An aqueous solution of \(10 \%\) NaCl (consider ideal behaviour of the solution) is cooled. It will allow some (a) \(\mathrm{NaCl}\) to crystallize (b) water to freeze (c) water to solidify along with some \(\mathrm{NaCl}\) (d) precipitation of \(\mathrm{NaCl}\)

Short Answer

Expert verified
Upon cooling, NaCl will crystallize from the solution.

Step by step solution

01

- Understanding the behavior of aqueous NaCl solutions upon cooling

When an aqueous solution of NaCl is cooled, the solubility of NaCl decreases. This means that the salt will start to crystallize out of the solution as the temperature drops, because the solution can hold less dissolved salt at lower temperatures.
02

- Determine what crystallizes out upon cooling

Considering ideal behavior, and that the solution is initially completely liquid with both water and NaCl dissolved in it, as the temperature decreases, the first thing to crystallize out would not be pure water, because the presence of NaCl depresses the freezing point of water. Instead, it will be the NaCl that begins to crystallize out of the solution.
03

- Choosing the correct option based on the solubility concept

Based on the behavior explained in the previous steps, when an ideal aqueous NaCl solution is cooled, NaCl will crystallize out of the solution. Therefore, the correct option is (a) NaCl to crystallize.

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Key Concepts

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

Freezing Point Depression
Freezing point depression is a phenomenon that occurs when a solute is dissolved in a solvent, causing the freezing point of the solvent to decrease. This happens because the dissolved particles interfere with the formation of the solid lattice structure required for the solvent to freeze. For example, in the context of a saltwater solution, the presence of NaCl (sodium chloride) particles disrupts the hydrogen-bonding network between water molecules, preventing them from organizing into a solid and hence, lowering the freezing point of the solution.

Mathematically, the freezing point depression can be described by the equation \( \Delta T_f = i \cdot K_f \cdot m \) where \( \Delta T_f \) is the freezing point depression, \( i \) is the van 't Hoff factor (which represents the number of particles the solute breaks into), \( K_f \) is the freezing point depression constant of the solvent, and \( m \) is the molal concentration of the solute. In this scenario, with a 10% NaCl aqueous solution approximating ideal behavior, the expected result upon cooling is for the solution’s freezing point to be lower than that of pure water.
Solubility of Salts
The solubility of a salt in water refers to the maximum amount of salt that can dissolve in a given quantity of water at a specific temperature. This is an important concept when considering what happens to a salt solution as it cools. In general, the solubility of ionic compounds like NaCl decreases with decreasing temperature. This is due to the reduced thermal energy available to overcome the lattice energy of the solid salt, which is the energy holding the ions together in the crystal lattice.

As the solution is cooled, the decreased solubility leads to a point where the salt can no longer remain fully dissolved. At this point, crystals of the salt begin to form and precipitate out of the solution, which is known as crystallization. This process can be influenced by various factors, such as the presence of impurities, the rate of cooling, and the concentration of the salt in the solution.
NaCl Crystallization
Crystallization of NaCl, commonly known as table salt, from an aqueous solution is an essential process in the understanding of colligative properties and solution behavior. When the solution reaches the point where NaCl's solubility limit has been exceeded due to cooling, NaCl begins to emerge from the solution as solid crystals. This occurs because the dynamic equilibrium between the dissolved ions and the solid salt is disturbed, with the formation of solid salt favored.

The actual process of crystallization involves ions slowly organizing into a regular lattice structure, called a crystal lattice. It's crucial to recognize that only the NaCl will crystallize first in a solution behaving ideally, not the water, because the dissolved NaCl lowers the freezing point of the water component of the solution. This process of NaCl crystallizing out of a solution over water freezing is fundamental to understanding how solutes can affect the physical properties of solvents, like the depression of freezing points, in electrolyte solutions.

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Most popular questions from this chapter

Liquids \(\mathrm{A}\) and \(\mathrm{B}\) form an ideal solution. The plot of \(\frac{1}{X_{\mathrm{A}}}\) ( \(Y\) -axis) versus \(\frac{1}{Y_{\mathrm{A}}}\) \(\left(X\right.\) -axis) \(\left(\right.\) where \(X_{\mathrm{A}}\) and \(Y_{\mathrm{A}}\) are the mole fractions of A in liquid and vapour phases at equilibrium, respectively) is linear whose slope and intercept, respectively, are given as (a) \(\frac{P_{\Lambda}^{o}}{P_{\mathrm{B}}^{\circ}}, \frac{\left(P_{\mathrm{A}}^{0}-P_{\mathrm{B}}^{\circ}\right)}{P_{\mathrm{B}}^{\circ}}\) (b) \(\frac{P_{\mathrm{A}}^{\circ}}{P_{\mathrm{B}}^{\circ}}, \frac{\left(P_{\mathrm{B}}^{0}-P_{\mathrm{A}}^{\circ}\right)}{P_{\mathrm{B}}^{\circ}}\) (c) \(\frac{P_{\mathrm{B}}^{\circ}}{P_{\mathrm{A}}^{\circ}}, \frac{\left(P_{\mathrm{A}}^{\mathrm{o}}-P_{\mathrm{B}}^{\circ}\right)}{P_{\mathrm{B}}^{\circ}}\) (d) \(\frac{P_{\mathrm{B}}^{\circ}}{P_{\mathrm{A}}^{\circ}}, \frac{\left(P_{\mathrm{B}}^{0}-P_{\mathrm{A}}^{\circ}\right)}{P_{\mathrm{B}}^{\circ}}\)

The Henry's law constant for the solubility of \(\mathrm{N}_{2}\) gas in water at \(298 \mathrm{~K}\) is \(1.0 \times 10^{5} \mathrm{~atm} .\) The mole fraction of \(\mathrm{N}_{2}\) in air is \(0.8 .\) The number of moles of \(\mathrm{N}_{2}\) from air dissolved in 10 moles of water at \(298 \mathrm{~K}\) and 5 atm pressure is (a) \(4.0 \times 10^{-4}\) (b) \(4.0 \times 10^{-5}\) (c) \(5.0 \times 10^{-4}\) (d) \(5.0 \times 10^{-5}\)

The vapour pressure of water is \(12.3 \mathrm{kPa}\) at \(300 \mathrm{~K}\). What is the vapour pressure of 1 molal aqueous solution of a nonvolatile solute at \(300 \mathrm{~K}\) ? (a) \(1.208 \mathrm{kPa}\) (b) \(12.08 \mathrm{kPa}\) (c) \(2.08 \mathrm{kPa}\) (d) \(1208 \mathrm{kPa}\)

Two elements A and B form compounds having molecular formula \(\mathrm{AB}_{2}\) and \(\mathrm{AB}_{4}\). When dissolved in \(20 \mathrm{~g}\) of \(\mathrm{C}_{6} \mathrm{H}_{6}, 1 \mathrm{~g}\) of \(\mathrm{AB}_{2}\) lowers the freezing point by \(2.55 \mathrm{~K}\), whereas \(1.0 \mathrm{~g}\) of \(\mathrm{AB}_{4}\) lowers it by \(1.7 \mathrm{~K}\). The molar depression constant for benzene is \(5.1 \mathrm{~K}-\mathrm{kg} \mathrm{mol}^{-1}\). The atomic masses of \(\mathrm{A}\) and \(\mathrm{B}\) are (a) 50,25 (b) 50,50 (c) 25,50 (d) 75,25

The immiscible liquid system containing aniline-water boils at \(98^{\circ} \mathrm{C}\) under a pressure of \(760 \mathrm{~mm}\). At this temperature, the vapour pressure of water is \(700 \mathrm{~mm}\). If aniline is distilled in steam at \(98^{\circ} \mathrm{C}\), what per cent of total weight of the distillate will be aniline? (a) \(7.89\) (b) \(8.57\) (c) \(30.7\) (d) \(44.3\)

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