Part 1 (2 points) hi See Periodic Table See Hin A solution is created by dissolv
ID: 540650 • Letter: P
Question
Part 1 (2 points) hi See Periodic Table See Hin A solution is created by dissolving 13.0 grams of ammonium chloride in enough water to make 235 mL of solution. How many moles of mmonium chloride are present in the resulting solution? moles of NH CI Part 2 (2 points) 9 See H the solution. Molari When thinking about the amount of solute present in a solution, chemists report the concentration or molaritycf s calculated as moles of solute per liter of solution. What is the molarity of the solution described above? ? See Part 3 (2 points) 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 NHACI? mL of solutionExplanation / Answer
no of moles of NH4Cl = W/G.M.Wt
= 13/53.5 = 0.243moles
molarity = no of moles/volume of solution in L
= 0.243/0.235 = 1.034M
no of moles = molarity * volume in L
0.05 = 1.034*volume in L
volume in L = 0.05/1.034 = 0.048 L = 480ml
volume of solution = 480ml