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Chapter 3: Problem 41

The little-known rare earth element nauseum (atomic weight \(=172\) ) has the interesting property of being completely insoluble in everything but 25 -year- old single-malt Scotch. This curious fact was discovered in the laboratory of Professor Ludwig von Schlimazel, the eminent German chemist whose invention of the bathtub ring won him the Nobel Prize. Having unsuccessfully tried to dissolve nauseum in 7642 different solvents over a 10 -year period, Schlimazel finally came to the \(30 \mathrm{mL}\) of The Macsporran that was the only remaining liquid in his laboratory. Always willing to suffer personal loss in the name of science, Schlimazel calculated the amount of nauseum needed to make up a 0.03 molar solution, put the Macsporran bottle on the desk of his faithful technician Edgar P. Settera, weighed out the calculated amount of nauseum and put it next to the bottle, and then wrote the message that has become part of history: "Ed Settera. Add nauseum/" How many grams of nauseum did he weigh out? (Neglect the change in liquid volume resulting from the nauseum addition.)

Short Answer

Expert verified
The mass of nauseum Professor Schlimazel weighed out was approximately 0.155 grams.

Step by step solution

01

Calculation of number of moles needed

Since Molarity is defined as the number of moles of solute divided by volume of the solution in liters, we can rearrange the formula to find the number of moles. Thus, Moles = \(Molarity \times Volume\). Here, Molarity is 0.03 M and Volume is \(30 mL = 0.03 L\). Plugging the numbers: Moles = \(0.03 M \times 0.03 L = 9 \times 10^{-4} moles\)
02

Calculation of the mass of nauseum needed

Since we know the number of moles we require, and we鈥檙e given the atomic weight of nauseum, we can calculate the mass - \(Mass = Moles \times Atomic \space weight\). Here, Atomic weight = 172 g/mol let's plug the values: Mass = \(9 \times 10^{-4} moles \times 172 g/mol = 0.1548 g\)

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Key Concepts

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

Rare Earth Elements
Rare Earth Elements are a group of 17 elements on the periodic table. They include the 15 lanthanides, as well as scandium and yttrium. These elements are often found together in nature and are difficult to separate from one another. Despite their name, rare earth elements are relatively plentiful in the Earth's crust. However, they are not typically concentrated in ore deposits in amounts economically viable for extraction.

Rare earth elements have many critical uses and are essential in the production of many high-tech products. For instance:
  • They play a vital role in the production of magnets used in wind turbines and electric vehicles.
  • They are used in catalysts for petroleum refining and vehicle emission control systems.
  • These elements are also crucial for manufacturing smartphones and other consumer electronics.
While the term 'rare earth' might seem to imply scarcity, it's their dispersion that makes these elements "rare" from a commercial extraction perspective.
Solubility
Solubility is a property that describes how well a solute can dissolve in a solvent. It is typically expressed in terms of the maximum amount of the solute that can dissolve in a given quantity of solvent at a specific temperature.

Factors that can influence solubility include:
  • Temperature: Generally, solubility increases with temperature, though there are exceptions depending on the solute and solvent involved.
  • Pressure: For gases, increasing pressure often increases solubility in liquids.
  • Nature of the solute and solvent: Polar solutes typically dissolve well in polar solvents; non-polar solutes dissolve better in non-polar solvents.
In the story of "nauseum", an element perfectly insoluble in all but a very peculiar solvent (25-year-old single-malt Scotch), we see the intersection of chemistry and curiosity fascinate the scientific world. It underscores the importance of exploring a wide range of conditions and materials when conducting solubility experiments.
Atomic Weight
Atomic Weight, also known as relative atomic mass, is the average mass of an atom of an element, expressed in atomic mass units (amu). This value is weighted according to the abundance of the element's isotopes, and it allows chemists to calculate quantities in chemical reactions.

Understanding atomic weight is crucial for various calculations, including:
  • Determining the mass of a substance needed to achieve a desired molarity in solution.
  • Calculating the amounts of reactants and products in a chemical reaction.
  • Identifying and understanding the properties of elements.
For the rare earth element nauseum, with an atomic weight of 172 amu, atomic weight enables accurate calculation of the necessary mass for a specified solution molarity. Such calculations are essential for conducting precise scientific experiments and obtaining reliable results.

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Most popular questions from this chapter

A rectangular block of solid carbon (graphite) floats at the interface of two immiscible liquids. The bottom liquid is a relatively heavy lubricating oil, and the top liquid is water. Of the total block volume, \(54.2 \%\) is immersed in the oil and the balance is in the water. In a separate experiment, an empty flask is weighed, \(35.3 \mathrm{cm}^{3}\) of the lubricating oil is poured into the flask, and the flask is reweighed. If the scale reading was \(124.8 \mathrm{g}\) in the first weighing, what would it be in the second weighing? (Suggestion: Recall Archimedes' principle, and do a force balance on the block.)

A mixture of methanol and propyl acetate contains 25.0 wt\% methanol. (a) Using a single dimensional equation, determine the g-moles of methanol in \(200.0 \mathrm{kg}\) of the mixture. (b) The flow rate of propyl acetate in the mixture is to be 100.0 ib-mole/h. What must the mixture flow rate be in \(\mathrm{Ib}_{\mathrm{m}} / \mathrm{h} ?\)

Perform the following estimations without using a calculator. (a) Estimate the mass of water (kg) in an Olympic-size swimming pool. (b) A drinking glass is being filled from a pitcher. Estimate the mass flow rate of the water (g/s). (c) Twelve male heavyweight boxers coincidentally get on the same elevator in Great Britain. Posted on the elevator wall is a sign that gives the maximum safe combined weight of the passengers, \(W_{\mathrm{max}},\) in stones. (A stone is a unit of mass equal to \(14 \mathrm{lb}_{\mathrm{m}}\). It is commonly used in England as a measure of body weight, which, like the numerical equivalence between the \(1 \mathrm{b}_{\mathrm{m}}\) and \(\mathrm{Ib}_{\mathrm{f}},\) is only valid at or near sea level.) If you were one of the boxers, estimate the lowest value of \(W_{\max }\) for which you would feel comfortable remaining on the elevator. (d) The Trans-Alaska Pipeline has an outside diameter of 4 ft and extends 800 miles from the North Slope of Alaska to the northernmost ice-free port in Valdez, Alaska. How many barrels of oil are required to fill the pipeline? (e) Estimate the volume of your body \(\left(\mathrm{cm}^{3}\right)\) in two different ways. (Show your work.) (f) A solid block is dropped into water and very slowly sinks to the bottom. Estimate its specific gravity.

An inclined manometer is a useful device for measuring small pressure differences. The formula given in Section 3.4 for the pressure difference in terms of the liquid-level difference \(h\) remains valid, but while \(h\) would be small and difficult to read for a small pressure drop if the manometer were vertical, \(L\) can be made quite large for the same pressure drop by making the angle of the inclination, \(\theta,\) small. (a) Derive a formula for \(h\) in terms of \(L\) and \(\theta\) (b) Suppose the manometer fluid is water, the process fluid is a gas, the inclination of the manometer is \(\theta=15^{\circ},\) and a reading \(L=8.7 \mathrm{cm}\) is obtained. What is the pressure difference between points? and?? (c) The formula you derived in Part (a) would not work if the process fluid were a liquid instead of a gas. Give one definite reason and another possible reason.

The reaction \(A \rightarrow B\) is carried out in a laboratory reactor. According to a published article the concentration of A should vary with time as follows: \(C_{\mathrm{A}}=C_{\mathrm{A} 0} \exp (-k t)\) where \(C_{\mathrm{A} 0}\) is the initial concentration of \(\mathrm{A}\) in the reactor and \(k\) is a constant. (a) If \(C_{\mathrm{A}}\) and \(C_{\mathrm{A} 0}\) are in \(\mathrm{Ib}-\) moles \(/ \mathrm{ft}^{3}\) and \(t\) is in minutes, what are the units of \(k ?\) (b) The following data are taken for \(C_{\mathrm{A}}(t):\) $$\begin{array}{cc}\hline t(\min ) & C_{\mathrm{A}}\left(\mathrm{lb}-\mathrm{mole} / \mathrm{ft}^{3}\right) \\\\\hline 0.5 & 1.02 \\\1.0 & 0.84 \\\1.5 & 0.69 \\\2.0 & 0.56 \\\3.0 & 0.38 \\\ 5.0 & 0.17 \\\10.0 & 0.02 \\\\\hline\end{array}$$ Verify the proposed rate law graphically (first determine what plot should yield a straight line), and calculate \(C_{\mathrm{A} 0}\) and \(k\) (c) Convert the formula with the calculated constants included to an expression for the molarity of A in the reaction mixture in terms of \(t\) (seconds). Calculate the molarity at \(t=265 \mathrm{s}\).
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