A solution containing 80. g of KCl in 200. g of H 2 O at 50 °C is cooled to 20 °C. a. How many grams of KCl remain in solution at 20 °C?

Welcome back. Everyone refer to the information given in the table below. A solution was prepared by dissolving 350 g of lithium chlorine and 400 g of water at 60 Celsius. If the solution was cool to 30 degrees Celsius and then agitated the mass of lithium chlorine left in the solution. So we're given in our table our solubility tart for lithium chlorine, we're at 30 degrees Celsius. We have 62.7 g per 100 g of water. And then at 60 degrees Celsius, we have 96.9 g of lithium chlorate per 100 g of water. So given our solubility at 30 degrees Celsius for 62.7 g of lithium chlorine in 100 g of water, we're going to set this proportion equal to our unknown solubility in grams of lithium chlorine in 400 g of water at 60 degrees Celsius or sorry at 30 degrees Celsius we need to solve for X being our solubility of lithium chlorine. So we're going to need to cross multiply so that we have the following equation where we have X grams of lithium chlorine multiplied by 100 g of water set equal to 25,080 g of lithium chlorine hydrate. In our next line, we're going to divide both sides by 100 g of water. So that that term will cancel out on the left. So on the right, we can cancel out water. And in our last line, we would be left with our solubility of lithium chlorine at 30 degrees Celsius equal to about 251 g of lithium chlorine. And this would be our solubility of lithium chlorine in g of solution at 30 degrees Celsius. And also represents our mass left of lithium chlorine 251 g.

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9. Solutions

Solutions: Mass Percent

GOB Chemistry

9. Solutions

Solutions: Mass Percent

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Problem 27a

Textbook Question

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?

Verified step by step guidance

Step 1: Understand the solubility concept. Solubility refers to the maximum amount of solute (in this case, KCl) that can dissolve in a given amount of solvent (H₂O) at a specific temperature. Look up the solubility of KCl at 20 °C in a solubility table.

Step 2: Compare the solubility of KCl at 20 °C with the amount of KCl initially dissolved (80 g). If the solubility at 20 °C is less than 80 g, the excess KCl will precipitate out of the solution.

Step 3: Calculate the amount of KCl that remains dissolved in the solution at 20 °C. Use the solubility value from the table to determine how many grams of KCl can stay dissolved in 200 g of water at this temperature.

Step 4: Subtract the solubility value (grams of KCl that can dissolve at 20 °C) from the initial amount of KCl (80 g) to find the amount of KCl that precipitates out. The remaining amount in solution will be equal to the solubility value.

Step 5: Ensure the units are consistent throughout the calculation. Verify that the solubility value corresponds to the same amount of water (200 g) as given in the problem. If necessary, adjust the solubility value proportionally to match the 200 g of water.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Solubility

Solubility refers to the maximum amount of a solute that can dissolve in a solvent at a given temperature and pressure. For KCl in water, solubility changes with temperature; typically, as temperature decreases, solubility also decreases, meaning less solute can remain dissolved.

Temperature Effects on Solubility

The solubility of ionic compounds like KCl is significantly affected by temperature. Generally, higher temperatures increase solubility, allowing more solute to dissolve, while lower temperatures can lead to precipitation of the solute as it exceeds the solubility limit.

Saturation Point

The saturation point is the concentration of a solute in a solution at which no more solute can dissolve at a specific temperature. When a solution is cooled, if the concentration exceeds this saturation point, the excess solute will precipitate out of the solution.

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