A 3.236 g quinine tablet was dissolved in 0.100 M HCl and diluted to 1.00 L of s

Question

A 3.236 g quinine tablet was dissolved in 0.100 M HCl and diluted to 1.00 L of solution. Dilution of a 20.0 mL aliquot to 100 mL yielded a solution that gave a fluorescence reading at 347.5 nm of 348 arbitrary units. A second 20.0 mL aliquot of the original solution was mixed with 10.0 mL of 75 ppm quinine solution before dilution to 100 mL. The fluorescence intensity of this solution was 425. Calculate the percent quinine in the tablet. Please show ALL work. If work is not shown, I will not rate lifesaver

Explanation / Answer

let the initial concentration be x ppm.After dilution concentration=20*x/100=x/5;........for x/5 reading is 347.5 nm;......... For the second solution final concentration( =(20*x+10*75)/100=(x/5)+7.5) is 425;........So (x/5)/(x/5)+7.5)=347.5/425;......So x=168 ppm;=(168*1000)/1000000=168/1000=0.168 gm.... percent quinine in the tablet=0.168*100/3.236=5.19%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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