Question

13:17 51439 87% < t go TOⓇ cao , 2NW 220 Question 7 (22 marks) 2 Naz S-03 +Iman Naz Sobota I 03, + I fexcess +H it (5 marks) 5.8emmo Hulle (a) A certain mass of iodine (12) reacted completely with 48.0 mL of 2.25 M sodium thiosulfate' solution. Calculate the mass of jodine used for the reaction. (5 marks) (b) A student performed an experiment to determine the concentration of an unknown sodium thiosulfate solution. First, 5.88 mmol of potassium iodate was used to react with an excess of iodide and hydrogen ions to produce iodine. Second, titration of the 12 produced required 38.36 mL of sodium thiosulfate solution. Calculate the molar concentration of the sodium thiosulfate solution. It (c) A 1.168-g sample of an iron ore is dissolved in acid. The iron is then reduced to Fe2+ and titrated with 38.85 mL of 0.03395 M KMnO4 solution. Calculate the result of this analysis in terms of mass percentage of Fe3O4 in the sample. (6 marks) นาม่อาจ (d) The propyl acetate concentration in an alcoholic solution was determined by diluting a 20.0-ml sample to 200.0-mL. A 50.0-mL portion of the diluted solution was refluxed with 25.0 mL of 0.0975 M NaOH to give the acetate salt and propan-1-ol. After cooling, the excess hydroxide was back-titrated with 2.3 mL of 0.0478 M sulfuric acid for complete reaction, Calculate the mass of propyl acetate in the original 20.0 mL sample. (6 marks) ok of ZO 200ml Soml Hisou TNaOH 25 ml 6-09757

Answer 1

a) uemy of 2.25 m) nasos lo Bemmel 231.55 x 0.36 B3 g 167.55 2Na₂O3 + Iz Na, Suo6 +2Na I (2 :)) nala,&03 = 108mmol = 54mmol MI 0.5089g fezon 4. fez oy = 0.5089 xloo T.168 =143.66 2 2 m= nxMW = 54x03 x 23.99 m= 13.70 57g | Iz. 6) n KIOz= 5. Benmol [0] +5It 6Ht. 3I₂ + 3H₂O *3(Ig) = NIZ 3(5.86) from ① Asor NI ₂ mmol V propyl acetate moi ma Id) encen ont back tetrated with Hasou 2.3 mY of 0.0478md Hysay 0.10994 mmol H₂Soy H₂SO4 + 2NaOH → Na₂SO4 + 2H₂O Nooh = 2(NH, soy) = 2(0.10994) = 17.64 mmol Iz (ее) -0.21988 mmol total alao H added; 2.NI, = 2(17.64) = 35.28 25my of 0.0975 mol NaOH U M = 2 = 35.2BXlomo 2.4375 mmol NaoH 38.36x1032 M=0.9197M Na2O3 .. Nach used in neutralising Reacting) is (2.4375-0.21986) c) 38.85 ml of 0.03395 me kMindy non thanh lonat woh 1.3189 mmol Konoy Nester = n Nach = 2.21762 mmol 5fe2t + Mnou +5fe3+ + Mn2+ +utho M = n = 2.21762 famol = 0.0443514 50ml Meet Before dilution = 6.5948 mmol MiV = M₂ V2 stock ditute Mpe=1 feX MW = 6.sque xo x55.859 mol moi My Zoma 0.04435M X Zoome = 0.36839 Mstock = 0.4435M M=m modar mans Pegou =231.55 glmol fez= 55.65x3 = 167.55 g te foresent in 231.559 fez oy | m=0.9048gI frofplacetate 2.21762mmol v +8Ht the 5 nunar = 5(13164 que lo 5.859 MW XV #m=0.435x102x20x0 4 :: 167.55 g fe -0.36639 "