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Chapter 1: Problem 96

The average time it takes for a molecule to diffuse a distance of \(x \mathrm{~cm}\) is given by $$ t=\frac{x^{2}}{2 D} $$where \(t\) is the time in seconds and \(D\) is the diffusion coefficient. Given that the diffusion coefficient of glucose is \(5.7 \times 10^{-7} \mathrm{~cm}^{2} / \mathrm{s},\) calculate the time it would take for a glucose molecule to diffuse \(10 \mu \mathrm{m}\), which is roughly the size of a cell.

Short Answer

Expert verified
It would take approximately 0.88 seconds for a glucose molecule to diffuse a distance of 10 micrometers, the roughly size of a cell.

Step by step solution

01

Convert distance from micrometers to centimeters

To use the formula, the distance x needs to be in the same unit as the diffusion coefficient (centimeters). Given that 1 cm = 10,000 micrometers, the distance of 10 micrometers can be converted to centimeters by dividing by 10,000. So, \(x = \frac{10}{10,000} = 0.001\) cm.
02

Substitute the given values into the formula

Now that we have x in the appropriate unit, we can substitute the given values into the formula \(t=\frac{x^{2}}{2 D}\). We are given that D is \(5.7 \times 10^{-7}\) cm虏/s and we found that x is 0.001 cm. So, \(t = \frac{(0.001)^{2}}{2 \times 5.7 \times 10^{-7}}\).
03

Compute the result

Solving the equation gives \(t = \frac{0.000001}{2 \times 5.7 \times 10^{-7}} = \frac{0.000001}{1.14 \times 10^{-6}} = 0.87719\) seconds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Diffusion Coefficient
Understanding the diffusion coefficient is crucial in grasping how molecules spread over time. This coefficient, represented by the symbol 'D', is a measure of how quickly molecules diffuse through a particular medium. It is affected by factors such as the size and shape of the molecules, temperature, and viscosity of the medium.

The diffusion coefficient has units of area per time (typically cm虏/s), expressing the rate at which the concentration of a substance changes over distance with time. For example, in our glucose diffusion scenario, a higher 'D' would mean glucose molecules spread more rapidly. When calculating diffusion times, it's essential to have the diffusion coefficient to use the equation \( t=\frac{x^{2}}{2 D} \), where 't' is the time and 'x' is the distance traveled.
Converting Units in Chemistry
Chemistry calculations often require unit conversions to ensure that all quantities are expressed in the same system of units. This consistency is vital for the accuracy of chemical equations and formulations. In the context of diffusion time calculation, it is essential to convert the distance unit 'micrometers' to 'centimeters', because our given diffusion coefficient is in cm虏/s.

To convert micrometers to centimeters, we use the fact that 1 cm equals 10,000 micrometers. Therefore, we divide the micrometer value by 10,000 to express it in centimeters. Proper unit conversion can be the difference between a correct answer and a significant miscalculation, making it a foundational skill in chemistry.
Molecular Movement
Molecular movement is the random motion of molecules as they spread from areas of higher concentration to areas of lower concentration, a process driven by their intrinsic kinetic energy. This phenomenon is known as diffusion and is inherently connected to the nature of molecules being in constant, random motion. The speed and pattern of this movement can be influenced by various factors, such as temperature, size of the molecules, and the medium in which they are moving.

Understanding molecular movement allows us to predict how substances disperse in different environments. For example, glucose molecules might take more time to diffuse across a cell than smaller molecules due to their larger size. Hence, molecular movement is a key aspect of many biological and chemical processes, including the diffusion example addressed in our problem.

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Most popular questions from this chapter

Classify each of the following as an element, a compound, a homogeneous mixture, or a heterogeneous mixture: (a) water from a well, (b) argon gas, (c) sucrose, (d) a bottle of red wine, (e) chicken noodle soup, (f) blood flowing in a capillary, (g) ozone.

Chlorine is used to disinfect swimming pools. The accepted concentration for this purpose is 1 ppm chlorine, or \(1 \mathrm{~g}\) of chlorine per million grams of water. Calculate the volume of a chlorine solution (in milliliters) a homeowner should add to her swimming pool if the solution contains 6.0 percent chlorine by mass and there are \(2.0 \times 10^{4}\) gallons of water in the pool. ( 1 gallon \(=3.79 \mathrm{~L} ;\) density of liquids $=1.0 \mathrm{~g} / \mathrm{mL} .

A pycnometer is a device for measuring the density of liquids. It is a glass flask with a close-fitting ground glass stopper having a capillary hole through it. (a) The volume of the pycnometer is determined by using distilled water at \(20^{\circ} \mathrm{C}\) with a known density of \(0.99820 \mathrm{~g} / \mathrm{mL}\). First, the water is filled to the rim. With the stopper in place, the fine hole allows the excess liquid to escape. The pycnometer is then carefully dried with filter paper. Given that the masses of the empty pycnometer and the same one filled with water are \(32.0764 \mathrm{~g}\) and \(43.1195 \mathrm{~g},\) respectively, calculate the volume of the pycnometer. (b) If the mass of the pycnometer filled with ethanol at \(20^{\circ} \mathrm{C}\) is \(40.8051 \mathrm{~g},\) calculate the density of ethanol. (c) Pycnometers can also be used to measure the density of solids. First, small zinc granules weighing \(22.8476 \mathrm{~g}\) are placed in the pycnometer, which is then filled with water. If the combined mass of the pycnometer plus the zinc granules and water is \(62.7728 \mathrm{~g}\). what is the density of zinc?

The density of methanol, a colorless organic liquid used as solvent, is \(0.7918 \mathrm{~g} / \mathrm{mL}\). Calculate the mass of \(89.9 \mathrm{~mL}\) of the liquid.

What is the mass of one mole of ants? (Useful information: A mole is the unit used for atomic and subatomic particles. It is approximately \(6 \times 10^{23}\). A \(1-\mathrm{cm}\) -long ant weighs about \(3 \mathrm{mg}\).
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