution

**Given Data:**

- Thickness of stratum corneum, $ L = 1.5 $ mm $ = 1.5 \times 10^{-3} $ m  
- Concentration at point 1 ($ x_{A1} $): 12.5 wt% $ = .125 $  
- Density at point 1 ($ \rho_1 $): 950 kg/m³  
- Concentration at point 2 ($ x_{A2} $): 4.5 wt% $ = .045 $  
- Density at point 2 ($ \rho_2 $): 980 kg/m³  
- Diffusivity ($ D $): $ .600 \times 10^{-9} $ m²/s  
- Area of patch: $ 5 $ cm $ \times 5 $ cm $ = 25 $ cm² $ = 2.5 \times 10^{-3} $ m²  
- $ \rho_A = x_A \cdot \rho $

---

#### 1. **Calculate the concentration of drug at both surfaces in kg/m³:**

At point 1 (inner surface),  
$ \rho_{A1} = x_{A1} \cdot \rho_1 = .125 \times 950 = 118.75 $ kg/m³

At point 2 (outer surface),  
$ \rho_{A2} = x_{A2} \cdot \rho_2 = .045 \times 980 = 44.1 $ kg/m³

---

#### 2. **Use Fick's First Law for steady-state diffusive flux:**

$$
J_A = -D \frac{\Delta \rho_A}{L}
$$

Where:
- $ J_A $ is the mass flux (kg/m²·s)
- $ D $ is the diffusivity (m²/s)
- $ \Delta \rho_A = \rho_{A2} - \rho_{A1} $
- $ L $ is the thickness (m)

Since the drug diffuses from high to low concentration, the flux direction is from point 1 to point 2. The negative sign accounts for diffusion direction.

So,
$$
J_A = -D \frac{\rho_{A2} - \rho_{A1}}{L}
$$
$$
J_A = -.600 \times 10^{-9} \frac{44.1 - 118.75}{1.5 \times 10^{-3}}
$$
$$
J_A = -.600 \times 10^{-9} \frac{-74.65}{1.5 \times 10^{-3}}
$$
$$
J_A = -.600 \times 10^{-9} \cdot (-49766.67)
$$
$$
J_A = 2.986 \times 10^{-5} \text{ kg/m}^2\cdot\text{s}
$$

---

#### 3. **Calculate the total mass delivered in 1 day through the patch:**

Total mass delivered ($ m $):
$$
m = J_A \cdot A \cdot t
$$
Where:
- $ A = 2.5 \times 10^{-3} $ m²
- $ t = 1 $ day $ = 86400 $ seconds

$$
m = 2.986 \times 10^{-5} \cdot 2.5 \times 10^{-3} \cdot 86400
$$
$$
m = 2.986 \times 2.5 \times 86400 \times 10^{-8}
$$
$$
m = 7.465 \times 86400 \times 10^{-8}
$$
$$
m = 645,816 \times 10^{-8}
$$
$$
m = 6.46 \times 10^{-3} \text{ kg}
$$

---

### **Final Answer**

The steady-state diffusive flux of the drug across the skin barrier is $ 2.99 \times 10^{-5} $ kg/m²·s. The total mass of drug delivered through a 5 cm by 5 cm patch in 1 day is approximately **6.46 g** (rounded to two decimal places).

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