Chapter 14, Problem 50Carbon tetrachloride (CCI4) is diffusing through benzene (CgHs), as the drawing illustrates. The concentration of CCI, at the left end of the tube is maintained at 1.65 x 10-2 kg/m3, and the diffusion constant is 19.5 x 10-10 m2/s. The CCI4 enters the tube at a mass rate of 5.56x 1013 kg/s. Using these data and those shown in the drawing, find (a) the mass of CCI4 per second that passes point A and (b) the concentration of CCl, at point A.5.00 x 10-3 mCross-sectionalarea = 3.00 x 10-4 m2

We know that mass is a scalar quantity and is conserved, the mass flow rate is the same at all points as described by the equation of continuity. Therefore, the mass flow rate at which CCl4 enters the tube is the same as that at point A. The concentration difference of CCl4 between A and the left end of the tube ∆C can be calculated by using Fick's law of diffusion. a) The mass flow rate of CCl4 as it passes point A is the same as the mass flow rate at which CCl4 enters the left end of the tube. Therefore the mass flow rate of CCl4 at point A is 5.56 x 10-13 kg/s. b) We will calculate the value of ∆C with the help of Fick's law, we obtain: ∆C=mLDAt =mtLDA =5.56×10-13kg/s×5.00×10-3m19.5×10-10m2/s3.00×10-4m2 =4.75×10-3kg/m3 Now, CA=Cleft end-∆C =1.65×10-2kg/m3-4.75×10-3kg/m3 =1.175×10-2kg/m3 Thus, the concentration of CCl4 at A is 1.175 x 10-2 kg/m3.

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Carbon tetrachloride (CCl) is diffusing through benzene (CgHe), as the drawing illustrates. The concentration of CCl, at the left end of the tube is maintained at 1.65 x 102 kg/m, and the diffusion constant is 19.5 x 1010 m2/s. The CCI, enters the tube at a mass rate of 5.56 x 10°13 kg/s. Using these data and those shown in the drawing, find (a) the mass of CCl, per second that passes point A and (b) the concentration of CCl, at point A. 5.00 x 10m Cross-sectional area 3.00 x 10m?

Carbon tetrachloride (CCl) is diffusing through benzene (CgHe), as the drawing illustrates. The concentration of CCl, at the left end of the tube is maintained at 1.65 x 102 kg/m, and the diffusion constant is 19.5 x 1010 m2/s. The CCI, enters the tube at a mass rate of 5.56 x 10°13 kg/s. Using these data and those shown in the drawing, find (a) the mass of CCl, per second that passes point A and (b) the concentration of CCl, at point A. 5.00 x 10m Cross-sectional area 3.00 x 10m?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

Chapter 14, Problem 50
Carbon tetrachloride (CCI4) is diffusing through benzene (CgHs), as the drawing illustrates. The concentration of CCI, at the left end of the tube is maintained at 1.65 x 10-2 kg/m3, and the diffusion constant is 19.5 x 10-10 m2/s. The CCI4 enters the tube at a mass rate of 5.56
x 1013 kg/s. Using these data and those shown in the drawing, find (a) the mass of CCI4 per second that passes point A and (b) the concentration of CCl, at point A.
5.00 x 10-3 m
Cross-sectional
area = 3.00 x 10-4 m2

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