posted by lara on .
0.1802 g sample of a chlorocarbon compound was analyzed by burning in oxygen and collecting the evolved gases in a solution of sodium hydroxide. After neutralizing, the sample was treated with 39.50 mL of a 0.025 M AgNO3. This precipitated the Cl- as AgCl and left an excess of silver nitrate. This excess required 27 mL of 0.006 M KSCN in a Volhard titration. Calculate the %w/w Cl (MM = 35.45 g/mol) in the sample. [Answer in 2 decimal places]
Cl- + AgNO3 → AgCl + NO3-
AgNO3 + KSCN → AgSCN + KNO3
millimols AgNO3 initially = 39.50 mL *0.025 M = ?
mmols KSCN used to back titrate = 27.0 x 0.006 M = ?
Subtract initial mmols - back titrated mmols = difference which is the amount AgNO3 used; i.e., it's the mmols AgCl produced.
Convert to mols, change to grams Cl, calculate % w/w by
%chloride = (g Cl/mass sample)*100 = ?
how do you convert mmols to mols?
Divide by 1000.
mols = millimoles/1000