a Secure https://sessi ion.masteringchemistry.com/myct/itemView assignmentprobie

Question

a Secure https://sessi ion.masteringchemistry.com/myct/itemView assignmentprobiemip-95393524 K Chapter 9 Core Chemistry Skill: Using Concentration as a Conversion Factor A solution of rubbing alcohol is 77 8 % (v/v) isopropanol in water How m Express your answer to three significant figures. any milliliters of isopropanol are in a 82.3 ml sample of the rubbing alcohol solution? View Available Hirn Millliters of isopropandl mL Submit Part C How many lters of a 3.50 MK SO4 solution are needed to provide 85 B g ot KaSO4 (molar mass 17401 g/mol)? Recall hat M is equlvalent to mol/L Express your answer to three significant figures. iView Available Hint(s Volume of K SO Submit

Explanation / Answer

part B

%(v/v) = a definite ml of isopropanol present in 100 ml solution.

77.8%(v/v) = 77.8 ml isopropanol present in 100 ml solution.

volume of isopropanol present in 82.3 ml solution = 82.3*77.8/100 = 64.0 ml

              = 64.0 ml

part C

no of mol of K2SO4 must provide = w/M = 85.8/174.01 = 0.493 mol

volume of K2SO4 must take = n/M = 0.493/3.5 = 0.141 L

answer; 0.141 L

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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