A concentration difference of a certain solute of 1.0 × 10^-² kg/m³ is maintained between the ends of a tube with a length of 3.5 m and a cross-sectional area of 0.25 m².When 0.0040 g of the solute is introduced to the tube,it takes 350 minutes for this solute to diffuse through the solvent to the opposite end of the tube.What is the diffusion constant for the solute?

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Quizplus · Accepted Answer

The diffusion constant can be calculated using Fick's first law, $J = -D \frac{\Delta C}{\Delta x}$, where $J$ is the flux, $D$ is the diffusion constant, $\Delta C$ is the concentration difference, and $\Delta x$ is the distance over which diffusion occurs. However, to find $D$, we use the rearranged form of the equation that relates the amount of substance ($Q$), the area ($A$), the concentration difference ($\Delta C$), the length ($L$), and the time ($t$):$$D = \frac{Q \times L}{A \times \Delta C \times t}$$Given:- $Q = 0.0040 \, g = 4.0 \times 10^{-6} \, kg$ (since $1 \, g = 10^{-3} \, kg$)- $L = 3.5 \, m$- $A = 0.25 \, m^2$- $\Delta C = 1.0 \times 10^2 \, kg/m^3$- $t = 350 \, minutes = 350 \times 60 \, s = 21000 \, s$Plugging these values into the equation:$$D = \frac{4.0 \times 10^{-6} \times 3.5}{0.25 \times 1.0 \times 10^2 \times 21000}$$$$D = \frac{14.0 \times 10^{-6}}{5250}$$$$D = 2.666... \times 10^{-9} \, m^2/s$$Rounding to the correct significant figures, we get:$$D = 2.7 \times 10^{-9} \, m^2/s$$Therefore, the correct answer is A) $2.7 \times 10^{-9} \, m^2/s$.

A Concentration Difference of a Certain Solute of 1