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Ch.9 Solutions
Chapter 9, Problem 52

If you had only 23 g of KOH remaining in a bottle, how many milliliters of 10.0% (m/v) solution could you prepare? How many milliliters of 0.25 M solution?

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1
Calculate the mass of KOH needed for the 10.0% (m/v) solution. A 10.0% (m/v) solution means 10.0 grams of solute per 100 mL of solution.
Determine the volume of the 10.0% (m/v) solution that can be prepared with 23 g of KOH. Use the proportion: 10 g KOH / 100 mL solution = 23 g KOH / x mL solution.
Calculate the number of moles of KOH in 23 g. Use the molar mass of KOH (approximately 56.11 g/mol) to convert grams to moles.
Determine the volume of the 0.25 M solution that can be prepared with the moles of KOH calculated. Use the formula: Molarity (M) = moles of solute / liters of solution.
Convert the volume from liters to milliliters for the 0.25 M solution.

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Key Concepts

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

Mass/Volume Percent (m/v)

Mass/volume percent (m/v) is a concentration expression that indicates the mass of solute in grams per 100 milliliters of solution. In this case, a 10.0% (m/v) KOH solution means there are 10 grams of KOH in every 100 mL of solution. This concept is essential for calculating how much solution can be prepared from a given mass of solute.
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Molarity (M)

Molarity (M) is a measure of concentration defined as the number of moles of solute per liter of solution. A 0.25 M KOH solution contains 0.25 moles of KOH in every liter of solution. Understanding molarity is crucial for converting between mass and volume when preparing solutions, as it allows for the calculation of how much solution can be made from a specific amount of solute.
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Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between reactants and products in chemical reactions. It involves using balanced chemical equations to calculate the amounts of substances consumed and produced. In this context, stoichiometry helps determine how much of each solution can be prepared based on the available mass of KOH.
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