Use the following table for problems 9.25 to 9.28: A solution containing 80. g of NaNO₃ in 75 g of H₂O at 50 °C is cooled to 20 °C.b. How many grams of solid NaNO₃ crystallized after cooling?

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1

Identify the solubility of NaNO₃ at both 50 °C and 20 °C using the provided table.

Determine how much NaNO₃ can remain dissolved in 75 g of H₂O at 20 °C.

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Calculate the difference between the initial amount of NaNO₃ (80 g) and the amount that can remain dissolved at 20 °C.

The difference calculated in the previous step represents the amount of NaNO₃ that crystallized out of the solution.

Express the final answer in grams of NaNO₃ that crystallized.

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Key Concepts

Solubility and Temperature

Solubility refers to the maximum amount of solute that can dissolve in a solvent at a given temperature. For many substances, solubility decreases as temperature decreases, meaning that a solution can hold less solute when cooled. Understanding this relationship is crucial for predicting how much solute will crystallize out of a solution when it is cooled.

Saturation Point

The saturation point is the concentration of solute in a solution at which no more solute can dissolve at a specific temperature. When a solution is cooled, it may exceed this saturation point, leading to the formation of solid crystals. Knowing the saturation point of NaNO₃ at both 50 °C and 20 °C is essential for determining how much NaNO₃ will crystallize upon cooling.

Crystallization Process

Crystallization is the process where solute particles come together to form a solid phase from a solution. This occurs when the solution becomes supersaturated, often due to a decrease in temperature. Understanding the conditions that lead to crystallization helps in calculating the amount of solute that will precipitate out of the solution as it cools.

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Textbook Question

How is volume/volume percent concentration defined and for what types of solutions is it typically used?

Textbook Question

In a laboratory experiment, a 10.0-mL sample of NaCl solution is poured into an evaporating dish with a mass of 24.10 g. The combined mass of the evaporating dish and NaCl solution is 36.15 g. After heating, the evaporating dish and dry NaCl have a combined mass of 25.50 g. (9.4)a. What is the mass percent (m/m) of the NaCl solution?

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Textbook Question

Use the following table for problems 9.25 to 9.28:

A solution containing 80. g of KCl in 200. g of H₂O at 50 °C is cooled to 20 °C.a. How many grams of KCl remain in solution at 20 °C?

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Textbook Question

How is volume/volume percent concentration defined and for what types of solutions is it typically used?

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Verified step by step guidance

1. Identify the solubility of NaNO₃ at both 50 °C and 20 °C using the provided table.
2. Determine how much NaNO₃ can remain dissolved in 75 g of H₂O at 20 °C.
3. Calculate the difference between the initial amount of NaNO₃ (80 g) and the amount that can remain dissolved at 20 °C.
4. The difference calculated in the previous step represents the amount of NaNO₃ that crystallized out of the solution.
5. Express the final answer in grams of NaNO₃ that crystallized.

Key Concepts

Solubility and Temperature

Solubility refers to the maximum amount of solute that can dissolve in a solvent at a given temperature. For many substances, solubility decreases as temperature decreases, meaning that a solution can hold less solute when cooled. Understanding this relationship is crucial for predicting how much solute will crystallize out of a solution when it is cooled.

Saturation Point

The saturation point is the concentration of solute in a solution at which no more solute can dissolve at a specific temperature. When a solution is cooled, it may exceed this saturation point, leading to the formation of solid crystals. Knowing the saturation point of NaNO₃ at both 50 °C and 20 °C is essential for determining how much NaNO₃ will crystallize upon cooling.

Crystallization Process

Crystallization is the process where solute particles come together to form a solid phase from a solution. This occurs when the solution becomes supersaturated, often due to a decrease in temperature. Understanding the conditions that lead to crystallization helps in calculating the amount of solute that will precipitate out of the solution as it cools.