Solutions (Chemistry), Study notes of Chemistry

Download Solutions (Chemistry) and more Study notes Chemistry in PDF only on Docsity! = ® Solutiona 9 22 Solkbhim.3 1A a ems Feneous Mixture of Twe ev mere BubRtency | Rame ov ol 4fe vent Prysical Prager , ay SotthHim ! 2 yA 4 wl a pare af Salute and Solveyt Components , “4 Nel + LO ——p Balt Soluttom v J Solute Sol vet present i” lavger amount , Dinara Soluticy 13> =A Seluttm coms ming enly one Rolute digsalved iM a Salvent {SB calle loimiar Solum Importance of Solutions x Ditferent 3 olustioms nave aU ff-e rent pyeperH”) and hence are put +o different Used. “9 @ «) Bras —> Cut =y Wi) eran silence Cu, ty Mr (rp BrMze — 5 Cursgny @ © lpm af Aluoricle jms wm Water prevents ho-0 $4 decay. a A\uevide ions causer the tooth \s ar fo become motted. Arig a cent redt od of 4 luowcle joa (nad) i a ore ch iy yer poison. Clarificahion af solvHons i> (0 m the basis ot, Physical Bfodk af Solute and Bolvent: Various types of solutions §,No. | Solute | Solvent | Type of Sol. Examples “SOLID SOLUTIONS (Solid Solvent) 1 Solid Solid Solid in Solid Alloys (brass, German silver, bronze, 22 carat gold etc.) 2. Liquid Solid Liquid in Solid | Hydrated salts, Amalgam of Hg with Na 3 Gas Solid Gas in Solid Dissolved gases in minerals or H, in Pd ‘LIQUID SOLUTIONS (Liquid Solvent) 4. Solid Liquid Solid in Liquid | Salt or glucose or sugar or urea solution in water 5. Liquid Liquid Liquid in Liquid Methanol or cthanol in water 6. Gas Liquid Gas in Liquid Acrated drinks, O, in water ‘GASEOUS SOLUTIONS (Gaseous solvent) 1. Solid Gas Solid in Gas Todine vapours in air, camphor in N, gas 8. Liquid Gas Liquid in Gas Humidity in air, chloroform mixed with N, gas 9. Gas Gas Gas in Gas Air (0, + N,) (i) Defending uP™ Rye awpunt a) solute pre cent My ape c liad aneunh @® Unsatorvetech — Solution, © eatureted — goto © Bupevsetureter] Foludiony of <olven. Solurd oy L Vv Solute Solve aA Wy o We Wer Wy = 7 = MM? a Nyper crt moje MA Deere mele + drastton x %" 2 _ ) Molar- Mey ey My .0 7 A ey Mole - frecHon (0¢) ‘9 It fs Pehined 3 The weatte of dhe Numbey <f, mols a a Component +o cthe tote number a2 moles ap alk the Comyorrenta — fev Q Binary Selution o ° 4 Solution = Soluke Salven vy TY ge _meler oa oy “ip, MA Mole Fraction of selvent tn) = “A A “Na +p Wat Ne Natne “Ata, Marne > (ete!) Eye eT . © : A Strrengih ‘= Me Strength ah, Qa Solutioy Sg -- Aebined a he Amount af Hie Solude jy grams preetnt jm one Litre Awe) af the Solution, Btyreng th =. Mey of the Solude ln qrams Yo huame the Solution in Ge oy a) . Lojhyes B.3) ® melas molarity of @g soluHon 18 dof ined! yo A+A?he Number af me (2 af dhe Solute CU azolved per Utre af, cSoludion, Dt tg demoted by m, Melacity = MA = ), % ) lar? = Wr x Jt Me y oo Ve | if vig i mite cw? they Molar %y > Mt 2 . -1y x 1 @ i ; (ou) of A Sofediod Jz 0 Mo\elu _ Mo vi ne Ael,j d ay te mu mbev ad yr o/eh af . me ot S Auge Al a30lvecd iy | Ka ag the solv J+ jg denoted b4 ™. Number of, mio ley af Pre folute Malet may af the solvent in kg | l VYN4. _ Mx ola = — lo mC (Qo fre mars af solvent 16 MM grams, Molatity = We y tl ie wi) olati = We x x {00 " Y My Wig) [ooo Moladi = be x ooo d My W (kg) GF Normeatity to Neovmattt af a Solubtm 1B Aehired %& the umber a} g ram equrvalents af the Solute ou 2836lved per Life af given SolUrioyy, Th 12 devodeol by Ni. Pee m"s Hare; Vy = Vy V9 M\vy + MyVg MY 5 Mp MU = m, © +¥2) mM = MW MV e Vi Vy The mM o\ Q wo co | Woot) vK\ choy te acd or7 between cw o ypeactants ‘8. FF Reladimstsp between Maelatly and molarity . Molar ity (M) Meany M mole of solutt are . preeevb ty lsoom| a coluslo, tP chamoth 2 ot, then wees af cout = |900 And May cxf Solute = MM 2melebhe - 2x2 - Wy rs Mags of Solvweyd = Mat af SolukHion — mang Soluce W | = (loved — MMy)) 4 mee Wmiol-ty af} Solute K% lov® meus of colveud 14 graves Ww M y |ooo | cood — MMg n of ethylene glyco! (C,H,O,) and water In a solution contalning i\ > Ex A Calculate the mole fractio 0% of C,H,O, by mass. Solution. 20% of C}H,O, by mass means that 20 g of C,H,O, arc present in 100 g of the solution, i.e Mass of solute (C}H,O,) = 20 ¢ (W2); Mass of solvent (H,0) = 100 - 20 g = 80 g@41) 2or/- Molar mass of C)H,O; = 62 g mol"; Molar mass af HO = 18 g mol"! (m) Wa 220 . ls = Fo No. of moles of C,H,O, = 2° - 9.329 a No, of moles of HO = “2 = 4.aaa (=) w = \wo m, & ve mS | Mole fraction of C,H,O, in the solution = —"CzHs02_ 0322 voes , % = NegHg0, * "4,0 0322+ 4404 , —hL 24, = |-Ftg , Wy te =I, Mole fraction of H3O in the solution = | - 0-068 = 0-932 Calculate the molality and mole fraction of 2:5 g of ethanole acid (CH,COOH) ln gmol"!, Kx RAS 75 g of benzene. We wv! Solution. Mass of solute (CH;COOH ) = 2-5 g Mass of solvent (C,H,) = 75 g = 0-075 ke Molar mass of C,H, ey g mol"! Molar mass of CH,COOH = Calculation of molality ; Moles of the solute (CHRO = — 258 = 00917 2.) g& mol My Molality = Moles of tescie = 20417mol _ 0:556mol kg”! /Malolit 2 Ve Massof thesolventinkg 0075kg wR) ‘Calculation of mole fraction : Moles of solute (cH,c00Hn) = 00417 (calculated above) Moles of solvent (0 4) = 158 et ons" 78g mol~ 1 y ’ 2 n = Mole fraction of CH,COOH in the solution= ——CHsSOOH 00417 = gga tg a a EN Mcucoou + "cy, 0-0417+0-961 ay Cx 229 Co cafate Melati hy a) a: 4 =f ethansic ado m to 9 of, bengeme . Exc ° oft) Ang Molay wie, of Cy Hy Sy CAtycomtr) Aw. 4d) FAXIg) = 6sg/mel GlIVyp = 7 59 May S olute = asa mags of Wr x loon Mz Wy \ Molly — 9:5 56 mil Molatity = ae x ier = 08 wel kg Melatify = OSS 6m TQ- 21. Calculate the mass percen, of ber ‘CCI,) if 22 g of benzenc Is dissolved in 122 g of cart tetrachloridee ? nd carbon tetrachloride (CC1)IF22 8 Ans. Mass of solution = Mass of benzene 4 Mass of carbon tetrachloride = 22 gr l22p= lg Mass Percentage of benzene = Mass of benzene x100 = 228 * 100 = 15-28% Mass of solution ~ 44g Mass Percentage af CCl, = Mass of cc, Mass of solution 100 - Mass percentage of henvenc = 100 — 15.28 = R4.72% EQ 22. Calculate the mole fraction of benzene in Solution containing 30% by muss in carbon tetrachlaride, Ans. 30% of benzene in Carbon tetrachloride by mass Means that, Bots, Wy = Be, lor = Fo Mass of benzene in the solution = 30 g: Mass of solution = 100 ¢ (w) Mass of carbon tetrachloride = 100 - 30 g = 70 8+ Molar mass of benzene (CylI,) = 7K y mol (™) Molar mass of CCly= 124.4% 35.5 = 154 ¢ mot! (in) 122 p 100 = = «100 = B472% M44 Alternatively, MASS percentage of CCl, = No. of moles af benzene =— Mass -— 20k = 0385 =) Molar mass 78g mol Mao No. of moles of CCl, =—Mass_ _ 0g = 0455 <t) Y) Molar mass 154 ¢ mol-! NM IG 7 = ” nn" . Mole fraction of benzene = Moles of benzene = = 0385 _ ' Hh Total moles in the solution _ ORG . Mole fraction of CCl, = | — 0.458 = 0-542. T@- 23, Calculate the molarity of each of the following solutions (a) 30 g of Co (NO)),.6H,0 in 4-3 L of solution (b) 30 mL of 0-5 M H,SO, diluted to Si) mi. (Atomic mass of cobalt = 58:7) Ans. (a) Molar mass of Co(NO,),.6H,O = 58-7 + 2 (14 + 48) +6 x 18) mol! y Ww, 238 °g = 58-7 + 124 + 108 g mol"! = 290.7 g mol"! (ms) =-— Mass _ 30 ¢ _ Ww No. of moles of Co(NO,), 6H,0 = M8 mas 007 g port = 01082 (a Volume of solution = 4-3 L . No. of moles of solute 01032 mole Molar ids 2 hoy x | =. Eee - 00: Molarity of solution Volume of solution in L 43L 24M , my MY (b) 1000 mL. of 0-5 M H,SO, contain H,S0,= 0-5 mole osm O-. 5 mole 30 mL of 0-5 M H,SO, contain H,S0, = a x30 mole = 0.015 mole Tw econ) Volume of solution = 500 mL = 0-500 L oy MY = Mg Vo_ . No. of moles of solute — QUIS mole where re { sol = ee eM. = oS Molanty of solution Volume of solution in 0500 L. My oe a Vy > Boml V_ 2 Soom m= ?. Sohubi Ui ty 29 Selubi Lity of @ subsdem(e ix Re Maeartnuwm amount af the Substance at COM be dizselued In a SpeciGied ewmoum#p of A Solvent at given Heurperadure, Feetors on ul dh Zolublity olepenals, @ Nature of Solute aucl Solvent © “Temperature © Paeesure . ooluli Livy ad a Sella Mm a Rigg (3 debined ay Hie Maximum amouyt of He | Colid Golute) qram vlach can Aissolve ty Jrog of the —Riquid Getvemt) +o Pow the Baturoted olution at that Particulay tow. | foctor4 offetrg Selb tity af Q Solfol mM 4 Riguiel « (G) Nadu ye of Soludt aud Solvem} - ike ai g3alven Aulee) Ho + Nac — Soluble frlay Polae Giunic) Hyo + $9 ——9 Nob - Soluble fola hen -polay ® fleet of Fev pevertare mM the Soluled ty a Bolid In a Riauid, C Endotermlc disselution: ! eee es Seite of Selvend +Heeft ———> Product Fee lg'y le chadeliey'y primey ple T &® Reaction wil slut ™ fev werd directory Sdlubi L+y @ “4 Solute 2 AN, 4 kee, , Wad, KCI eFC (ti) Crothermic dizsolution + + —_—_—_ procluct +- Heat prind pr Solute t+ Selven Hier’ Appey “4 be an kt 1 +, Reaction wilt A pack werd Civectinn: Solubr 2h ) es Solu 2 41, VR (n"ho ete Gi) “Those Sub gta Cay whose Solubrlity a} 023 Mot tcrease oy olecyease Continuously “4 MBS UH © Solulntit, creases upto BAUeC Hit ofecvests Masa, + [etre 206s) +t Saye Het y 5 Due +o Any drat ton energy Ls retest | Transition Trenpe rature !> The Te wmaperatu re at which me Form af the substance changes to auothey fs Catled +transition fom perature , Above 32°C Os NA 5% 10H pg Sey Pela 324°C © APA @ ct ab, Preszure to * Pressure Do No significant effect oy) the Fbulor Ls Selicls \y riquids, feesms9 Selid aud Riquds aye Megtigibly Pecked log Pissu re wy Mey Aye highly tn compressible, Crrapisicad Se pMecenteadi ar of, Hevnvyls Jad Pa = Was FH ta Warten setaudkre 40.3 \ ye 1.07 nN, | ans TRA Formaldehyde | 208 | LR tp x | nos famaa | Perinaltehyac p | 3% 10 ob, | 208 | wag | Methane 208 | Ogis eo, | 303 | 46.82 | Vinyl chloride 298 | aon ‘ I Fyom +e table, the Fo] lowing VEAULFA way be dyaun. a) kw 18 @& 4unttio af, matere af gos va / _[ ra () KH & satubltity < Tenge a dimitetions ah Henry's law'o @) The prersure “Shout be lo andl the “Pempe ractere Should tre high Le the. gos Clrautol behave Like ay dead oad () The 93 Shut not urclergo Compound doormat witty dhe solvent. «4 CoeHy + be, —? Cae, +40 NA +Ho —— 9 WOH Hu +Ho — 9 Hea + Ci @a) etc. Ap ications of Henry's lawl > oC “Im the. produeti ov) af Carbeated beverages: Te Wereapre oe solubility of Co, ty ; . okt Ari ley, Secla watey ay usel] ax beey ete the Lo wt le AYE Seatecl trdey igh prcerurr, ~ wong ov) Opent na, Abe bottle Cor bubble Aveo whey fle bottle 13 opened to ady the partial Pretsure of Co, above the Golubiiom ‘clecreasrey pg 2 pesult Solubl Lite | Aecyeapey avd hence Coy bubbley oap - : @ Im Aeee cen Siving . 1 fo Scuve diver . \ Se Oya Solve, © 1 Seq —wotey . | TOS . fe Pe . Setublalty off @ ats | . & , Oy, Ny { used for eee nedabo lism ’ Bubbly af My —7 Cause Benes ™m blewd ov “pe Comprarsion 38 reR yer, ‘To avotd henels Mees civ ie olbleded with He - - . He — Listy 4 Naw Divers cylinder Ri 4» Bay, U9 SERY, He (3 Ute fo ge dUce re Solubl rity 4 Nz 1M bowel thot Cauzes — bevoly 00 Planati'ey) + He (3 Mesd Li Hasire Iw fPorettc dbl, bg a Vapor frepune:—> ef a Aine [Selotton j4 The Pieasuve auertd b the Vapouns iy quill brium uith +e Aa quird | Solutio et a ParHelew deupe redure . r Va pour - Pressure at the Liquid /folution , factors ffe Cty Uapeuy P ress ure is Vapour - pressure of \ ao _ | a lene. Bp hyde racking Vapour Pressure al A 4c al S olubonsit R aoult's lau.\> Poy a Soluttiny al eS Volatile Alqwds, the pardtret ve pour Pressure 4 ~ \ ‘ af gach Component im fhe golutisyy th iret, Proper at to 4a mole fraction. foot 13 ter component |, fr ts avel Py > PP ty for Component 2, Ps — Pe Hy Fret =A A cer Ning to “Dattoys [aw al Pardok 7 Pressure), Oregas = P + Pe reper 2 Pr Pym | 7H, = | er . 58 i ° brh Py d->te,) TV y= 1-9 Propet = Pr Ps, amet Ha . 8 © 9 Cree = Py + Ce A, Trowe = Pie + Pe Guoe) Tore = Pir, +i — fo pe = PR (PPE) >«, © lave. Gorapuri ca — ye preren bation for aut FF al y= 4 ° WM,2 0 WH, wole- freti™ o 9 Pi, “b Pe ate fr = Pit Pa = § = 5 = > | « —~] 3 Mele - Azosti@M A=! — ey «x, =o Pieter To er. Gre ple cal ¥ pPreeen padres 13 Wire volatile teen Component 2 Covmpoveu + \. Noe idead Solutions GP, ae Pron Lhe Pe Ak Gi) bH #0 (iti) bv #O Ht Pay Nos — Stead SolusptoyS Show . wry OS pesittve - Levi acto @ , g-o~ O [om| asm] frasume Mteractiond | low] Selvenb- Solvent aye Weakly te Solute -Solstt ‘teraction, OQ pv @® © db. @® @ tr > Pity 2 Pa > Pam rr > Arh) Neg adive - deviaHim PAG 5 L Ate BB lewl emi tral jasume aduayt -¢olvent Wterertions age Bryon ger they Sselude — Solutt mrerections . sv © @® SH & @) if < Ps, , ix fe. pr <i + fe) ! Crean Le fev SolupPiong Show) devi choy Ayre given loelow , @ “1 9 Pos Hye pP Qo K-eh + csy CH, — Uc, + CH Yer 4, © cH op, TAD Q Chow + tho © Chis cH ot + 140 &@ CCl, + CHCl @ CCl, 4 OI Gy cel, + oy Crawgleh for seiatiods Sheeting vegeta doviatioms aye given below’ Oct G OCH + Systrociasy © cy Ban, + o° GQ Ha + wate @ Halo, + Ho . @ CHa De, +4 cH cl, Seplavotion! > at fra cotyepic Ory Constanynd be\ina aii xt x23 2D Are the +4 98 “fh Wriptares which Iyave a cel init compo sittor1. and boiling ke a pare Dgqurd, ‘ps called 4 cons row bo! ling minture oY 7 Azeotrspls Wier aye on simply ay 4 eotrope. these are formed by Non - ileal Solutary @ Minima boil Age otyopes 13 ave “forred) by tno5e Liquid pairs which shows posit ve devieldm Ayom ideof behaviouy, Suda gee trop have beiling polyts lower Proetion of Seluge ju the gelutin, eS pp = FP The yedutioy fy the Us pouy pressure of solvent} bP, 18 give as, PPr = Pi - Po Pp _ 4 P*, APL = Pj Cla) = C/A. PhS PHP = PP oth f,— Pi - r * P- Ph Me ° Naten I 7 f Y oy o nh) Ma. wt © ”) Mor valaHip yA] = Pure - Selvent det p= { Om <Scapin fevcloney aff Solvent. particle 1s due 7 mm Yrlatle Setute # OstiusetA ard weatl<ev wetrod (3 Upech to clutermine H~ rvelatve lowering 4 Vapouy pressure, (2) Elevation iM) bei ling Porm + (6%) bith actlitim af nm- velatle solute, Vapour pressure alec reasey auch Denice boiling point — ucreadey . <> | re] Hoe fread M — oo xy not Nat. oe c\ ™ M . “eH H Try = loo . * Ho oo «= b= No (Cet) 6TF = (200 —— Ay, = Kx We tolnere, W, = Mas al slut Ma = Molay Mads of Solute mars of solvent wy = oll k Propartion att ty cmetant |kvewn % b ring h = Molatk point elevatHm c™Mistent oy i rc tenstayh elevator coistenyy ov (- bu blios cop! a L KL => RTs = Mm, « T, ot [ “bat “non £ v [000 Ay } 000 Avep H My; luheve, Te = botiin . ~ 4 peiut =f pure solvent 4, = latent he ad of, Va Pouy hsation per wm “4 the Solve Duy H = Oe tet iv vapourisadiery Per mole g a\lvern Mm = Melaw- mary =f dye solvent: R = ge cositt eB BM Tete! [18 Jv witht ssa are iy y f + It cadole Yb | ; 4 ey oles d CG v Oven H are Wy Calori 8 ) Cyaplricah ve prepeutation fer Lo ck-F3} oO”) Hreeging Point - Lote f “ Sia wt ceive - a io) wy “x J gow Newt ; Le e ofa : aa 2 ys a : 2 Do > -ébT aac : > sO | vu _— Tee vate Asmosixg ancl osmohe Prepure to Osmo 313 12 The Process of} Plow of Solvent Woleculen rem pure Solvent do Selubim ow Prom selwtt™ af ower comceutratin +e hot eecontrratrny Phydugt, a Solttio™) af a nad oe ¢ i. evwy-ealal Qem e (1035-wares) . . _ — Ce [7 OF ) vesia - ref le watd we oh Wey lyst eb se cvecl by Abbe Nollet. 2 membrane 1% callecl osvqes/'s , Boe matural Feil permed le Membranes aye A ANN a} bolaclaley _ Cell “haa “3am hy move less &, Osmesig = ——___> Membraye ete, OswoHc - Pr epsuve Cr) ‘5 The Pressure veq uiveol “pe yur atop the Alow af solvevt cue +s Osmosis 3s called osmotic pressure’ &1) af tHe EoltHom.- Ty = RT a oe RT OS CRT V = Mer = WeRT fF HM RT a 41 A Vv My _ UW RT = Ww RT = A RT TS Se vom " ™ KT OSwetic Pressure —s Ig of eterminedk by Aidde rep welds hag good oe Rk Berkeley and Havtley's Metyod, (43) Tsotowic solution? Two solutions are Coie) joodonic i He exey{ the same osmottC pre-sure cy Same Cmiteuryatrans) ba]. Sele pure Nact ta psetonic with “Pc z wacl Auman RB lS, fer o:st | “fenetl 4 @Q Endosmoesi EO SNE TS © Exosmects , ® Hypertonic solution © Hypodonic sctution, @ Hemolysi3 @ Plarwolysis (raphering af te) creation Cell : welling of ail. Shrinkelvg aR Cel} Some Pheromena oy Hie brie ab ocmosig O Rows Monger) Shy iMk Uyto prekle Whey plesed yo brime. Revexse ©swosis, Pesatinatim of cee & Wetty 1% ome by TRVEYSE OSwWos''s, “Water outlel sem Dex eunation- Plan} Abnormal Molay Mattes! bue fo arsottatton a di zsouasing a Woleculey, the mMelav ways of 9 substance Colcatatecl yom ifs col lig active property [8 eifyey lowey oy higher tian +he expe ctec} CY VYormol palue , abnovwal molay mess, rf Such molay mass jz Calleo/ Coll lgaHve PMOperty x a d Molar = meq ok the solute DiBedation - kc) eq 5 \t +c fAisz0ctation . ‘ a Oh <> €4,Coot Cy — ON A CH, Cool € , loo Mp ‘ , a LF CH, +: @ Account foxy We 2x0 bet of, UZ50 0 att oy NO VBA Yan't Hoff ivdyaduces! a -Pactov 9 4, Known as the vaid Hoff Pactoy. f= Novmeal Molav mags co 0 Abnemeal Molar med = “Tothag number «fh Mole of Particles al tev A330 chadtreyy / leo Clady) « Yo Juumbe af, tel) al? particles lbefoye AY8oct ation / Wyss 0 crating Me & —_ Yio ys Ww dis Me —_— oA Mo WC a Assoui ation Di 450 chin . 4 ~ p A<)) fa Mec A >| Mo » Me 42> Mce< Mo Mo Me > Mo Moi fie? equations Poy Co\lig ative properties al-tey jw cls om of va + Hofefn Factey e Relative ower ing al vapour pyenur of Salven t x ”) Meh 8 el a, a rir Do Th = ukypn eo elevating af berlg Point, AM, = 4 REM » “De pression -) Treeying pom+, Tp =) ky ™m von, &T . chat - im, &L © OSsmotre Pressure ap SoltH™. TT = Cue 3 holy fey ASIC ahion Dis Foc{ ack ~, ga > Ng CL —» 2¢ a a [Avy > nA © , —_— WX )— & A, |} —e Meo =] Ne =I Me = 1X +n “Me = I T%, “Mo = | tof Cy-t) = _ ou _ ; mee Te Ct) As me Leste eS te Ce) fz +x Cua) A= Me. }+ & Cn) depos X Cnty Ne ——— x on 4) = fe = I+ Chm) a My = Le Me Me & ~ Me-mMo “hep Q. 16) An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normakboiling‘point ofthe solvent. What is the molar mass of the solute? : (Ans, : 41.35 g mol-?] Q. 17) At 25°C, the saturated vapour pressure of water is 3.165 Kpa (23.75 mm Hg). Find the saturated vapour pressure of a 5% aqucous solution of urea at the same temperature, [Molar mass of urea = 60.05 g mol~*}. [Ans. : 3.115 Kpa] Q. 18) Calculate the mass of a noa-volatile solute (molar mass 40g mol~") which should be dissolved is. 114g octane to reduce its vapour pressure lo 80%, JAns. : 8.0 Oe ELEVATION IN BOILING POINT Q. 19) For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent. determine the vapour pressure (mm of Hg) of the solution. [Given : Ky for water = 0.76 kg mol-?| [Ans.: 724 mm of Hg] [Hint : AT, = Ky.m > 0.76 x mex 2K My =9.5 g mol"? Prac Pa 25 18 P’, 95° 100 760- Pa _ 25 18 TaD 55% too 7 Pa = 724 mm of Hg] Q. 20) An aqucous solution of glucose of molar mass 180 2 mol™! boils at 100,01°C, The molal boiling point elevation constant for water is 0.5 K kg mol7!, What is the number of glucose molecule in the solution containing 100 g of water, [Ans. = 1.2 + 107! molecules] Q. 21) An aqueous solution containing 3.12 g of barium chloride in 250 g of water is found to be boil at 100.0832°C. Calculate the degree of dissociation of barium chloride. [Given molar mass BaCl; = 208 g mol", Kb for water = 0.52 K/m] [Ans, : 83.3%] Q. 22) 18g of glucose, CsHi20¢ (Molar mass=:180'g mol~*) is dissolved in Ikg of water in a sauce pan. At what temperature will this solution boil? [Kz for water = 0.52 K kg mol}, boiling point of pure water = 373.15 K) lAns. : 373.202 K| DEPRESSION IN FREEZING P POINT Q. 23) 15.0 g of an unknown molecular substance was dissolved in 450 9 of water. The resulting solution was found to freeze at -0.34°C, What is the mol of this substance. (Ky for water = 1.86 K kg mol"), Q.24)2¢ of C6HsCOOH dissolved in 25 g of benzene shows depression in freezing point equal to 1.62 K. Molar [reezing point depression constant for benzene is 4.9 K kg mol'. What is the percentage association of acid if it forms a dimer in solution? [Ans. : 99.2%] ar muss: Q. 25) Three molecules of a solute (A) associate in benzene to form species A3. Calculate the freezing point of 0.25 molal solution. The degree of association of solute A is found to be 0.8. The freezing point of benzene is 5.5°C and its Kr value is 5.13 K/m. [Ans. : 4.9°C] Q. 26) 75.2 g of Phenol (CsHsOH) is dissolved in 1 kg solvent of Kr= 14 Knr", if the depression in freezing point is 7K, then find the % of phenol that dimerises. AE Q. 27) Calculate the amount of NaCI which must added to one kg of water so that the freezing point is depressed by 3K. Given Ky= 1.86 K kg mol '. Atomic mass : Na = 23, Cl = 35.5). Ans. : 0.81 mol NaCl] - Q. 28) 0.6 mL of acetic acid (CH3;COOH), having density 1.06 g ml .' is dissolved in 1 litre of water. The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the van’t Hoff Factor (/). Q. 29) Two elements A and B from Compounds having formula AB, and-ABa. When dissolved in 20.0 g benzene (CgHe), 1 g and AB2 lowers the freezing point by 2.3 K whereas 1.0 g of AB, lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol"!. Calculate atomic masses of A and B. [Aas. : A = 25.584. B = 42.644] OSMOTIC PRESSURE + Q. 30) A 5% solution of sucrose (Cj2H220)1) is isotonic with 0.877% solution of urea. NH2CONH)?) Calculate the molecular mass of urea. [Ans. : 59.99 g mol-t| Q. 31) Osmotic pressure of a 0.0103-molar solution of an electrolyte was found to be 0.75 atm at 27°C. Caleulate'van’t Hoff factor. [Ans : 7 = 3] Q. 32) At 300K, 36 g of glocuse,(CgH 20s) present per litre in its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of another glucose solution is 1.52 bar at the same concentration, calculate the concentration of the other solution. [Ans. : 0.0610 mol I. '] Q. 33) 100 mg of a protein is dissolved in just enough water of make 10.0 ml. of solution. If this solution has an osmotic pressure of 13.3 mm Hg at 25°C. what is the molar mass of the protein? [R = 0.0821 I. atm mol 'K ! and 760 mm Hg =| atm]. JAns. : 1398045 2 mol | Q. 34) Determine the osmotic pressure of a solution prepared by dissolving 2.5 = 102 2 of K2SO, in 2L of water at 2.5°C, assuming that it is completely dissociated. [R = 0.0821 L atm k™, Molar mass of KySO4= 174 g mol] [Ans. : 5.27 x 107 atm] GSkamgle “Solved (ES) 3 2-2 Calculate the molarity of a sohition containing 5 g of NaQH in 450 ml. Wy, = 84 solution, Solution Moles of NaOH = 5g = 0.125 v= wesem ao +64) 30g mol =0. mol Me path = . Su Volume of the solution In Iltres = 450 mL 71000 mL L” * Using equation (2.8). Mala oh = ton yx loos 0.125 mol x 1000 mL L"! “i ve Molarity = = me © y 450 mL =0.278M = 0.278 mol L"' = 0.278 mol dm* Es ~-8} If, gas is bubbled through water at 293 K, how many millimoles of N, gas would disolve in 1 tre of water ? Assume that N, exerts a partial pressure of 0-987 bar. Given that Henry's law ‘oustant for N, at 293 K is 76-48 kbar. fay = OPI hav Ke = FEU 8 bar . , _ Pu, _ O987bar _ -5 Solation. According to Henry's law, Py, =KyX4n.i oy, = x = 56480 ber t2* 10 7 locce n n Mig 7 Loe Enmoles of N, are present in 1 L of water (i.e., 55:5 moles), ye = Ty sgs gg NCD - he ae 1 on "555 =129x10°S, or n= 1.-29x10-5 x 55.5 moles = 71-595 x 10-5 moles = 0-716 millimoles €S~ S$” _ Vapour pressures of chloroform (CHCI,) and dichloromethane (CH,Cl,) at 298 K are 3, 12C12) 700 mum Bg and 415 mm Hg respectively. Calculate () the vapour pressure of the solution prepared by mixing'25-5 & of CHCl, and 40 g of CH,C), at 498 K, and (#) the mole fraction of each component in vapour phase. Solution. (i) Calculation of vapour pressure of the solution co Mass of CHC], =25-5g - U cries =~ 2oeomny, Mass of CH,Cl, = 40 g 3 Molar mass of CHCl, = 12 + 1 +3 x 35-5 = 119-5 g mol! PCa. =v Molar mass of CH,Cl, = 12 +2 +2 x 35-5 = 85 g mol! Bol, = ATS ea _ 255g 40g ” f CHCl, = =0213mole ; Moles of CH,Cl, =__®>_ _ _ Moles o 5 195g mol HCl, 85g mor 0-470 mole 0213 i =————_ =0312 Mole fraction of CHCI, @cua, ) 021340470 Mole fraction of CHCl, (%cqz,cy, ) = 1 - 0312 = 0-688 Frou = Pci, + Pon,cl, = Xcuci, * P°cuc, +*cH,c1, * Pouch, = 0-312 x 200 + 0-688 x 415 = 62-4 + 285-5 = 347-9 mm quabyey” (i) Calculation of mole fraction of each component in vapour phase fem 1 - u As calculated above, Porc), = 624mm, Pey,cr, = 2855mm, p, = 347.9 mm J = di Tete . : Pout 62-4mm Mole fraction of CHCl, in the vapour phase = CHC, _ 624mm _ 44g t —_ 3 Pour phase (cuca) Prom 347-9mm grty2>) Mole fraction of CH,Cl, in the vapour phase Ocu,c,) = 1=Yopic,, = 1-018 = 0-82 7a = I-91 SEAID 0°6 mL of acetic acid (CH,COOH) having a density of 1-06 g mL~ is dissolved in 1 litre of water. The depression in freezing point observed for this strength of the acid was 0-0205°C. Calculate the van't Hoff factor and the dissociation constant of the acid. K, for water = 1-86 K kg mol~), (_NCERT Solved Example _, Pb, Board 2011) Solution, Calculation of van’t Hoff factor (i) aah 06 mL of acetic acid means solute (w,) = 0-6 x 1-06 g = 0-636 g we dxv. 1 Litre of water means solvent (w,) = 1000 g ; (AT observed = 0-0205°C 1000xKy x, _ 1000gmol~! x 186 Kkg mol-!x 0636g “. Observed molar me oo = Ww XAT, 1000 g x 0-0205K = 57-7 g mol DTg = kex be y 10a 7 bw 4 Calculated molar mass of CH,COOH = 60 g mol (jn ¢) ° Wg) x van't Hoff factor () = Caleulatedmolarmass __60gmol-! _ (a = a) Observed molarmass 57-7 gmol~! 104. Mo Calculation of degree of dissociation () from van’t Hoff factor (i) If is the degree of dissociation of acetic acid, then CH,COOH === CH,COO- + Ht Me =! Initial moles 1 vez Ith Moles at eqm. 1-« o , Total=1+0 -. ™o “ i= 8a 140 or a=1-i=1-04-1.0=0-04 yr Mme Calculation of dissociation constant. If we start with C mol L~ of acetic acid, then CH,COOH == CH,COO- + Ht Initial conc, Cmol L" Cone, at eqm. C-Ca Ca Ca =C(1-a) - 2 - Dissociation constant (K,) ~ ICH;COO“IIH*] __Caca _ Ca? & = e-ologweld! [CH,COOH] C(Il-a) 1l-a But C=0636 g1-! = 263681" _porgg mol t-l 2, K_ - £000106)(004)2 60g mol"! “e004 = 1-76 x 10-5, L@ Q.28, They FLEA - Gusto 7 AaPOUr pressures out ae comperaa ten ee A and B are 450 and 700 mm Hg at 350 K respectively. Find composition of the vapour pl quid mixture if total vapour pressure is 600 mm Hg. Also find the Ans. Here, Px’ = 450 mm, Pp? = Le O29 . 700 mm, Pro = 600 mm Applying Raoult’s law, PREXAXPA?! Pp =XpX Pp? = (1 — 4p) Po” fe i Prout = Pa + Pp =XaPa’ + (1-44) Pp’ = Pp’ + (Pa - Pp) *, Substituting the given values, we ‘“ AtPa=xsPa a) Pp = Pp + Pa Pp) ta Xp = 50 Thus, composition of the liquid mixture will be *,(mole fraction of A) =0-40, x, (mole fraction of B) = 1 - 0-40 = 0-60 “ Pa=X,* py? =0-40 x 450 mm = 180 mm, pg = Xg X Pg? = 0:60 x 700 mm = 420 mm 600 = 700+ (450-700)x, or 250x,=100 or - 040 Mole fraction of A in the vapour phase = a -—180_ = 7m eft Pat+P, 1804420 Tv Mole fraction of B in the vapour phase = | - 0-30 = 0-70, Woe icdi Vapour pressure of pure water at 298 K is 23-8 mm Hg. 50 g of urea (NH,CONH,) Is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering. Ans. Here, p=238 mm w,= 50g, Mp (urea) = 60g mol!; w, = 850 g, My (water) = 18 g mol"! Our aim is to calculate p, and (p” - p,)/p* Applying Raoult’s | pe=Pp_ Mm IM, ___—50/ 60 pplying Raoult’s law, P hy th wj/M,+W2/M, — 850/18+ 50/60 083 __ _ 083 _ gor “47224083 4805 Thus, relative lowering of vapour pressure = 0017 Substituting p* = 23-8 mm, we get 238 — Py 017 op 28 Pt = 0017 0 238 p, = 0.017 x 23-8 = 0-40 P 238 or p, = 23-8 -0-40 = 23-4 mm Thus, vapour pressure of water in the solution = 23-4 mm, JH % 2.10. Bolling point of water at 750 mm Hg is 99-63°C. How much sucrose Is to be added to 500 g of water such that it bolls at 100°C ? Molal elevation constant for water Is 0:52 K kg mot". (Assam Board 2012) Ans, Elevation in boiling point required (AT,) = 100-99-63°= 037° ,o% > Th —Th Te Q. 2. Ans. Mass of solvent (water), w, = 500 g Molar mass of solvent, M, = 18 g mol’; — Molar mass of solute, CyH0) = 342 g mol! 1000 K, w, ~ tw. Joos. Applying the formula, My =——Tp=* s ¢ OTS WHR TE aCe) _Maxmxat, _ 3428 mol~! x 500 g x037 K . “2° —1000xK, 1000 g ke! x052 K kg mor-l “21-7 g, Calculate the mass of ascorbic acid (vitamin C, C,H,0,) to be dissolved in 75 lower its melting point by 1°5°C. Ky= 39 K kg mot. Bot acetic acta to Lowering in melting point (AT) = 1:5° ; Mass of solvent (CH,COOH), w, = 75 g Molar mass of solvent (CHyCOOH), M, = 60 g mol"! Molar mass of solute (C,H,O,), M, = 72 + 8 + 96 = 176 g mol"! For acetic acid, Ky= 3-9 K kg mol! - ker vax [eee Applying the formula, My => OL" , ote Me ity) wos : M. ‘ot! i ee w, = Mex, XAT, _ _ 076g mol ) (5 g) 5K) = =5 2° ToonxK, 000g ke) G9 K kg mol“) 77g F-PQ 212. Calculate the osmotic pressure in pascals exerted by a soluti polymer of molar mass 1 85,000 in 4 orwa oA ra lon prepared by dissolving 1-0 g of Ans, m= CRT == RT yp i =neT Here, number of moles of solute dissolved (n) = 10g d —— = —_—__. II 185,000 g mol-! ~ 185,000 ™” V=450 mL =0-450L, T= 37°C = 37 +273 =310K R= 8.314 kPa LK"! mol"! = 8.314 x 10? Pa L K-! mot"! Substituting these values, we get “ 1 1 185,000 mol x, MSL A yh a) We Re _" a n at *8314%10? Pa L K~! molm!x310 K = 30-96 _NCERT [EXERGISES] | Q. 2.1. Define the term solution. How many types of solutions are formed ? Write briefly about each type with an example. Ans, 2yem note) Q. 2.2. Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small.. What type of this solid solution is likely to be ? Ans. Interstitial solid solution. Q. 2.3. Define the terms : (#) Mole fraction (/) Molality (i) Molarity (iv) Mass percentage. Ans. 4pom “net - Q. 2.4. Concentrated nitric acid used in the laboratory is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1-504 g mL-!? , Ans, 68% nitric acid by mass means that . Mass of nitric acid = 68 g, Mass of solution = 100. ¢ Molar mass of HNO, = 63g mol? /HiNo, = (aia ame = Tle tug = £3) [ 68 g HNO, -= mole =1079 mole, Density of solution = 1-504 g mi! Volume of solution = “mil. = 665 ml. = 00665 L , dz “KK, 2 ve & vy a 1:0t?, = (623M, wralon hy = WD 666 », G013=1004) _ 2g 18g mol"! _ 2x18 1013 1"! = 41-35 g mot hd - X—— g mo W13bar M,”98g M2 = “GR -X gang Q, 2.16. Heptane and octane form ideal solution. At373 K, the vapour pressares of the two liquid components ‘ are 105:2 kPa and 46:8 kPa respectively. What will be the vapour pressure of a mixture of 26-0 g of | heptane and 35-0 g of octane ? Pos 10S'2 , PS = UR, UA a 26 Ans. Molar mass of heptane (C,H,,) = 100 g mol-!; | Molar mass of octane (C,Hy,) = 114 g mol"! tu, = 35 | 260 g heptane = en =026 mol; 350g octane = ———S— ae = 031 mol | oe, Parmee) MEME sry x (heptane) = = ‘ =1-0-456 =0:544 2 | Plane) = eos; 70456 3 x (octane) be = Yo ° A> fh Thy p (heptane) = 0-456 x 105-2 kPa = 47:97 kPa; p (octane) = 0-544 x 46-8 kPa = 25-46 kPa Pro = 47-97 + 25-46 = 73-43 kPa, , p> = Prep + Prey Q. 2.17. The vapour pressure of water is 12°3 kPa at 300 K. Calculate the vapour pressure of 1 molal solution of a solute in it. Imelat = Imole Ans. | molal solution means 1 mol of the solute in 1 kg of the solvent (water). TK wa ?} = on Mol . = = = riety G Solum p = feed. ‘ol fraction of solute 1n555 00177 wren m) = soe ev 123 - Mert, Now, a = py he, Ps = 0017 or p,= 12408 kPa. f Q. 2.18. Calculate the mass of a non-volatile solute (molar mass 40 g mol-!) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%, (CBSE Sample Paper 2017) Ans. Reduction of vapour pressure to 80% means that if p°= 100 mm, then p, = 80 mm. Applying complete formula Mas ey D P°-P,_ My _HIM, mi = mY Eg Hg) Pp mtn, w/M,+w,/M, ’ Wyo Ma . - /40 , 100-80 _ Wo (Mol. mass of octane CgHyg = 114 g mol-!) 100 114/114 + w,/40 20 w,/40 1 Ww, W, = s—4—_ oor 4-2 |= ogg yy, =10 * 100 14+w,/40 3{ i] 40 geue a a i Note that complete formula is required because concentration of solution is greater than 5%, Complete formula can also be applied in the form P°-P, _ WIM, 100-80 w,/40 lw o-i1=-2 2 or =—— or —=-2 = P, » w/M, 80 «114/114 4 49 % Wa=l0g Alternatively, suppose mass of solute dissolved = w g Moles of solute = 08 114 Moles of solvent (octane) = t= 1 mole (Mol. mass of CgHjg = 114 g mot!) “ Mole fraction of solvent = Ts wid For a non-volatile solute, . Vapour pressure of solution = Mole fraction of solvent in the solution x Vapour pressure of pure solyent ve 80= 1 w _ 100 w 10, 2,1 lewap or tO or a ar ae | or w=10g Q. 2.15. A solution containing 30 g of a non-volatile solute exactly in 90 g water has a yapour pressure of 28 kPa at 298 K. Further 18 g of water is then added to the solution, the new vapour pressure becomes 2-9 kPa at 298 K. Calculate () molar mass of the solute, () vapour pressure of water at 298 K. Ans. ' (i) Suppose the molar mass of the solute = M g mol"! 30 908 bits » P. = 218K fe solute) = — moles, I = =5 moles = m( ) M 7% (solvent, H0) 18 g mol"! . VB o_. 9, . M PrP ig, BRB _3O/M_ 4, y_28__30/M_ P ntn, p? 5+ 30M . p? 5+*30/M = 2 iM or 28 30/M__5+30/M-30/M_— 5, po _5+30/M 4, 6 AO p° 5+30/M 5+30M 5+30/M 28 5. _M After adding 18 g of water, n (H,0), ie,.m=Gmoles , my, = WHS = = 6. . B29 30/M_ 5 29 _30/M. Po = tS KR. “—p? 6 +30/M p? 6+30/M or 2221 -—30/M_ _ 6+30/M—-30/M _ 6’ P= 6 +30/M 15 id P 6+30/M 6+30/M 6+30/M 29 6 M 29 1+6/M 5° 6 ividi 5 . (ii), == 9} 1+— |= 28) 1+— 0 ee (i) by eqn. (if), we get 28 145/M or 29/ +3) ( +s) 4.5/5 : or Wr, 4 294245 294268 gf 2301 of M=230 M M M i =23i Pe 5 29 e. 29 2-8 = 3°53 kP, (i) Putting M = 23 in eqn. (i), we get 2 ta a or Pp: 7% a, Q. 2.20. A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose in water if freezing point of pure water is 273-15 K. Ans, 5% solution by mass means 5 g of solute is present in 100 g of solution , W2= Sgr = 985 ‘ “Mass of solvent (water) = ; g icon pnaletity = ra wx tore 7 OR ia Haren) Molality of sugar solution = 34a 95 = 0154 (mol. mass of sugar C,,H,,0,, = 342) oy G-S 25 AT, for sugar solution = 273-15 - 271 = 2:15°; AT;=K-xXm -. K; = 0154 . 5 _ 1000 ; Molality of glucose solution = Tn0% 95 = 0-292 (mol. mass of glucose C,H)20, = 180)2t_ 215 Wa = S91 29% :. AT,(Gh = =——— = [y(Glucose) = Kyx m 0154 0-292 = 4.08 «. Freezing point of glucose solution = 273-15 — 4-08 = 269-07 K. Q. 2.21, Two elements A and B form compounds having molecular formula AB, and AB,. When dissolved in 20 g of benzene (C,H,), 1 g of AB, lowers the freezing point by 23 K whereas 1-0 g of AB, lowers it by 1:3 K. The molal depression constant for benzene is 5-1 K kg mot, Calculate atomic masses of A and B. (Uttarakhand Board 2012) 1000 K Ans. Applying the formula, My=———-£"2, Tg =kgx s x 580. . XAT, We wiCky) M,. = 1000x51x1 ‘AB 20x25 = 11087 g mol” ; Mas, = sexsi = 19615 g mol! Suppose atomic masses of A and B are ‘a' and ‘b’ respectively. Then | Molar mass of AB, = a + 2 b = 110-87 g mol"! a) Molar mass of AB, = a + 4 b = 19615 g mot! Ban. (if) - Eqn. (/) gives2b= 85-28 or b= 42:64 | Substituting in eqn. (i), we get a 4-2 42-64= 110-87 or a= 25.59 Thus, Atomic mass of A=25:59u, Atomic mass of B = 42°64 u Q. 2.22. At 300 K, 36 g of glucose present per litre {n its solution has an osmotic pressure of 4-98 bar. If the osmotic pressure of the solution Is 1-52 bar at the same temperature, what would be its concentration ? w x &T 36 T= MT = Ta aS Ans. n= CRT ~. In the first case, 498 = > xRx300= 60R, Vw 364- of) In the second case, 152= CxR x 300 ¢ -y \Be Ge "9° Dividing (ii) by (0, we get C = 0-061 M. ~ Q. 2.23. Suggest the most important type of intermolecular interaction in the following pairs : (@ n-hexane and n-octane (i/) 1, and:CCl, (ii) NaCIO, and water (fv) methanol and acetone (v) acetonitrile (CH,CN ) and acetone (C3H,0). Ans. (i) Both are non-polar. Hence, intermolecular interactions in them will be London dispersion forces (if) Same as (i). (ii) NaClO, gives Na* and ClO ions in the solution while water is a polar molecule, Hence, intermolecular interactions in them will be fon-dipole interactions. . (iv) Both are polar molecules, Hence, intermolecular interactions in them will be dipole-dipole interactions, (v) Same as (iv). . Q. 2.24, Based on solute-solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH,OH, CH,CN Ans. (i Cyclohexane and 7-octane both are non-polar. Hence, they mix completely in all proportions. (i) KC1 is an ionic compound while n-octane is non-polar. Hence, KCI will not dissolve at all in n-octane. (tii) CH,OH and CH,CN both are polar but CH,CN is less polar than CH,OH, As the solvent is non- polar, CH,CN will dissolve more than CH,OH in n-octane. Thus, the order of solubility will be KC] <CH,OH < CH,CN < Cyclohexane, Q. 2.25. Amongst the folluwing compounds, identify which are Insoluble, partially soluble and highly soluble in water (1) phenol, (Ll) toluene, (lil) formic acid, (iv) ethylene glycol, (¥) chloroform, (vi) pentanol. Ans. (i) Partially soluble because phenol has polar -OH group but aromatic phenyl, C.H.- group. (if) Insoluble because toluene is non-polar while water is polar, (il) Highly soluble because formic acid can form hydrogen bonds with water. (iv) Highly soluble because ethylene'glycol can form hydrogen bonds with water. (¥) Insoluble because chloroform is an organic liquid. (vi) Partially soluble because -OH group is polar but the large hydrocarbon part (CsH,,) is non-polar, Q. 2.26. If the density of some lake water is 1:25 g ml.~! and contains 92 g of Nat fons per kg of water, calculate the molality of Na*ionsin the lake. ., - “/m Ans. No. of moles in 92 g of Nations = 928 = 4 mole (as atomic mass of Na = 23) 23 g mol" -, enaletity “es As these are present in 1 kg of water, by definition, molality = 4m. ” var(kg) Q. 2.27, If the solubility product of CuS is 6 x 10°, calculate the maximum molarity of CuS in aqueous solution. . ‘ Ans. Maximum molarity of CuS in aqueous solution = Solubility of CuS in mol L-! ing Henry’ pa P= 100mm _ ing 10-3 Applying Henry's law, p=Kyx, wehave x= K, "27x10 mm x ie, mole fraction of methane in benzene = 1-78 x 10°, a; 1000 of tus 2. 2.36. 100 ¢ of uid A (molar mass 140 g mol-!) was dissolved in g quid B (molar mass 180 g ). The vapour pressure of pure Liquid B was found to be 500 torr. Calculate the vapour Pressure of pure liquid A and its vapour pressure in the solution If the total vapour pressure of the solution is 475 torr. 100 5 = we " Ans. No. of moles of liquid A solute) = ———2_ = = mole . ~y ~ meen) 140g mol! * 7 f Ma . we No. of moles of liquid B (solvent) =— 1000 g — = 3 role rp = ni nA 180g mol"! 9 & at ¥ . 5/7 5/7. _ 5-63 _ 45 ATM *. Mole fraction of A in the solution Gy) = So Ee XE 5/7+50/9 395/63 7, 395 395 “. Mole fraction of B in thé solution Gp) =1>0-114=0-886 oy =. l=>ta} . - ‘ ° cB — Also, given Pg’ = 500 torr ——— . Applying Raoult’s law, Pa=*%aPy° = 0-114 x PA . o(B) Ps =p Py = 0-886 x 500 = 443 torr Prot = Pat Pp LAN 475-443 475 =0-114 Py’ +443 or PA°= “a = 280-7 torr | Substituting this value in eqn. (i), we get Pa= 0-114 x 280-7 tor = 32 torr. ). 237. Vapour pressure of pure acetone and chloroform at 328 K are. 741:8.mm Hg and 6328 mm Hg respectively, Assuming that they form ideal solution over the entire range of composition, plot Protals Peblorotorm 204 Pacetone 25 a function of x. . The experimental data observed for different compositions of mixtures is : nestone 100 XX, ctone 0 118 234 36-0 50-8 58-2 64-5 721 Pacetoodtam Hg 0 = 549 110-1 202-4 322-7 405-9454 521-1 Peblocotorn/tam Hg 6328 5481 469-4 3597 257-7 193-6 161-2 120-7 Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution. ADS. Xpctone 0-0 0-118 . 0-234 0-360 0:508 0-582 0-645 0-72] P, /mm Hg 0 54.9 110-1 202-4 322-7 405.9 454-1 521-1 Pehiorofon/m Hg 632-8 548-1 469-4 359-7 257-7 193-6 161-2 120.7 Prout 632-8 603-0 579-5 3621 580-4 * 599.5 615-3 641.8 700;--— - pe 4 : =e 600 SS fox ro |Pioa Lot __| = sooh-—!S jor : ff 200}-— yar PN to0}----|_ 2% PS ole“ “| 0 Of 02 03 04 O85 06 O7 08 Mole fraction of acetone (acetone) so As the plot for Pyox, dips downwards, hence the solution shows negative deviation from the ideal behaviow. 2.38, Benzene and toluene form ideal solution over the entire range of composition. The vapour pressures Q*” of pure benzene and toluene at 300 K are 50-71 mm Hig and 32-06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene. Ans. Molar mass of benzene (CsH,) = 78 g mol-!, Molar mass of toluene (CgH,CH,) = 92 g mol! z. No. of moles in 80 g of benzene = 898 __ _ 1.926 mole we = Bor Wir = Ne 78 g mol-! No. of moles in 100 g of toluene —__!©8 __ _ 1089 rote 92 g mol-! ~ ». In the solution, mole fraction of benzene = 1026 __ 1026 _ nage ) rg *- a 2 n fc 1026+1087 2113 mole fraction of toluene = 1 — 0-486 = 0-514 ‘ ; P Benzene = 50-71 mm, Pp rojyene = 32-06 mm Applying Raoult’s law, — Pyeasene = “Benzene * P’Benzene = 0°486 x 50-71 mm = 24-65 mm - Premere Ferma = “Toluene * P*totuene = 9-514 x 32-06 mm = 16-48 mm 24-65 24-65 2. Mole fraction of benzene in the vapour phase = ___PBenzene _ = = = 185 = Ho Pouuene + Proms 65+1648 4113 ), 2.39. The alr is a mixture of a number of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K, if the Henry’s law constants for oxygen and nitrogen are 3-30 x 107 mm and 6:51 x 107 mm respectively, calculate the composition of these gases in water. Ans. Total pressure of air in equilibrium with water = 10 atm ~<-—--- < ——, As air contains 20% oxygen and 79% nitrogen by volume, ~. Partial pressure of oxygen (Po,) = ax 10 atm = 2 atm = 2 x 760 mm = 1520 mm Partial pressure of nitrogen (py, ) = 3x10 atm = 7-9 atm = 7-9 x 760 mm = 6004 mm 7 Kj (O,) = 3:30 x 107 mm, Ky (N,) = 651 x 107 mm : = — Po, __1520mm__ 5 Applying Henry's law, Po, = KyX2o, or Yo, = Ky = 330x107 mm 4-61x10 PN 6004 mm = = =e C= 922x105, Py, = Ku %*n, OF Ne 651x107mm 2.40. Determine the amount of CaCl, (= 2-47) dissolved in 2:5 litre of water such that its osmotic pressure is 0-75 atm at 27°C. n __uxXV 0-75 atm x 2-5 L = 0-0308 mol Ans, m=/CRT=iRT oF "= TR yp 747 x00821L am K!molx300K 40+2%35-5=l1llg mol"! .*. Amount dissolved = 0-0308 x 111 g = 3-42 g. ure of a solution prepared by dissolving 25 mg of K,SO, In 2 litre of dissociated. (Pb. Board 2011) tely Kay > vers 502° oe } Molar mass of CaCl, = 2.41, Determine the osmotic press water at 25°C, assuming that it Is comple! We=l Ans, K,S0, dissolved = 25 mg = 0.025 5 8% ' > Vol tion =2L, T=25°C= ? au \ Mord folume of solution . 4x 16 =174 g mol he “Whe = 3, = > Molar mass of K,SO, = 2 x 39 + 32 + As K,SO, dissociates completely as K,S0,— 2K* +803", ie., ions produced = 3, -. 1=3 0.025 1 Loney n Wo) pm = 3x—Ce 8 x —x 00821 L atm K"! molt! x 298 K “ mei CRT <i CRT ix y XRT 3 174g mol = 5-27 x 10° atm. SOLUTIONS PREVIOUS YEARS’ QUESTIONS Very Short Answer Type Questions [1 Mark] 1. Predict whether van’t Hoff factor, (i) is less than one or greater than one in the following: (i) CH,COOH dissolved in water. (ii) CH,COOH dissolved in benzene. [AI Chennai] Ans. (i) It will be more than 1 because CH;COOH will dissociate into CH,COO™ and H,O*. (ii) CH;COOH dissolved i in benzene will form dimer i.e. undergo association, therefore, ponone nD i<d cH, 52 C-CHs 2. Give reasons for the rollowing Aquatic animals are more comfortable in cold water than in warm water. [AI Chennai] Ans. It is because cold water has more dissolved oxygen than warm water because solubility of gas in liquid decreases with increase in temperature. Short Answer Type Questions [2 Marks] 3. State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obeys Raoult’s law at all concentrations. {Delhi] Ans. (i) Raoult’s law: It states that partial vapour pressure of each component is directly proportional to its mole fraction when both solute and solvent are volatile. Pa %Xq_ => Pa =P4X4 and pgp &<xXzp > Pg = Ppxtp (ii) 1. AH, = 0 2. AV nix = 0 3. They follow Raoult’s law. 4. They can be separated by fractional distillation. [Any two] ( 2018 } Short Answer Type Questions [2 Marks] 11. Calculate ab freezing point of a solution containing 60 g of glucose (mot = 180 g mol) in 250 g of water. (K, of water = 1.86 K kg mot) ar mas, Ans. Mass of solute (Wg) = 60 g; Mass of solvent (W,) = 250 g (CBgy) Molar mass of solute (M,) = 180 g mol | K, of water = 186 K kg mol | Ws _ 1000 | = x —x | Oe EW, | 60 _ 1000 1 7.44 | AT, = 1.86 X—— x —— = 186x—x4 = 14 _ f 180 ~ 250 3X4 = > = 248K AT, = 2.48 K Freezing point of solution = Freezing point of water — AT, = 273.15 K - 2.48 K = 270.67 K Freezing point of solution = 270.67 K or — 2.48°C Long Answer Type [I] Questions [3 Marks] 12. Give reasons for the following: { | | t | j ! (a) Measurement of osmotic pressure method is preferred for the determination of molar | masses of macromolecules such as proteins and polymers. (6) Aquatic animals are more comfortable in cold water than in warm water, (c) Elevation of boiling point of 1 M KCI solution is nearly double than that of 1M sugar solution. [CBSE) Ans. (a) It is because osmotic pressure is measured at room temperature and has appreciable value even for very dilute solution of macromolecules like proteins and polymers. (b) Refer Ans. to Q.2. (c) KCl. —> K+ + cr | KCl is strong electrolyte. It dissociate completely into ions. The number of particles are double as compared to sugar solution. Sugar is non-electrolyte, does not form ions. i = 2 for KCl andi = 1 for sugar, therefore, AT, is double in KCl. 2018 (Compartment) Short Answer Type Questions [2 Marks] : t's 13. Why a mixture of carbon disulphide and acetone shows positive deviation from aes law? What type of azeotrope is formed by this mixture? f Ans. It is because of weaker force of attraction between acetone and CS, after mixing, it shows positive deviation from Raoult’s law, It forms minimum boiling azcotropes with specific composition due to increase in vapour pressure duc to decrease in forces of attraction and therefore, decrease in boiling point, forms minimum boiling azcotropes. Long Answer Type [I] Questions [3 Marks] 14, Calculate the freezing point of an aqueous solution containing 10,5 g of Magnesium bromide in 200 g of water, assuming complete dissociation of Magnesium bromide. (Molar mass of Magnesium bromide = 184 g mol’, K, for water = 1.86 K kg mol!). [CBSE] Ans. W, = Mass of MgBr, = 10.5 g, W,, = Mass of water = 200 g Mg = Molar mass of MgBr, = 184 g mol, K, for water = 1.86 K kg mol! MgBr, —> Mg?* + 2Br- i=3[- Complete dissociation takes place] W, AT; =i x K, x —8 y, 1000 | Mp W, 200 Depression in freezing point = AT, = 3X1,86 105 x5 = 292.95 _ 159K f 184 184 Freezing point of solution = Freezing point of water — AT, = 273.15 K - 1.59 K = 271.56 K Short Answer Type Questions [2 Marks] 15. Define the following terms: [Delhi] (i) Colligative properties. (it) Molality (m) Ans. (i) Colligative properties: Those properties which depend upon the number of particles of solute present in solution and not on nature of solute €.g. osmotic pressure. (it) Molality (m): It is defined as number of moles of solute dissolved per kilogram AT,=3 x 1.86 x 105 ,, 1000 184 | | | (kg) of the solvent e.g. 0.5 m glucose solution contains 0.5 mole of glucose in 1 kg | of water. | 16. Define the following terms: | (®) Ideal solution (ii) Molarity (M) [Delhi] 17. 20. Ans. (i) Ideal solution: It is a solution which obey Raoult’s 1 concentration at a specified temperature. No heat is evoly in volume take place, e.g. Benzene and Toluene. aw : d over & Wide Ta ©d or absorbe, d.No a of “Nog a (ii) Molarity (M): It is defined as number of moles of solute dissolveg ge solution, e.g. 1 M urea solution contains 1 mole of urea in 1 litre of go) litte of Define the following terms: Ution, (i) Abnormal molar mass (ii) van’t Hoff factor (é) (i) Abnormal molar mass: The molar mass determined with the help of colle property when it undergoes either association or dissociation is called abnon molar mass of solute. mal (ii) van’t Hoff factor: It is defined as the ratio of normal molecular mass to the obse molecular mass of the solute. Ted What are colligative properties? Write the colligative property which is used to find the molecular mass of macromolecules. [Foreign Refer Ans. to Q.15(i). Osmotic pressure is used to find molecular mass of macromolecules. State Raoult’s law for a solution containing non-volatile solute. What type of deviation from Raoult’s law is shown by a solution of chloroform and acetone and why? [Foreign] Raoult’s Law: The relative lowering of vapour pressure is equal to mole fraction of solute if only solvent is volatile and solute is non-volatile. Chloroform and acetone show negative deviation from Raoult’s law. It is due to H-bond formation between CHCL, and CH,COCH;. It forms maximum boiling azeotrope. CH, 1 \peo--- HCL CH,” Na What is meant by elevation in boiling point? Why is it a colligative property? [Foreign] The increase in boiling point of a solvent on adding non-volatile solute is called elevation in boiling point. AT, = Boiling point of solution — Boiling point of solvent. It is colligative property because it depends upon molality of solute, i.e. number of particles of solute in solution and not on nature of solute. Long Answer Type [I] Questions [3 Marks] 21. Given: (Molar mass of sucrose = 342 g mol) A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water, if freezing point of pure water is 273.15 K. (Molar mass of glucose = 180 g mol!) (Den AT; = 273.15 — 269.15 = 4.0K W, AT, = K, x —% x 1000 Ms W, P,=450 mm, Py = 700mm, Pr = 600 mm Pp =Pyr, + Pory Pr =PA(l ~Xq) + PEXp [vx tty el A= 1 x5) 600 = 450 (1 -x,) + 700 x, 600 = 450 - 450 x, + 700 xy 250 xy = 600 - 450 = 150 . » e Stae 150 3 Mole fraction of ‘B’ in liquid phase = = == =06 gure Pnase = *8 = 950 5 Mole fraction of ‘A’ in liquid phase = x, = 1-x, = 1- 0.6 = 0.4 Long Answer Type [tl] Questions [5 Marks] 27. (a) Explain why on addition of 1 mol glucose to 1 litre water, the boiling point of water increases. (6) Henry’s law constant for CO, in water is 1.67 x 10° Pa at 298 K. Calculate the number of moles of CO, in 500 ml of soda water when packed under 2.53 x 10° Pa at the same temperature. [All India} Ans. (a) When 1 mole of glucose is added to 1 litre of water, vapour pressure of solution decreases because surface molecules consist of both glucose and water molecules, escaping tendency of water into vapours decreases, therefore it is to be heated at higher temperature so that it boils i.e. boiling point is increased. (6) Ky = 1.67 x 10° Pa, agg, = 2, V = 500 mL, Pygs = 253 X 10° Pa Peas = Ky X gas 2.53 x 10° Pa = 1.67 x 108 Pa X Xga5 5 Xan = 253x110" _ 4514 x 103 1.67 x 10° n, n, 3 _ x isiaxioe = 2H > n= 1x10 x50 m M0 500 bas 18 18 3 "co, = 757X10™ = 42.06 x 10? moles = 42.06 millimoles. 2 18 Ngo, = 4.26 X 10? moles = 0.0426 moles 28. (a) Define the following terms: (i) Ideal solution (ii) Osmotic pressure. (6) Calculate the boiling point elevation for a solution prepared by adding 10 g CaCl, to 200 g of water, assuming that CaCl, is completely dissociated. (K, for water = 0.512 K kg mol"!; Molar mass of CaCl, = 111 g mol!) Ans. (a) (i) Ideal solution: Refer Ans. to Q.16 (i). (ii) Osmotic pressure. It is extra pressure to be applied on solution to Stop osmosis is known as osmotic pressure. (b) Wz = Mass of solute (CaCl) = 10g Mz, = Molar mass of CaCl, = 111 g mol W, = Mass of water = 200 g K, = 0.512 K kg mol! CaCl, —> Ca?* + 2C1 i=3 [- it is completely ionised] AT, = iK, x me W, Elevation in boiling point = AT, = 3 x 0.512 x a ane AT, = SAOTDPIOS = 18 = 069K Boiling point of solution = Boiling point of water + AT, = 373.0 K + 0.69 K = 373.69 K Short Answer Type Questions [2 Marks] 29. (Gas (A) is more soluble in water than Gas (B) at the same temperature. Which one of the two gases will have the higher value of K,, (Henry’s constant) and why? (ii) In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes? [All India] Ans. (i) ‘A is more soluble than gas B. Higher Kj; for ‘A will be less than ‘B’ because gas the value of Ky, lower will be solubility. (#) Negative deviation shows formation of maximum boiling azeotropes. 30. State Henry’s law. Write its one application. What is the effect of temperature on solubility of gases in liquid? [Foreign] Ans. Refer Ans. to Q.5. The solubility of gas in liquid decreases with increase in temperature. Long Answer Type [I] Questions [3 Marks] 31. Calculate the boiling point of solution when 4 g of MgSO 4 (M = 120 g mol”) was dissolved in 100 g of water, assuming MgSO, undergoes complete ionization. (K, for water = 0.52 K kg mor) [All India] 32. In dilute solution, MgSO, —> Mg?* + SO? N . issociati van't Hoff factor, i = lumber of moles of particles after dissociation 2 =- =F 2 Number of moles of particles before dissociation 1 i = 2, Mg = 120g mol", W, = 4g, W, = 100g, K, = 0.52 K kg mol W,; AT, = ix —2x——xK, pW, 2x 4 x 1000 9.52 120 100 2 1.04 =x052=—— = 0.346 3 5 3 Boiling point of solution = Boiling point of solvent + Elevation in boiling point (AT7;) Boiling point of solution = 373 + 0.346 = 373.346 K. Calculate the freezing point of solution when 2 g of Na,SO,(M = 142 g mol!) was dissolved in 50 g of water, assuming Na,SO, undergoes complete ionization. (K, for water = 1.86 K kg mol”) [Foreign] In dilute solution, , Na,SO,—> 2Na* + SOF . Number of moles of particles after dissociation 3 van’t Hoff factor, i = - —____ == =3 Number of moles of particles before dissociation 1 i=3 W, AT, = iK,x —2 x 0 Mg Ws 2 . 1000 = 3 x 186 xX —-x—— 142 «(50 — 3X1.86%20 _ 1116 457% 71 71 Freezing point of solution = Freezing point of solvent — AT Freezing point of solution = 273.15 - 1.57 = 271.58 K. or 0°C - 157 = -157°C Long Answer Type [II] Question [5 Marks] 33. (a) Calculate the freezing point of solution when 1.9 g of MgCl, (M = 95 g mo!"!) was dissolved in 50 g of water, assuming MgCl, undergoes complete ionization. (K;, for water = 1.86 K kg mol”) (6) (Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?