10/22/17 5% 3rd attempt Part 1 (1 point) hl See Periodic Table See Hint A soluti

Question

10/22/17 5% 3rd attempt Part 1 (1 point) hl See Periodic Table See Hint A solution is created by dissolving 14.0 grams of ammonium chloride in enough water to make 355 mL of solution. How many moles of ammonium chloride are present in the resulting solution? 261 moles of NH CI Part 2 (1point) O See Hint When thinking about the amount of solute present in a solution, chemists report the conce is calculated as moles of solute per liter of solution. What is ntration or molarity of the solution, Molarity the molarity of the solution described above? 737 Part 3 (1 point) 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 NH4C mL of solution 2nd attempt SUBMIT ANSWER + VIEW SOLUTION 01:15 > OOF 15 QUESTIONS COMPLETED

Explanation / Answer

1)

Molar mass of NH4Cl,
MM = 1*MM(N) + 4*MM(H) + 1*MM(Cl)
= 1*14.01 + 4*1.008 + 1*35.45
= 53.492 g/mol


mass(NH4Cl)= 14.0 g

number of mol of NH4Cl,
n = mass of NH4Cl/molar mass of NH4Cl
=(14.0 g)/(53.492 g/mol)
= 0.2617 mol
Answer: 0.262 mol

2)
volume , V = 355 mL
= 0.355 L


Molarity,
M = number of mol / volume in L
= 0.2617/0.355
= 0.7372 M
Answer: 0.737 M

3)
number of moles = M*V
0.0500 = 0.7372 M *V
V = 0.0678 L
V = 67.8 mL
Answer: 67.8 mL

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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