Question

Question 2 (2 points) A 3.2100-g sample of a salt was analyzed for its chloride content as follows: the entire sample was dissolved with distilled water in a 500.0-ml volumetric flask. A 100.00-mL aliquot of the solution required 24.89-mL of a 0.03769 M silver nitrate solution to reach equivalence. A blank correction was not required. Calculate the % (w/w) chloride in the sample. The net-ionic equation for the reaction is Ag*(aq) + CI (aq) - AgCl(s). Provide your answer to two places after the decimal and without units. Your Answer: Answer

Answer 1

Moles of AgNO3 = molarity * volume in liters = 0.03769 * 24.89/1000 = 0.0009381

All Cl- will be consumed by Ag.

So, moles of Ag+ = moles of Cl- = moles of AgNO3

Moles of Cl- = 0.0009381

100 ml of aliquot contains = 0.0009381 mole Cl-

So, 500 ml of sample will contain = 0.0009381 * 500/100 = 0.0046905 moles of Cl-

Mass of Cl- = moles * molar mass = 0.0046905 * 35.5 = 0.16651 grams

(w/w) % = 0.16651 * 100 / 3.2100 = 5.187 % = 5.19 % .....Answer