When chemists work with solid materials, we simply weigh out amounts of solid reagents and calculate mole amounts when thinking about stoichiometry. However, when we dissolve a solid (also known as a solute) in a solvent to form a solution, the solute becomes evenly distributed throughout the solution and we need to know how many moles of solute are present in a particular volume of solution. 1. A solution is created by dissolving 10.0 grams of ammonium chloride in enough water to make 205 mL of solution. How many moles of ammonium chloride are present in the resulting solution? moles of NH4Cl 2. When thinking about the amount of solute present in a solution, chemists report the concentration or molarity of the solution. Molarity is calculated as moles of solute per liter of solution. What is the molarity of the solution described above? M 3. To carry out a particular reaction, you determine that you need 0.0500 moles of ammonium chloride. What volume of the solution described above will you need to complete the reaction without any leftover NH4Cl? mL of solution
When chemists work with solid materials, we simply weigh out amounts of solid reagents and calculate mole amounts when thinking about stoichiometry. However, when we dissolve a solid (also known as a solute) in a solvent to form a solution, the solute becomes evenly distributed throughout the solution and we need to know how many moles of solute are present in a particular volume of solution.
1. A solution is created by dissolving 10.0 grams of ammonium chloride in enough water to make 205 mL of solution. How many moles of ammonium chloride are present in the resulting solution?
moles of NH4Cl
2. When thinking about the amount of solute present in a solution, chemists report the concentration or molarity of the solution. Molarity is calculated as moles of solute per liter of solution. What is the molarity of the solution described above?
M
3. To carry out a particular reaction, you determine that you need 0.0500 moles of ammonium chloride. What volume of the solution described above will you need to complete the reaction without any leftover NH4Cl?
mL of solution
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