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In water conservation, chemists spread a thin film of a certain inert material over the surface of water to cut down on the rate of evaporation of water in reservoirs. This technique was pioneered by Benjamin Franklin three centuries ago. Franklin found that \(0.10 \mathrm{~mL}\) of oil could spread over the surface of water about \(40 \mathrm{~m}^{2}\) in area. Assuming that the oil forms a monolayer, that is, a layer that is only one molecule thick, estimate the length of each oil molecule in nanometers \(\left(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{~m}\right)\).

Short Answer

Expert verified
Each oil molecule is approximately 2500 nanometers long.

Step by step solution

01

Understand the Problem

The problem involves estimating the length of a single oil molecule within a monolayer spread over a given area of water. We are given the volume of oil used and the area it covers.
02

Calculate the Volume of the Monolayer

The volume of oil that forms the monolayer is given as \(0.10 \text{ mL}\). We need to convert this volume from milliliters to cubic meters since area is in square meters. \[0.10 \text{ mL} = 0.10 \times 10^{-3} \text{ cubic meters} = 1.0 \times 10^{-4} \text{ cubic meters}\]
03

Relate Volume and Area to Molecule Length

Assuming the oil spreads uniformly to form a monolayer, the volume of the oil (\(V\)) is also equal to the area of the spread (\(A\)) multiplied by the height (or length) of one molecule (\(L\)): \[V = A \times L\]\[1.0 \times 10^{-4} = 40 \text{ m}^2 \times L\]
04

Solve for the Length of the Molecule

Rearrange the formula from the previous step to solve for \(L\):\[L = \frac{V}{A} = \frac{1.0 \times 10^{-4}}{40}\]\[L = 2.5 \times 10^{-6} \text{ meters}\]
05

Convert Length to Nanometers

Convert the length from meters to nanometers, knowing that \(1 \text{ nm} = 1 \times 10^{-9} \text{ meters}\):\[L = 2.5 \times 10^{-6} \text{ meters} = \frac{2.5 \times 10^{-6}}{1 \times 10^{-9}} \text{ nanometers}\]\[L = 2,500 \text{ nanometers}\]

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Key Concepts

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

Monolayer Formation
A monolayer is a single, one-molecule-thick layer that covers a surface. This concept is important, especially in chemistry and water conservation. When chemists think about monolayer formation, they aim to create a uniform and even layer of molecules. This layer is only one molecule thick over a substance, such as water.
Monolayers are used in water conservation to help prevent water loss from reservoirs and other water bodies.
Here are some key points about monolayers:
  • Monolayers ensure that there is a straightforward, thin sheet of molecules covering a large surface.
  • The molecules are closely packed, covering the area uniformly without leaving gaps.
  • In the historical context, Benjamin Franklin was among the first to investigate the benefits of oil monolayers for reducing water evaporation.
Monolayers are a fascinating way to utilize chemistry for beneficial purposes, such as conserving water by minimizing evaporation.
Evaporation Reduction
Reducing evaporation, especially from large water bodies like reservoirs, is critical in water conservation. Evaporation reduction techniques ensure that precious water resources are not lost to the atmosphere unnecessarily.
By forming a monolayer, oils can create a physical barrier that hinders water molecules from escaping into the air.
In this context, the application of an oil monolayer provides several benefits:
  • The oil molecules, spread as a thin film, form a barrier above the water's surface.
  • This barrier minimizes the direct interaction of water molecules with the air, restricting their evaporation.
  • The method is both cost-effective and easy to implement in large-scale water bodies.
The science behind evaporation reduction through monolayer formation highlights the ingenuity in simple chemical applications to solve larger environmental challenges.
Oil Molecule Size Estimation
Estimating the size of oil molecules is integral to understanding how effectively a monolayer can form. Given an oil's volume and the area it covers when spread as a monolayer, we can deduce the molecule's thickness or size.
The process typically involves these steps:
  • Calculate the total volume of the oil used and convert this to appropriate units (e.g., cubic meters).
  • Consider the area the oil covers, usually given in square meters.
  • By assuming a monolayer formation, relate the volume of the oil to the covered area to estimate the length of a single molecule.
Understanding the dimensions of oil molecules not only aids in estimating the efficacy of a monolayer in reducing evaporation but also gives insight into molecular behavior on surfaces.
The simple proportionality between volume, area, and molecule length makes it easier for students to grasp the linkage between macroscopic observations and microscopic properties.

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Most popular questions from this chapter

The unit "troy ounce" is often used for precious metals such as gold \((\mathrm{Au})\) and platinum \((\mathrm{Pt})(1\) troy ounce \(=\) \(31.103 \mathrm{~g}\) ). (a) A gold coin weighs 2.41 troy ounces. Calculate its mass in grams. (b) Is a troy ounce heavier or lighter than an ounce \((1 \mathrm{lb}=16 \mathrm{oz} ; 1 \mathrm{lb}=453.6 \mathrm{~g}\) )?

A student is given a crucible and asked to prove whether it is made of pure platinum. She first weighs the crucible in air and then weighs it suspended in water (density = \(0.9986 \mathrm{~g} / \mathrm{mL}\) ). The readings are \(860.2 \mathrm{~g}\) and \(820.2 \mathrm{~g}\), respectively. Based on these measurements and given that the density of platinum is \(21.45 \mathrm{~g} / \mathrm{cm}^{3},\) what should her conclusion be? (Hint: An object suspended in a fluid is buoyed up by the mass of the fluid displaced by the object. Neglect the buoyancy of air.)

Carry out the following conversions: (a) \(185 \mathrm{nm}\) to meters (b) 4.5 billion years (roughly the age of Earth) to seconds (assume 365 days in a year), (c) \(71.2 \mathrm{~cm}^{3}\) to cubic meters, (d) \(88.6 \mathrm{~m}^{3}\) to liters.

The density of water at \(40^{\circ} \mathrm{C}\) is \(0.992 \mathrm{~g} / \mathrm{mL}\). What is the volume of \(27.0 \mathrm{~g}\) of water at this temperature?

Aluminum is a lightweight metal (density \(\left.=2.70 \mathrm{~g} / \mathrm{cm}^{3}\right)\) used in aircraft construction, high-voltage transmission lines, beverage cans, and foils. What is its density in \(\mathrm{kg} / \mathrm{m}^{3} ?\)

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