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The solution of sugar in water contains (a) free ions (b) free atoms (c) free molecules (d) free atoms and molecules

Short Answer

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(c) free molecules

Step by step solution

01

Identify the Composition of the Solution

A solution of sugar in water typically consists of sugar molecules dissolved in water. Sugar, in its most common form, is a compound made of molecules (C12H22O11 for sucrose).
02

Analyze the Dissolution Process

When sugar dissolves, the sugar molecules separate and become dispersed throughout the water. Unlike ionic substances, sugar does not dissociate into ions. It simply mixes with the water to form a homogeneous mixture.
03

Determine the Type of Particles Present

Since sugar dissolves as whole molecules and does not ionize, the solution contains free molecules of sugar. Water molecules are also present around these sugar molecules.
04

Conclusion on Particle Type

The solution contains free molecules of sugar. There are no free ions or separate atoms formed in this process, as sugar simply disperses as molecular entities.

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Key Concepts

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

Composition of Solutions
When we talk about the composition of solutions, we're focusing on what exactly makes up the solution. In our case of a sugar solution, we have two main components: sugar and water. Sugar, specifically sucrose, is a disaccharide with the chemical formula \( C_{12}H_{22}O_{11} \). This means that each sugar molecule is made up of 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. By understanding the composition, we know that the sugar is present in molecular form, not as individual atoms or ions.

Water acts as the solvent in this solution, providing a medium for the sugar molecules to disperse. The fundamental aspect of the composition is that the sugar and water do not react chemically to form a new compound but stay as individual entities. Hence, when sugar is added to water, we end up with a mixture where sugar molecules are distributed within the water. This understanding of the solution's composition is crucial because it defines how the molecules interact and exist together in the mixture.
Molecular Dissolution
Molecular dissolution describes the process of molecules separating from each other when a solute dissolves in a solvent. For sugar, this means that the sucrose molecules "loosen" from their solid state and mix uniformly with water molecules. This is different from ionic dissolution, where a compound like salt would break down into its ionic components, such as Sodium \(Na^+\) and Chloride \(Cl^-\).

In the case of sugar dissolution, the covalent bonds within the sucrose molecules remain intact. They don't split into ions; instead, they become evenly distributed throughout the solution. The water molecules assist this process through a phenomenon known as hydration, where water molecules surround and interact with sugar molecules.

Due to this molecular interaction, the sugar spreads out evenly among the water molecules, becoming a part of the homogeneous mixture. This form of molecular dissolution is key because it helps maintain the integrity of the sugar molecules while still allowing them to mix thoroughly with water.
Homogeneous Mixtures
A homogeneous mixture is one in which the components are evenly distributed throughout the solution. In the case of our sugar solution, once the sugar has dissolved, you can't see the individual sugar particles because they are spread out uniformly amongst the water.

Such mixtures are characterized by a single-phase appearance, meaning the properties and composition are consistent throughout. No matter how much you stir or how long it sits, the sugar solution will not separate into different layers or visible particles. This consistency is essential for many applications, especially in cooking and manufacturing, where uniform distribution of ingredients is necessary.

Moreover, homogeneous mixtures like sugar water make it impossible to distinguish between the solvent and solute with the naked eye or even under a microscope. This is why homogenized mixtures are often referred to as "solutions" in scientific terms.
Non-Ionic Solutes
Non-ionic solutes are substances that do not dissociate into ions when they dissolve in a solvent. Sugar is a classic example of a non-ionic solute. When sugar dissolves in water, the molecules maintain their entire structure and form none of the ionic bonds or charged particles seen in ionic solutions like saltwater.

This behavior affects how the solution conducts electricity. Ionic solutions can conduct electricity due to the presence of free-moving ions, while a sugar solution will not, because it lacks those charged particles.

Understanding whether a solute is ionic or non-ionic is essential in fields like chemistry and engineering because it affects both the physical properties of the solution and how it can be used. With sugar being non-ionic, it creates a very stable solution that doesn’t change the basic chemical nature of the water – it simply suspends the sugar molecules uniformly.

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Most popular questions from this chapter

A \(5.25 \%\) solution of a substance is isotonic with a \(1.5 \%\) solution of urea (molar mass \(=60 \mathrm{~g} \mathrm{~mol}^{-1}\) ) in the same solvent. If the densities of both the solutions are assumed to be equal to \(1.0 \mathrm{~g} \mathrm{~cm}^{-3}\), molar mass of the substance will be (a) \(115.0 \mathrm{~g} \mathrm{~mol}^{-1}\) (b) \(105.0 \mathrm{~g} \mathrm{~mol}^{-1}\) (c) \(210.0 \mathrm{~g} \mathrm{~mol}^{-1}\) (d) \(90.0 \mathrm{~g} \mathrm{~mol}^{-1}\)

In the freezing point depression experiment, it is observed that (a) at the freezing point the solvent begins to freeze. (b) relative lowering of vapour pressure of solution \((\mathrm{X})\) is more than \((\mathrm{Y})\). (c) the decrease in freezing point of \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) (X) in \(\mathrm{C}_{6} \mathrm{H}_{6}\) is lower than that of \(0.1 \mathrm{M} \mathrm{CH}_{3} \mathrm{COOH}\) (Y) in water. (d) the decrease in freezing point in the two solutions is same.

A \(0.2\) molal aqueous solution of a weak acid \((\mathrm{HX})\) is \(20 \%\) ionized. The freezing point of this solution is (Given \(\mathrm{K}_{\mathrm{f}}=1.86^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}\) for water \()\) (a) \(-0.45^{\circ} \mathrm{C}\) (b) \(-0.90^{\circ} \mathrm{C}\) (c) \(-0.21^{\circ} \mathrm{C}\) (d) \(-0.43^{\circ} \mathrm{C}\)

When attraction between \(\mathrm{A}-\mathrm{B}\) is more than that of \(\mathrm{A}-\mathrm{A}\) and \(\mathrm{B}-\mathrm{B}\), the solution will show deviation from Raoult's law (a) positive (b) negative (c) no (d) cannot be predicted

\(\mathrm{K}_{2} \mathrm{HgI}_{4}\) is \(50 \%\) ionized in aqueous solution. Find the value of \(i\).

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