Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Units of Concentration: Molarity and Parts Per... QUESTIONS AND PROBLEMS NAME DA

ID: 1042523 • Letter: U

Question

Units of Concentration: Molarity and Parts Per... QUESTIONS AND PROBLEMS NAME DATE 1. Simple Molarity Problems a. the moles of HF in 275 mL of g. the volume of 11.7 M KCIO, containing 16.5 g of KCIO 0.65 M HF ans. ans. b. the volume of 1.3 M sulfuric acid h. the moles of solute in 2.83 liters of containing 3.75 moles of sulfuric acid 0.18 M silver nitrate ans. ans. i. the molarity of a solution prepared by mL of water and diluting to 750 mL. c. the molarity of 135 mL of solution containing 0.12 moles of sodium hydroxide dissolving 61.8 g of copper (II) sulfate in 500 ans. ans d. the volume of 1.4 M zinc II chloride containing 23.7 g of zinc II chloride j. the volume of 1.2 M potassium hydroxide containing 3.5 moles of potassium hydroxide ans. ans. e. the grams of barium hydroxide in k. the grams of solute in 340 mL of 1.5 liters of 0.55 M barium hydroxide 2.5 M sulfuric acid ans. f. the molarity of 184 mL of a solution containing 71.3 g of phosphoric acid 1. How would you prepare 500. mL of 0.492 M potassium sulfate? ans

Explanation / Answer

(a)

Moles of HF = molarity * volume in L

n = 0.65 * 0.275

n = 0.179 mol of HF

(b)

Volume = moles / molarity

V = 3.75 / 1.3

V = 2.88 L

(c)

Molarity = moles / volume in L

M = 0.12 / 0.135

M = 0.889 M

(d)

Volume = ( mass / molar mass ) * ( 1 / molarity )

V = ( 23.7 / 136.3 ) * ( 1 / 1.4 )

V = 0.124 L

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at drjack9650@gmail.com
Chat Now And Get Quote