Determination of percentage of an unknown salt by titration A volume of 12.30 mL

Question

Determination of percentage of an unknown salt by titration A volume of 12.30 mL of 0.1000 M NaOH solution was used to titrate a o.595 g sample of unknown containing HCHs03 what is the molecular mass of HC 3H503? (report answers to 4 or 5 significant figures) L Submit Answer Tries 0/99 What is the percent by mass of HC3Hs03 in the unknown? Submit Answer Tries 0/99 In this problem what mass of sample in grams would be needed to deliver about 22.40 mL in the next trial, Submit Answer Tries o/99 In the second trial above, exactly 1.075 g was transferred into a flask to be titrated. If the initial buret reading is 0.10 m, predict what the final buret reading be. Submit Answer T Tries 0/99

Explanation / Answer

a) The formula for Lactic acid is C3H6O3

mass of Hydrogen = 1.00794 gm/mol

mass of Oxygen = 15.9994 gm/mol

mass of Carbon = 12.0107

Molar mass = 12.0107*3 + 1.00794*6 + 15.9994*3 = 90.0779

b)

Number of moles of NaOH = number of moles of lactic acid

Number of moles of NaOH = Volume of NaOH (in L) * molarity

=> 12.30/1000 * 0.1000

=> 0.00123 moles

mass of lactic acid in sample = 0.00123 mol *  90.0779 gm/mol = 0.1107958 grams

c) For next Trial, sample mass = 0.595 * 22.49/12.30 = 1.0879 grams

d) Volume of NaOH required = 1.075 * 12.30/0.595 = 22.2226 mL

Final Buret Reading = Initial Buret Reading + volume used

=> 0.10 + 22.226

=> 22.326 mL

Related Questions

Navigate

• Browse (All)
• Browse D
• Subjects

---

---

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