Chem- Pract - Answers Pls Use Your School End Point . 1 a ) Volume of pepette = 25. 00Cm ^ 2 Indicator Used = Methyl Orange Tabulate BURETTE READING | ROUGH | 1 ST| 2 ND | 3 RD | Final reading Cm ^ 3 | 25. 00| 34. 00| 29. 20| 24. 10| Initial Reading Cm ^ 3 | 0 . 00| 10. 00| 5 . 00| 0 . 00| Volume Of Acid Used In Cm ^ 3 | 25. 00| 24. 00| 24. 20|24. 10| Average Volume of Acid used =( 24. 00+ 24. 20+ 24. 10/ 3 ) cm^ 3 = 24. 10cm^ 3 1 bi) Con in mol/ dm 3 = conc in g / dm ^ 3 / molar mass con of B in g / dm ^ 3 = 4 . 80/ 0 . 25 = 19. 2 g / dm ^ 3 molar mass = Fe + 2 Cl = 56+ 2 ( 35. 5 )=127 g / mol conc in mol/ dm ^ 3 = 19. 2 / 127 = 0 . 15mol/ dm ^ 3 1 bii ) using CAVA / CBVB = NA / NB CA* 11. 80/ 0 . 15* 20= 1 / 5 CA= 0 . 15* 20* 1 / 11. 80* 5 CA= 0 . 05moldm ^ - 3 1 biii) mole = con ( mol/ dm ^ 3 ) x vol ( dm ^ 3 ) = 0 . 15* 20/ 1000 = 0 . 003 moles ====================== (2a) TEST: C + about 5cm3 of distilled water OBSERVATION: C dissolves to form a clear solution; Turns blue litmus paper to pink INFERENCE: C is a soluble salt; C is fairly acidic (2ai) TEST: 1st portion of C + NaOH(aq) in drops; then excess OBSERVATION: pale-blue precipitate forms in drops of NaOH(aq) which persists in excess of NaoH(aq) INFERENCE: cu²+ present. (2aii) TEST: 2nd portion of C + NH3(aq) in drops and then in excess OBSERVATION: pale blue precipitate forms in drops of NH3(aq). Ppt dissolves to form a deep blue coloration. INFERENCE: Cu²+ present. Cu²+ confirmed (2aiii) TEST: 3rd portion of c + AgNO3(aq) + HCL(aq) OBSERVATION: a creamy white ppt forms. INFERENCE: Halogen/Halides present. (2bi) TEST: D + about 5cm³ of distilled H2O OBSERVATION: D dissolves to form a clear solution. The test tube feels warmth INFERENCE: D is a soluble salt. (2bii) TEST: 2cm³ of D + HCL(aq) OBSERVATION: effervescence occurs, an odourless and colourless gas evolved. Precipitate dissolves in dil HCL(aq) INFERENCE: CO32-, so32-present CO32- present 3i)A white precipitate of barium sulphate will be formed by the instant reaction between barium chloride and sulphuric acid. BaCl2 + H2SO4 = BaSO4 + 2HCl 3b)Iron sulfide reacts with hydrochloric acid, releasing the malodorous (rotten egg smell) and very toxic gas,hydrogen sulphide. FeS + 2 HCl → FeCl2+ H2S 3c)When solid iron filings are added to dilute aqueous Hydrochloric acid, Iron (II)chloride or ferrous chloride is formed , with the liberation of Hydrogen gas. Fe (s) + 2HCl (aq) ……..> FeCl2(aq) + H2 (g)