Question

Concentration of NaOH = 1600 x10- Molar Sodium hydroxide Nach Trial 1 Trial 2 Trial Trial 4 2.60/1.20 1.40/1.30 28.20 29.10.26.1026.50 Volume of NaOH added Initial buret reading Final buret reading Moles of NaOH added Moles of HCl reacted Molarity of diluted HCI Molarity of undiluted HCI Average molarity of undiluted HCI Precision

Answer 1

Ans volume...of. Naoluadded = final buret reading Onctial buret necdcha=2V for_trial .., DV = 28:20 -2.60 = 25:60 mil frcial 2,_01, _29.10 - 1.20 = 27.90 ml _frecich. 3.1 DVD = 26.10.- 1.40 = 24.70 mL trial y ly = 26.50 -1.30 = 25.20 ML. concentration of NaOH = 0.1 M · Moles of Nach_added molarity, X volume in L so n OLX 0.0256 = 0.00256_moles na 20 LX0.02193 0.00279 moles n = 0:1 X 0.0941: = 0.00247 moles - ny = 0.1 X 2.0252_t = 0.00252 moles Moles of Hčl_reacted = Moles of Naoh added _duning_titration so if moles of Hd reacted are ning ng __ne then ni ani=0.00256 moles bil na ang = 0.002.79 molesi ila - inz: 0.00247 moles - hu- ny = 0:00.252 moles molarity_of_diluted HCl for defferent trecals Mis 0.00256 mol = 0.1024 M 0.025 L M2 -0.002 79. Mols 1.0-LIGMA 0.025 L + M2 = 0.00241 mol = 0.0988 M_ 0.025 - My = 0.00252 mol = 0.1008 m. Scanned with 0.025 L CamScanner

..molarity of undiluted acid for — trial 10 All W 25 17 x M = 250 mL X0-1024 M ml SO Mi = 1.025 M., trical 2 25ML X M2 = 250 ML X 0.1116 M. on Mal = 1.116 M. trials in 25 mL X Mg = 250ML x 0.0988 M ! 0,988 M 1, - trialy 25 ml x My! = 250 ML. X0:1008 M. the Mulet..,00,8 M. Average molarity of undiluted acid elom = Ml + M2' + M3+ My 1.054M precis con Average Absolute devcations for different trials |(1.059 -11025)|+|1.034–1:116)1*161.034-1-008) +(1:034 -0.998)| Scanned with CamScanner

O-009 † 0.082 10:0267 0.036 = 0.0382 sa orecision = 1:034 = 0.0382 M Ics Scanned with CamScanner