Question

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.a) A solution is created by dissolving 13.5 grams of ammonium chloride in enough water to make 315 mL of solution. How many moles of ammonium chloride are present in the resulting solution?b) What is the molarity of the solution described above?c) To carry out a particular reaction, you determine that you need 0.0500 moles of ammonium chloride. What volume (mL) of the solution described above will you need to complete the reaction without any leftover NH4Cl?

Answer 1

See the answers below

Step-by-step explanation:

13.5 g of NH4Cl was dissolved in 315 mL of water.

Moles of NH4Cl present in the solution = mass/molar mass

Molar mass of NH4Cl = 54.49 g/mol

a) Moles of NH4Cl = 13.5/54.49

= 0.25 mole

b) Molarity of solution = moles of solute/volume of solution

moles of NH4Cl = 0.25

volume of solution = 315 mL = 0.315 L

molarity = 0.25/0.315

= 0.79 M

c) moles required = 0.0500 mole

molarity = 0.79 M

molarity = moles x volume

volume = molarity/moles

= 0.79/0.0500

= 15.8 L = 15,800 mL