A student conducted a series of trials to investigate Fick's first law of diffusion using … — ACT

A student conducted a series of trials to investigate Fick's first law of diffusion using a synthetic membrane. The rate of diffusion of a solute, $J$ (in milligrams per second, $\text{mg/s}$ ), is directly proportional to both the surface area of the membrane, $A$ (in square centimeters, $\text{cm}^2$ ), and the concentration difference of the solute across the membrane, $\Delta C$ (in moles per liter, $\text{mol/L}$ ), and is inversely proportional to the thickness of the membrane, $x$ (in millimeters, $\text{mm}$ ).

The parameters for Trial 1 and Trial 2 are shown in the table below:

Trial	Membrane thickness, $x$ ( $\text{mm}$ )	Membrane surface area, $A$ ( $\text{cm}^2$ )	Concentration difference, $\Delta C$ ( $\text{mol/L}$ )	Diffusion rate, $J$ ( $\text{mg/s}$ )
1	$0.20$	$3.0$	$0.06$	$27.0$
2	$0.60$	$8.0$	$0.04$	?

Based on the table, what was the resulting diffusion rate of the solute in Trial 2, in $\text{mg/s}$ ?

Answer: 16 mg/s

Answer

The diffusion rate of the solute in Trial 2 is 16.0 mg/s.

By setting up the proportionality equation

J = k \frac{A \cdot \Delta C}{x}

, we find the constant

k = 30

using the parameters from Trial 1. Substituting the parameters from Trial 2 yields a diffusion rate of exactly 16.0 mg/s.

Step-by-Step Solution

Set up the algebraic relationship for the variables based on proportionality rules.

J = k \frac{A \cdot \Delta C}{x}

The rate of diffusion (

J

) is directly proportional to the surface area (

A

) and concentration difference (

\Delta C

), meaning they appear in the numerator. It is inversely proportional to membrane thickness (

x

), meaning it appears in the denominator. Here,

k

represents the constant of proportionality.

Calculate the constant of proportionality,

k

, using the values provided for Trial 1.

k = 30

Substitute

J_1 = 27.0

A_1 = 3.0

\Delta C_1 = 0.06

, and

x_1 = 0.20

into the equation:

27.0 = k \frac{3.0 \times 0.06}{0.20} \implies 27.0 = 0.9k \implies k = 30

Calculate the diffusion rate for Trial 2,

J_2

, by substituting the new parameters and the calculated constant

k

into the equation.

J_2 = 16.0

Substitute

k = 30

A_2 = 8.0

\Delta C_2 = 0.04

, and

x_2 = 0.60

into the equation:

J_2 = 30 \times \frac{8.0 \times 0.04}{0.60} = 30 \times \frac{0.32}{0.60} = 16.0

Key Concept

Direct and Inverse Proportionality

Question

Answer

Step-by-Step Solution

Key Concept

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