In Order To Answer The Following Question Pls Look At The Pic Below Part 1 (0.3

Question

In Order To Answer The Following Question Pls Look At The Pic Below


Part 1 (0.3 point) i See Periodic Table See Hint A solution is created by dissolving 11.5 grams of ammonium chloride in enough water to make 235 mL of solution. How many moles of ammonium chloride are present in the resulting solution? moles of NH CI Part 2 (0.3 point) See Hint the concentration or molarity of the solution. Molarity When thinking about the amount of solute present in a solution, chemists report t is calculated as moles of solute per liter of solution.What is the molarity of the solution described above? SUBMIT ANSWER 01:18 > OOF 18 QUESTIONS COMPLETED

Explanation / Answer

Q1

m = 11.5 g of NH4Cl present in

V = 235 mL of solution

get moles of NH4Cl

mol = mass/MW = 11.5/53.4915 = 0.21498

mol of NH4Cl = 0.21498

Q2

find molarity

molarity = mol of solute / Volume of solution

M = 0.21498 / (0.235) = 0.91480 M

Q3

n = 0.05 mol of NH4Cl required, then, find V

M = mol/V

V = mol/M = 0.05 /0.91480 = 0.05465 L

V = 0.05465*10^3 = 54.65 mL

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.