01 Question 1 point) a See page 320 When chemists work with solid materials, we

Question

01 Question 1 point) a See page 320 When chemists work with solid materials, we simply weigh out amounts of solid reagents and calculate mole amounts when thinking about stoichiometry. However, whenwe 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. v6th attempt Part 1 (0.3 point) hi see Periodic Table ’ See Hint A solution is created by dissolving 13.5 grams of ammonium chloride in enough water tomake 285 mL of solution. How many moles of ammonium chloride are present in the resulting solution? moles of NILAC Part 2 (0.3 point) See Hint 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? Part 3 (0.3 point) O See Hint 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 NH4CI? mL. of solution

Explanation / Answer

Part 1

molar mass of ammonium chloride = 53.491 gm/mol

then 13.5 gm of ammonium chloride = 13.5 / 53.491 = 0.2524 mole

0.2524 mole NH2Cl

Part 2

Molarity = no. of mole of solute / volume of solution in liter

no. of mole of NH4Cl = 0.2524 mole

volume of NH4Cl solution = 285 ml = 0.285 L

Molarity of NH4Cl = 0.2524 / 0.285 = 0.8856 M

Molarity of NH4Cl = 0.8856 M

Part 3

volume of solution in liter = no. of mole / molarity

no. of mole of  NH4Cl = 0.0500 mole

molarity of NH4Cl = 0.8856 M

Volume of NH4Cl in liter = 0.0500 / 0.8856 = 0.0565 liter

0.0565 liter = 56.5 ml

56.5 ml volume required.

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