/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 Glycerin in water diffuses along... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 74

Glycerin in water diffuses along a horizontal column that has a cross- sectional area of \(2.0 \mathrm{~cm}^{2}\). The concentration gradient is \(3.0 \times 10^{-2} \mathrm{~kg} / \mathrm{m}^{4}\), and the diffusion rate is found to be \(5.7 \times 10^{-15} \mathrm{~kg} / \mathrm{s}\). Determine the diffusion coefficient.

Short Answer

Expert verified
The diffusion coefficient for glycerin in water is \(9.5 \times 10^{-9} \mathrm{~m}^{2}/\mathrm{s}\).

Step by step solution

01

Understand the Given Values

Identify and note down the given values from the problem. The cross-sectional area (A) is \(2.0 \mathrm{~cm}^{2}\), but it needs to be converted to \(\mathrm{m}^{2}\). So, \(A = 2.0 \times 10^{-4} \mathrm{~m}^{2}\). The concentration gradient (dC/dx) is \(3.0 \times 10^{-2} \mathrm{~kg} / \mathrm{m}^{4}\). The diffusion rate is \(5.7 \times 10^{-15} \mathrm{~kg} / \mathrm{s}\). We use the positive value for the rate because we are concerned with the magnitude only.
02

Rearrange the Formula for D

Rearrange the diffusion rate formula to solve for the diffusion coefficient (D). It becomes \(D = \mathrm{Rate} / (A \times dC/dx)\).
03

Substitute the Values

Substitute the given values into the rearranged formula. \(D = (5.7 \times 10^{-15} \mathrm{~kg} / \mathrm{s}) / ((2.0 \times 10^{-4} \mathrm{~m}^{2}) \times (3.0 \times 10^{-2} \mathrm{~kg} / \mathrm{m}^{4}))\).
04

Perform the Calculation

Perform the division to get the value of D. The units will be \(\mathrm{m}^{2}/\mathrm{s}\), which is the unit used for the diffusion coefficient.

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Most popular questions from this chapter

Sucrose is allowed to diffuse along a \(10-\mathrm{cm}\) length of tubing filled with water. The tube is \(6.0 \mathrm{~cm}^{2}\) in crosssectional area. The diffusion coefficient is equal to \(5.0 \times\) \(10^{-10} \mathrm{~m}^{2} / \mathrm{s}\), and \(8.0 \times 10^{-14} \mathrm{~kg}\) is transported along the tube in \(15 \mathrm{~s}\). What is the difference in the concentration levels of sucrose at the two ends of the tube?

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