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It is interesting how the Fahrenheit temperature scale was established. One report, given by Fahrenheit in a paper in \(1724,\) stated that the value of \(0^{\circ} \mathrm{F}\) was established as the freezing temperature of saturated solutions of sea salt. From the literature we find that the freezing point of a \(20 \%\) by mass solution of NaCl is \(-16.46^{\circ} \mathrm{C} .\) (This is the lowest freezing temperature reported for solutions of NaCl.) Does this value lend credence to this story of the establishment of the Fahrenheit scale?

Short Answer

Expert verified
Yes, the value provides some credence to the story as the freezing point is close to 0°F.

Step by step solution

01

Understand the problem

We need to determine whether the freezing point of a 20% NaCl solution supports the claim that 0°F was set based on this freezing point. The reported freezing point is \(-16.46^{\circ}\mathrm{C}\).
02

Convert Celsius to Fahrenheit

To compare the freezing point with 0°F, we need to convert \(-16.46^{\circ}\mathrm{C}\) to Fahrenheit using the formula: \(F = \frac{9}{5}C + 32\).
03

Apply conversion formula

Substituting \(-16.46\) for \(C\) in the conversion formula: \[ F = \frac{9}{5}\times(-16.46) + 32 = -29.628 + 32 = 2.372\, ^{\circ}\mathrm{F}. \]
04

Analyze the result

The converted Fahrenheit temperature is \(2.372^{\circ}\mathrm{F}\), which is not exactly 0°F, but close enough that Fahrenheit might have used it as a reference point.

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Key Concepts

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

Celsius to Fahrenheit conversion
Temperature conversion is a crucial skill in science, especially when comparing information recorded in different scales. The most common conversion is between Celsius (°C) and Fahrenheit (°F). The formula to transform Celsius to Fahrenheit is:
  • \( F = \frac{9}{5}C + 32 \)
This mathematical expression allows one to switch from the Celsius scale used widely around the world, to Fahrenheit, which is primarily used in the United States. For instance, in our exercise, we needed to compare the freezing point of a salt solution in Fahrenheit. The Celsius temperature of \(-16.46^{\circ}\mathrm{C}\) was converted using the formula, resulting in \(2.372^{\circ}\mathrm{F}\).
Remember:
  • Multiply the Celsius temperature by \(9/5\) (or 1.8).
  • Add 32 to the result to finalize the Fahrenheit equivalent.
Understanding this conversion helps in comparing temperatures across different regions and countries efficiently. It's always useful for students and professionals alike to be comfortable with these conversions to better manage and interpret scientific data.
freezing point of salt solutions
The freezing point of salt solutions, such as those made with sodium chloride (NaCl), can differ significantly from pure water. When salt is dissolved in water, it lowers the freezing point. This phenomenon is known as freezing point depression.
  • For a 20% NaCl solution, the freezing point is substantially lower.
  • The reported freezing point is \(-16.46^{\circ}\mathrm{C}\).
This property is vital in understanding how salt can be used in practical applications. For example, salt is often used on icy roads in winter to melt ice by lowering its freezing point.
By examining the specific case of the 20% NaCl solution, students can see how solute concentration directly impacts the temperature at which a solution freezes. It's an important concept in chemistry relates to colligative properties, which depend on the number of particles in a solution rather than their identity.
NaCl solution properties
Sodium chloride, commonly known as table salt, possesses unique properties when dissolved in water. In solution, NaCl dissolves into sodium (Na⁺) and chloride (Cl⁻) ions, interacting with the water molecules.
  • This ionization greatly affects the freezing and boiling points of the solution.
  • It increases the boiling point and decreases the freezing point.
These changes are attributed to the colligative properties of the solution, which depend on the dissolved particles rather than the type of particle. The effects are common and well understood in various applications like food preservation, where salt is used to inhibit the growth of spoilage organisms by lowering the freezing point and reducing water activity.
Understanding NaCl's properties is significant for many chemical processes and applications. It offers a real-world example of how solute particles interfere with solvent molecules, demonstrating fundamental concepts in chemistry.

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Most popular questions from this chapter

A tree is \(10.0 \mathrm{m}\) tall. (a) What must be the total molarity of the solutes if sap rises to the top of the tree by osmotic pressure at \(20^{\circ} \mathrm{C} ?\) Assume the groundwater outside the tree is pure water and that the density of the sap is \(1.0 \mathrm{g} / \mathrm{mL} .\left(1 \mathrm{mm} \mathrm{Hg}=13.6 \mathrm{mm} \mathrm{H}_{2} \mathrm{O} .\right)\) (b) If the only solute in the sap is sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11},\) what is its percent by mass?

Arrange the following aqueous solutions in order of (i) increasing vapor pressure of water and (ii) increasing boiling point. (a) \(0.35 \mathrm{m} \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\) (a nonvolatile solute) (b) \(0.50 \mathrm{m}\) sugar (c) \(0.20 \mathrm{m} \mathrm{KBr}\) (a strong electrolyte) (d) \(0.20 m \mathrm{Na}_{2} \mathrm{SO}_{4}\) (a strong electrolyte)

In a police forensics lab, you examine a package that may contain heroin. However, you find the white powder is not pure heroin but a mixture of heroin \(\left(\mathrm{C}_{21} \mathrm{H}_{23} \mathrm{O}_{5} \mathrm{N}\right)\) and lactose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) .\) To determine the amount of heroin in the mixture, you dissolve \(1.00 \mathrm{g}\) of the white powdery mixture in water in a 100.0 -mL volumetric flask. You find that the solution has an osmotic pressure of \(539 \mathrm{mm}\) Hg at \(25^{\circ} \mathrm{C}\). What is the composition of the mixture?

The dispersed phase of a certain colloidal dispersion consists of spheres of diameter \(1.0 \times 10^{2} \mathrm{nm}\) (a) What are the volume \(\left(V=4 / 3 \pi r^{3}\right)\) and surface area \(\left(A=4 \pi r^{2}\right)\) of each sphere? (b) How many spheres are required to give a total volume of \(1.0 \mathrm{cm}^{3} ?\) What is the total surface area of these spheres in square meters?

You dissolve \(45.0 \mathrm{g}\) of camphor, \(\mathrm{C}_{10} \mathrm{H}_{16} \mathrm{O},\) in \(425 \mathrm{mL}\) of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH} .\) Calculate the molality, mole fraction, and weight percent of camphor in this solution. (The density of ethanol is \(0.785 \mathrm{g} / \mathrm{mL} .)\)

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