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Today Chemistry Question And Answer by Nobody: 8:36am On Apr 04, 2012 |
Question1) burette of 50cm capacity pipette, either 20cm or 25 (using d usual apparatus for titratione) Reagents for qualitative work (i) red and blue litmus paper; (ii) aqueous ammonia (iii) dilute hydrochloric acid (iv)dilute sodium hydroxide solution (v)barium chloride solution (vi)dilute trioxonitrate (V) acid (vii)silver trioxonitrate (V) solution [AgNO3] (viii)lime water methyl orange indicator Question 2) A. 150cm of tetraoxosulphate (VI) solution in a corked flask or bottle,labelled 'A' containing 2.8cm of concentrated H2SO4 (about 98% w/w)per dm of solution. B. 150cm of NaOH solution, in a corked flask or bottle labelled 'B' containing 3.9g of NaOH per dm of solution. C. 10cm of (NH4) 2SO4 solution in a bottle labelled 'C' containing 66g of (NH4)2SO4 per dm of solution. D. 10cm of FeCl3 solution in a bottle labelled D containing 40g of FeCl3 per dm of solution. ANSWER > QUANTITATIVE ANALYSIS < TEST Solution C + aqueous NaOH + Heat ----------------------- Portion of Solution C + BaCl2 + dil. HCl. ------------- Portion of Solution D + NaOH in drops. Then in excess. ---------- Portion of Solutio D + NH3(aq) in drop then iq excess. -------- Portion of Solution D + HNO3 + AgNO3 solution. --------- Mixture frm above + HNO3 in excess. ------ OBSERVATIONS Evolution of colourless gas with characteristics choking smell which turns moist red litmus paper blue and form white foam with conc. HCl acid vapour. ----------------- white ppt insoluble in dilute HCl -------- Reddish brown ppt insoluble ---------- Reddish brown ppt insoluble -------- White ppt insoluble -------- White ppt dissolved to give a colourless solution. ----------------- INFERENCE NH3 gas from NH4+ is present. ------- SO42-, SO32-, CO32- are present -------- Fe3+ is present ------- Fe3+ is present -------- Cl- ion is present ------ Cl- ion is confirmed. (N.B u write a test scroll to its observation and then its inference. 'cos it is to be written in tabular form i.e make a 3 columns: then write TEST| OBSERVATION|INFERENCE.) , ==================== > VOLUMENTRIC ANALYSIS < A solution acid H2X 2.45g in 500cm3 of solution. Solution B contain 3.90g of NaOH in 1dm3 of solution. Put A into the burrette and titrate with 20cm3 or 25cm3 portion of B using methyl orange as indicator. Record the value of your pipette. Tabulate your readings and calculate the average titre value of acid used. From ur results and information provided. Cal. the: i) Concentration of solution B in mol/dm3 ii) Concentration of A in mol/ dm3 iii) Molar Mass of acid H2X and hence determine the relative mass of X. H2X + 2NaOH===> Na2X + 2H2O. Burrete reading (cm3) Final Initial Volume Rough | 1st | 2nd | 3rd 28.50 | 26.40 | 25.60 | 26.50 1.00 | 2.00 | 1.00 | 2.00 27.50 |24.40 |24.60 | 24.50 (N.B folow d order i.e final reading, initial reading and volume of A used). Average Volume of Acid used = (24.40+24.60+24.50) /3 = (73.50/3) =24.50cm3. A=H2X B=NaOH 2.45g ==> 500cm3 xg ==>1000cm3 :. x= (2.45 *100)/500 =4.90g/dm3. i.) Cb = mas concentration/molar concentration =(3.90/40)mol/dm3 Cb=0.0975mol/dm3. ii.) CaVa/CbVb =na/nb ; Va =24.50cm3, Cb=0.0975mol/dm3, Vb=25cm3, na=1, nb=2. (Ca*24.50)/(0.0975*25) Ca= (0.0975*25)/(24.50*2). Ca =2.4375/49.00 Ca=0.0497mol/dm3. iii.) Ca = mass concentration / molar concentration 0.0497=4.90/molar mass molar mass =4.90/0.0497 molar mass=98.6g/mol H2X =98.6 (1*2) + X =98.6 X= 98.6 -2 = 96.6 NOTE:- 1. Even if you don't Undestand the Question and Answer, Don't worry yourself, Just chill until you see the Question, Its gonna be easy with the above Post. 2. Number 3 Question is not Included because we can't get it.. Try and Get it done yourself in the Exam Hall. |
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