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You are given two colorless solutions, one containing \(\mathrm{NaCl}\) and the other sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right) .\) Suggest a chemical and a physical test that would allow you to distinguish between these two solutions.

Short Answer

Expert verified
Use conductivity to differentiate NaCl and sucrose: NaCl conducts electricity, sucrose does not. Add AgNO₃ for a chemical test: NaCl forms a white precipitate, sucrose does not.

Step by step solution

01

Understand the Nature of Solutions

The two solutions are colorless and contain different solutes: NaCl is an ionic compound, whereas sucrose is a covalent compound. Their properties can be exploited to differentiate them.
02

Physical Test - Conductivity Test

NaCl, being an ionic compound, dissociates into ions in solution, making the solution conductive. Sucrose, a covalent molecule, does not dissociate into ions, so its solution remains non-conductive. To perform this test, measure the electrical conductivity of both solutions using a conductivity meter. The NaCl solution will show a significant conductivity while the sucrose solution will not.
03

Chemical Test - Precipitation Reaction with Silver Nitrate

Add a few drops of silver nitrate solution (AgNO₃) to each solution. A white precipitate of silver chloride (AgCl) will form in the solution containing NaCl due to the reaction: \[\text{NaCl (aq)} + \text{AgNO}_3 \rightarrow \text{AgCl (s)} + \text{NaNO}_3 (aq)\]In the sucrose solution, no precipitate will form as sucrose does not react with AgNO₃.

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Key Concepts

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

Ionic Compounds
Ionic compounds are formed when atoms exchange electrons, leading to the formation of positively and negatively charged ions. These charged ions are held together by strong electrostatic forces, resulting in a crystalline solid structure.
Ionic compounds, such as sodium chloride (\( \text{NaCl} \)), typically have high melting and boiling points. When dissolved in water, they dissociate into their respective ions and create a solution that can conduct electricity.
  • They are usually soluble in water but not in non-polar solvents.
  • Commonly used in various applications like electrolytes in batteries or deicing roads.
Recognizing an ionic compound is crucial as it explains the solution's conductivity properties, distinguishing it from non-conductive covalent solutions.
Covalent Compounds
Covalent compounds are formed by the sharing of electrons between atoms, leading to the creation of molecules. These compounds primarily involve interactions between non-metal atoms. For instance, sucrose (\( \text{C}_{12} \text{H}_{22} \text{O}_{11} \)) is a covalent compound consisting of carbon, hydrogen, and oxygen.
The resulting molecules do not carry a charge, so covalent compounds generally do not conduct electricity when dissolved in water.
  • Covalent compounds often have lower melting and boiling points compared to ionic compounds.
  • They can be found in various states: gases, liquids, or solids at room temperature.
Covalent solutions are non-conductive, setting them apart from ionic solutions.
Conductivity Test
A conductivity test can help distinguish between ionic and covalent compounds in solution. By measuring a solution's ability to conduct electricity, you can determine if ions are present.
In this test:
  • An ionic compound, like \( \text{NaCl} \), will increase the solution's conductivity because of its dissolved ions.
  • A covalent compound, such as sucrose, will not dissociate into ions, resulting in low or no conductivity.
This simple test is often used in laboratory settings to quickly and efficiently identify the presence of ionic bonds within a solution.
Precipitation Reaction
A precipitation reaction occurs when two solutions react to form an insoluble solid, known as a precipitate. This chemical reaction is used to identify specific components in solutions. For example, when \( \text{AgNO}_3 \) is added to a \( \text{NaCl} \) solution, silver ions (\( \text{Ag}^+ \) ) react with chloride ions (\( \text{Cl}^- \) ), forming silver chloride (\( \text{AgCl} \) ), a white precipitate.
  • Precipitation reactions help identify the presence of ions that form insoluble compounds.
  • They are utilized in various applications, such as water treatment and testing water hardness.
The absence of a precipitate indicates that no such ion interaction occurs, as seen with covalent compounds like sucrose.
Chemical Properties of Solutions
The chemical properties of a solution depend on the solutes it contains. Ionic solutions are well-known for their conductive properties due to the free movement of ions.
Covalent solutions, on the other hand, usually involve entirely different interactions:
  • They might participate in unique reactions, like sugar dissolving in water without breaking into ions.
  • Their reactions often involve changes to the molecules themselves rather than forming a different state like a precipitate.
Understanding these chemical properties is valuable in various scientific fields and practical applications, from industrial processes to biological systems.

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Most popular questions from this chapter

The following "cycle of copper" experiment is performed in some general chemistry laboratories. The series of reactions starts with copper and ends with metallic copper. The steps are as follows: (1) A piece of copper wire of known mass is allowed to react with concentrated nitric acid [the products are copper(II) nitrate, nitrogen dioxide, and water]. (2) The copper(II) nitrate is treated with a sodium hydroxide solution to form copper(II) hydroxide precipitate. (3) On heating, copper(II) hydroxide decomposes to yield copper(II) oxide. (4) The copper(II) oxide is combined with concentrated sulfuric acid to yield copper(II) sulfate. (5) Copper(II) sulfate is treated with an excess of zinc metal to form metallic copper. (6) The remaining zinc metal is removed by treatment with hydrochloric acid, and metallic copper is filtered, dried, and weighed. (a) Write a balanced equation for each step and classify the reactions. (b) Assuming that a student started with \(65.6 \mathrm{~g}\) of copper, calculate the theoretical yield at each step. (c) Considering the nature of the steps, comment on why it is possible to recover most of the copper used at the start.

You have \(505 \mathrm{~mL}\) of a \(0.125 \mathrm{M} \mathrm{HCl}\) solution and you want to dilute it to exactly \(0.100 \mathrm{M}\). How much water should you add?

Determine the resulting nitrate ion concentration when \(95.0 \mathrm{~mL}\) of \(0.992 \mathrm{M}\) potassium nitrate and \(155.5 \mathrm{~mL}\) of \(1.570 \mathrm{M}\) calcium nitrate are combined.

A 22.02-mL solution containing \(1.615 \mathrm{~g} \mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is mixed with a 28.64-mL solution containing \(1.073 \mathrm{~g}\) \(\mathrm{NaOH}\). Calculate the concentrations of the ions remaining in solution after the reaction is complete. Assume volumes are additive.

Calculate the concentration (in molarity) of an \(\mathrm{NaOH}\) solution if \(25.0 \mathrm{~mL}\) of the solution is needed to neutralize \(17.4 \mathrm{~mL}\) of a \(0.312 \mathrm{M} \mathrm{HCl}\) solution.

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