Lab 4 Freezing Point Depression

Lab 4 Freezing Point Depression - Section 0 06 EXPERIMENT 4...

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Unformatted text preview: Section 0 06 EXPERIMENT 4 FREEZING POINT DEPRESSION PRE—LABORATORY QUESTIONS 1 The following preparatory questions should be answered before coming to lab. They S are intended to introduce you to several ideas that are important to aspects of the ' experiment. You must turn in your work to your instructor before you will be allowed to begin the experiment. l. a. What is a colligative property? : 1 .w ‘__ y r v 3 A flyyfiffifig O O ‘3 _ .1.» my: 55,. V, . ff; gig. @ijgfly‘g! if. Cry, 5’ a y. , 4 N. f 1 ,. I A My!” V57 «1 ’ )0; 733.11,st 4;: r " b. Which of the following is not a colligative property?. i) Boiling point elevation ii) freezing point depression iv) vapor pressure lowering v) osmotic pressure. 2. a. Define molality. .! , V r , ,4! {V (m {é'f'1"?r'cfi*; : #ewza #74? a £29 MS 01 My e a x _. ; - t a 4-», c, -. Co's" 5. ‘ fl/ is: a: a. s ,4 if? i"? g“? E - V ' ,, ““v ‘ x .'r 2.4 I c I! v m .7 . r 2 n a papery/1,4» —~ (WLO I ,r ,3;— “ My /67 5/1. CW1 ’: WTfi-wwufrr—mg gym 2r 7, f K: p- s b. Calculate the molality of a solution prepared dissolving g of magnesium nitrate Mg(N03)2 in 600 mL of water. (density of water = 1.00 g/mL). { ,, “ 3‘ O'w’i/j/giawfi 153‘” “W Mflwril ’ a ChemlSlS—«Academic Year 2010/2011 4—] Mb» 3. Three aqueous solutions of glucose ( C6H1205) were prepareddin proportions asbshown “N in the table below. Complete the table. Exp Mass of Moles org—Mass Volume Volume Molality Molarity, Density # glucose glucose of H20 of H20 solution solution Solution solution 5' ~ I j A 3.58:, eowm‘“ of??? ’23 25.00mL 27.5mL ; 3m . [’Mgfflé B 9.65% JENNA .029? 25.00mL 32.5mL gamma 237/1" #010,411 C 13.6% ,(Zfiml «02957 25.00mL 35.0mL 6'97“ $09MJ—iloffii Show sample calculations for Exp .#C. I D 66 1 $5 0 q l fileigis: 9. “it”... 3 I 8 rpm I A'> Mafia: 77'5ng [MI ‘02:; mare 3x l0?;i§c3 a . w / ":§ ./ ' I H ‘ 7 I08 : { #10 , ZSMV :2. x ’14“ - 026 5:, InfizéMLfiLfix “wan 02¢, \ MM. 19”?" ? iwif; [9613 c C k?\ 0% I»; Mean $275.1: & ;010 M4,! _/ WIGZ‘IL’XI‘LLE: ' r p :e 3' m "(jg lg W7 W: .3” (5/1105: ‘ om“‘*"‘9é ‘ .O p, l k ‘ a?! j Malar,%wéi rogqme‘ : 33!; m Dwezegz g. a flair 63.; flab/74.5,"; Omani-g: ___""§§___w*2l% /¢’?/?fiezl . Z75”- 22"“; C} Wig/€56 1%ng j . ,zg *stzfl. z? :— i.[§§;§;é [9%. he; 39¢; W“; #10; ngu?‘ / f iém : Mg [’rvtw 33563:? ‘5 Mammy LEE. C ,azskg “" 077 M Mathews, n/L‘ : 6,077 I“ 5 “316123 4. Would you expect the molarity and molality to be the same for a 0.05 % solution of glucose? Explain. 3 6€CQLL$Q gar sségq‘ WW l 5‘" gfitcfisf' (sea berm? ii We 51mg 2 7-3 1’2. 3' ~‘ x,- ‘rii'eA . i r? 0L Wri“? ’«W 1 17 $9 a: ' *5 was ‘ a}: afé. £439 W a {v in e”? ewe p/ L ‘/ 9gu,,&u.§ gm, MC g/LLQ.9§€_ Cheml 5 lS—Academic Year 2010/201 l 4___2 Section 6 Name RESULTS PART ONE: FREEZING POINT OF PURE WATER Temperature ‘j‘u 1“ PART TWO: FREEZING POINT OF WATER/SUCROSE SOLUTIONS Mass of sucrose Mass of water Freezing point of solution Molality of solution Freezing point depression €41,471? 29'7ch x “m” ~ n g L! a 3 _ “may? 4; ; 3542a. F u : r. 5 I, 34‘, r; ' £934“ 3‘ Bf“ Mass of sucrose Mass of water 3 q . . . , «7 We!” Freezmg pomt of solutlon " ~ =~ Molality of solution / ' I 1’, Cl W‘ Freezmg p01ntdepress1on ’ Z. . , I\ >-_ g ’7 ’ 695’s u i .. ,3 = = , o 0’ " “ W“ ,. ‘5 d x w e {3/ 7 Z /‘ . . ' , 1' [‘ILMM 0 ’ (L 1 « e g I 5 , ”( ,“RI-z‘gj . “44",, '53 :21 mag) r 'Olfiofig 1 ChemlSlS—Academic Year 2010/2011 4 7 PART THREE: FREEZING POINT DEPRESSION OF WATWUM CHLORIDW SOLUTIONS 50114—1 7,3?»ch 1321 new? m, 0* ’ erg? My? ‘, ’WW 2,4: «4:,» F. 10/614 M v :7 Mass of sodium chloride Mass ofwater 3 6 6 9‘. e Freezing point of solution “” 4'0 C, at! 4 Molality of solution % ' 2’ ‘ ’ M a Freezing point depression H ' b 5; Mass of sodium chloride Mass of water Freezing point of solution Molality of solution I /, Z 6’ (a m “f; Freezing point depression ‘ 2’ ’ 2 ~”‘ ,— mgz ,I’Wm/ 4 :15. ~ , t. ngn .ggq VM; aging“ I n J ‘ g if g 1 , «7: ‘3 Vim»: r /, 2 ‘24; tee/‘2le ’ SIGN-OUT: ll equipment is put away, and the student bench is clean and dry. Date Cheml 5 l SfiAcademic Year 2010/2011 4—8 Section @635 POST LABORATORY QUESTIONS 1. In the experiment, as the molality of the non-volatile solute increased (either sucrose or sodium chloride) what happened to the freezing point depression? rm 3 ; § ' g’\ ‘ t V ‘ a, 4;, , , ,3 any “ _ 3 #13 ,7 wt} if,» t f “VA”; W04 wart an: I x-“f‘” é: ‘ r * “f C5 ’ f 2. a. Write a balanced chemical equation that shows what occurs when solid sucrose is dissolved in water. , W. v “fat: 4 S33} 2‘ {/4 b. Write a balanced chemical equation that shows what occurs when solid sodium chloride is dissolved in water. iwlif-LJ ,t l ,-, a j if? , 3. When the molalities were approximately the same, which non-volatile solute (sucrose or sodium chloride) caused the greatest depression in the freezing point? {\laé 4. Colligative properties are directly related to the number of dissolved solute particles in the solvent. The mathematical equation for the calculation of the freezing point depression of a solution containing a non—volatile solute is ATf = where ATf is the freezing point depression, Kf is the freezing point depression constant, In is the molality, and i is the van't Hoff factor, which is the effective number of particles in the solution (obtained from the balanced chemical equation). a. What is i for the sucrose solutions? 1 b. What is i for the sodium chloride solutions? 4:2 0. What would i be if MgClz was used instead of NaCl? Chem1515—-Academic Year 2010/201 l 4 9 AT: l' M Wfiwwwwmmm N . 5. a. Based upon tin/e average results of your experimental‘dat’a: What is Kf for‘water? d éuamafi’i 1 4.00 C : jv IQ 'C 673m> Vi gal/+511 q-QOC : '2' K4; ' 2? Ilia M a «4.96: 1...: as 5 96 t \ u: 5C Kg: ‘Sfigm -’ Zég’ifigwx} 0' A 1-K4(/,H‘Im) oath-“#2.: 171°C- : z. K411 i-ngizum C - , f 21%“: K4“ 1.49m ’23; X 1 K9: 3 .N : A11, 9% Zitlgbm ) 4% Erma r: I ’I;'- Valm a ct bed/woo : [.85 (/M 6. Complete the following table. Expected Molar Mass of Mass of freezing molality Substance Mass solute= g water goint °C solution > Fructose 180 Z’Q’7 g 150 g .e-o. 219$°c 0.11 Glycerol, CgHgo3 92 2.8 g 120 g — . 9') °£ . 2 s Sodium sulfate ’5 142 5.6 g /7/ «o. 415’ ’c 0.23 Potassium chlorideZ 3.2 g 150 g -1.06 °C a 25) Unknown compound 98 2.8 g 140 g —- 0. 3'8 °C. - 20 forms two ions in solution 7. A solution of 3.6 g of an acid (monoprotic )with a formula weight of 80.9 dissolved in 250 g of water freezes at —0.66 °C. Would you classify the acid as a nonelectrolyte, a weak electrolyte, or a strong electrolyte? (Show your work)” If ‘ AT: xix/K: m :> J : fl {em ’3 AT: ‘ ,ee°c 1Q: /.5e°% , MD” MA: gIbEXJ—p‘X/X/iz_ '/8M 1 i ‘P A ~ ‘ 5“ E ChemlSlS—-Academic Year 2010/2011 4 10 W a (1 a $ 5% ...
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This note was uploaded on 01/14/2012 for the course CHEM 1515 taught by Professor White during the Fall '11 term at Oklahoma State.

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Lab 4 Freezing Point Depression - Section 0 06 EXPERIMENT 4...

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