HW14K-G1-Solution Conc

HW14K-G1-Solution Conc - HW Set 14 - KEY Solution...

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Unformatted text preview: HW Set 14 - KEY Solution Concentrations CHM 2045 HW13K-G1,L~S 1 What is the percent sucrose (cane sugar), by weight, in a syrup prepared by . dissoiving 756 9 sucrose in 1.20 kg water? .. meo =. OvVSék WOO: 53.57 0/0 NAM/gab d WWW» 0.76 kcki—LIOkz o . o _ iubiiit of KN03 in water at 20 C 15 31.§g KN03 per 100 m1 H20. The 2. ggfisggy 0f Wa%er at 20°C is 0.998 Q/mi. What is the per cent KN03 , by weight, in the saturated soiution? loo/mt HA0 x 0.99%”3/M = 99.7%.}[10 : 3!.61 x100 ” 1 Do % Kmoa 99.?{o6i- 331.67; '7' 14> tion is prepared by mixing the foliowing numbers of moies of hydrocarbons: 3' Q 3;];01e C7H16 , 1.80 mole C8H18 and 2.16 moie C9H20 . what is (a) the méie fraction and (b) the mo1e per cent of each component in this soiution? x a W C7Hié z lit/7W :: (3‘qu a) CjHIE W 4M1“, Algw +1.3'0M +2./6M X = M4 OtHI?’ : LEO/"41‘? z: 0.35“! Cus m( 11.32", /.1<3’mwl¢4 +I-30M+3./6M _. We‘lsz : 21é/iwdw = C9.’1I2. Xe“ H“ d m.— Wv; Anew»; + 130M +2./6M+&o A WOW mix = MW x loo C HIE ) Xe“ x ,oo = (3.2414 x I00 s. 25/1104. 7 7 Is .... D 0% Ht? 3 H x = X "‘ /O '1’ I? = 0le x loo = 91.1% 09 H10 ) chHzp x W /00.O% 2—2 4. Ca1cu1ate the mo1e fractions of ethy] a1coh01, CZH50H, and water in a so1ution made by disso1ving 9.20 g of the a1coho1 in 18.0 g of water. 1M W CXHSmOH = 9.20 «t X #6qu =2 0.100 M mm—QM, H900 3 [37.0%/ X __’_M._.._. =‘—- LOO/Yuk 110%! C‘H‘OH O ZOOM + Loo/mash X “LO " I‘OO‘“° El :. o 100M + 1 00M 5. a) Ca1cu1ate the m01a1ity of a so1ution made by disso1ving 262 g of ethylene g1yc01, CZHGOZ , in 8000.g of water. MOLHéol : 262%x ‘M =1 9.11M 62.060; W :1 WWW? = "m..." I? z 0.5mm M C>.SYl%'nw b) What is the mo1a1ity of P—dich1orobenzene in a so1ution prepared by disso1ving 5.50 g C6H4C12 in 65.09 benzene, 06H6? [H‘ifif W :1 GOSH/“Wet = osvs‘mwh —.-. O.§7§m (1065.0 k% c) What is the mo1a1ity of a so1ution obtained by disso1ving 15.29 urea, 00(NH2)2 , in 100.0 m] ethano1 at 2000 ? The density of ethano1 at 20°C is 0.789 g/ml. /n~+£ML MALa.. =1 IS fl.<3- X lrh~vic :1 C).2.§Tb rh~+£L 60.!06 W M = [00.0011 X (9.0%")7EL = 775.3 46/ W “ o.l§3M " $11M 7‘ 3le ODj‘E? 2-3 6. What are the mo1arities of the so1utes 1isted be1ow when disso1ved in water? a) 1.50 mo1e CZH5OH in 4.80 1 of solution M: M = M=O.3llfi£§5m MM Algal 4 0.2311 M b) 12.5 g CH30H in 50.0 m1 of so1ution W argon = 12.5%, X %_ -_—. 0.3%; M BQ.@¢K mm = 0.15%; M -.-. item 0117321“! 0.03001 ,L c) 10.0 m1 pure g1ycero1 (C3H803), with density = 1.26 g/m1, in 250.0 m1 so1ution. :. 10.0/w? X [.26 = [1.6 MW 7:83 ‘2; %W=/16 XI/mh zOIm/rwh /?w¢1~¢. ‘3 qq [It WM .. 0 WM = (>st M (9.sz M o 2301 A d) 125 g CO(NH2)2 , 98.6% pure, in 500.0 m1 so1ution. WW 000mg: =- Ixsfck x 0.9m; = 1735.05 now/€44 Cowl—Dz = 113% ‘X £032.12: :3 Log/rka CL! 3 ' .3 2.C>S‘/n~+£L. _ Mag” WI ‘“ ""O”jw~€-‘ “’L “0 "I 7. Ca1cu1ate the weight of each of the fo11owing needed to prepare 2000 m1 of a 1.50 M so1ution: (\ E ’2‘ I 3;! 0’ T, C) J O3 C) C) "X E .0 H «r ~12 \l o? m 2 .t: ~../ 0' "Cl C) 4: Cu Nazca \ 3.00 maven/3, >4 106;: = 3!?06 Nazca?) 2—4 8. What weight of caicium bromide, CaBrz , is needed to prepare 150 m1 of a 3.50 M solution? Describe the physicai procedure in the 1ab for this preparation. $§agam»¢§we yam/W4, CuCSra. = M X V = 3.50 mifié X 0.1501 '-‘- O«§1§M H News, C3371 o.§fl.§mwb. X 100.3. = US$46 Cofly‘z hm; hm Io§¢6 meg at}. W 7§m~€ W bu<EIEL ’aaaljjtiuh.owJoLubVMpujaL obvfitfiflfiql u»&IZL JLrCXjLJf Ida.me Isa/we. 9. Caicuiate the number of moles and number of grams of soiute in each of the foiiowing soiutions: a) 350.m1 of 1.22 M K25 solution Lil/wile. x 03sz = 0.99:; Mkls 0.1+r7 mwh x no = 117.0 KlS flat ?5 b) 2.400 1 of 4.500 M H2804 H.500 Mfk x 2.9001 :2 [ox/M Hgsoq 3 IOXM x 93.13: = Loé X10 :3 Hgso‘, rhwvflL c) 0.250 1 of 8.22 x 10'4 M HC1 tux toflmzin; x 0.1301 = 2.oex lo~ym+ic H01 1 —-'$ 2.05 x :o‘"rmu x 35.51%: = iflwo :3 H06 M 2—5 10. Two sucrose so1utions are mixed. One has a vo1ume of 480 m1 and a concentra- tion of 1.50 M C 2H220 1 ; the other, 620 m1 and a concentration of 1.20 M 012H22011 . What is the concentration of the fina1 so1ution? m [Sr/WW = Isomfbg x 0%on = Que/who. m Q’mm == [.10 mzeux x 0.6101 = 037%, W W 2 OHIO/ruin +- 0.7LIHIM'0‘VL = I-‘ié‘lrym-Qm onto/L + 0.6101 1.100 X W = 1.33% 00/1353 M i 11. Concentrated hydroch1oric acid is 37.0% H01 by weight and has a density of 1.18 g/m]. What is the mo1arity of concentrated H01? FOIL W)W.OL— LOGO/QWOELWHCLI loco/nil x 1.13/“2'336 = limo tad/we; Wyatt, 11M: [moogxosvo :: ygvagna Whoa :- Hen% >< 1M = 12.53me 156.3?3r W :2, name-4 .; ilffimw’ka 011/ 11c: M Ich>C>,€ .XL 12. What v01ume of a 1.000 M KOH so1ution must be diiuted with water to prepare 4.00 1 of 0.512 M KOH? Describe to 1ab procedure. MIXVI : szvl Loco M >c V, :: 0.5711"! X H.001 Vi = W = 2.051 Loco/v; . [1juukqnyufifi nrwikmwua,[email protected] aohibfikéc : 13W H10 2.051 net [.000 M mm :2 V541. == Wool—2.0331 :1 1.9596 H10 2—6 Water is evaporated from 70.0 mi of 0.450 M M9804 solution unti] the solution voiume becomes 45.0 m]. What is the moiarity of M9504 in the resuiting soiution? M. x V, 3 HLX V:L O‘Pl/ W M1 __._ CDHSO M x 70.0/n1 = C1700 M “3&1: ‘iSICD/hvi Caicuiate the equivaient weights of the foiiowing acids or bases given the foiiowing information or equations. a) H25 , diprotic : c) H3A504 —-——-=-- 2H+1 + HAso4'2 - nmwhfit ou¢ba£L ' . ' a4 a, a ' QALJ.- HJAL.$t.Lo W W mm Met/M HHS? Qg/M x M :1. vofitgg; 7. d) CsOH Jfi‘ 4a? #1903. x lmwh : , ‘ [41’ H9 9% +1 e) CH3COZH ———-—=»— H +CH3C02 60.003, X 112$; = 60.06 Wt- Wr 33:» f) Mg(0H)2 /h~£L 2.43, «fit 9) Cat:2 + 2H+1 —-—-—-=sv Ca+2 + (:sz 611.06 x 1M :. 32.02, fie at»? 3?; 15. 16. 2-7 Perform the foiiowing conversions taking advantage of the correct appiication of unit anaiysis. a) 2.0 M H2504 to N "f-ON HLEXDH , giffiL 3 2.C>ATZ§4 x txgfii- == itcp fit :— (WW. 4.130., W3 '5 ' CD-6<> x 113255 =2 (3.21) gggfig HE’PO“ ’ flit ’ § 341» 1. b) 0.60 N H3P04 to M on, C).QL> #4 c) 1.6 x 10‘3 N Sr(OH)2 23- P1 -3 -3 I.(3)<IO 33: x 11%“‘ ,L ESr'OQHAa. ) ¥%i§§L d) 2.2 x 10-2 M CH3C02H to N I 2 2.>< [0'1 I 00 H 14 - . MM x :2 CH5 1. , ) 7,513: e) 0.20 N HCN to M tiCnJ I ' >——1'$=) [L How many grams of soiute wouid be required for the preparation of the foiiowing solutions? Describe the steps of the actual 1aborat0ry procedure. a) 860 m1 of 5.60 N HBr 5W Me.- = Nxv = 960353, x OflséO/Q :2 432.4% H‘5§Q*Jfit X ‘%C>‘q.:3: ‘3 ESfiCD. $5 tiESr )3. b) 4.80 1 of 0.500 N Ca(OH)2 EM Odom; :- N xV = (9.30an x H.301 = 2.41043, 2.510).? x $7.03: : $5.9 ex 43.. c) 1560 mi of 4.40 x 10'4N H23e . ~44 , J/ Law H.356, = Nxv = ALL/o X10 3, x 1.5é/(=é.%6x1043_ -—‘Y __ —2. 636 X10 4*; Hagfi ~ QJYXIO 45/ 2-8 a) 1.86 mi of 2.80 x10‘3 N H3A$04 (WW\ . ‘3 '6 EM H5450” :7— N x v = 25ro><lo ? x 0.001%! = 57.21 X/O )7. _, -e b.11X/O 4% x £17.33; 2 2.07 We”? “’3 For each of the above preparations, add enough disti11ed water to the masses of soiutes ca1cu1ated to yie1d the totai v01ume stated for each soiution. 17. What voiume of 0.105 N KOH so1ution is required to neutraiize a so1ution that contains 1.120 g suifuric acid, H2504 ? [.110 H50. .x 154 :: 0.017;ng H50 0.02.sz KOH 06 1 119.0% “a? 1 " NW, x V = KOH on, V =; :53 fiUDH :; CD.C>25L5§§ ffik :: c3_1_r%'11 ,KJDH n; C).IC>S'%%% 18. How many miiiiiiters of 0.150 N NaOH wou1d be required to neutraiize 35.0 mi of 0.130 N H2504 ? MJVX V’s, 7': mmxv} OJL 19. A 15.0 mi sampie of an acid requires 37.3 mi of 0.303 N NaOH for neutraiiza- tion. What is the normaiity of the acid? MA’XV‘ = meV& on, “gt .3 {UJ, x \LL 2 C).E>C)E;td X 357.330w2 1: C>_-;gfg> h, Nfl£y+ V1 l§.Om-«L 20. An antacid tabiet containing magnesium hydroxide requires 30.5 mi of 0.450 N HC1 for neutra1ization. How many grams of Mg(0H)2 did the tabiet contain? EM no. a No. x V. = 02982)? x 0.03096 : 0.01374HC‘1. 001$? JAEHQQ 0.0137 4%. Mfisfonk E ' wit. M C 2. = . - W a; om 5‘3 379:“ x 12%;;5 ... 19.7. (511%, MW (“105031—951 = 29.1% x 0.01373? : ofloock 2-9 21. A 0.250 g samp1e of impure oxa1ic acid (H20204, diprotic), requires 30.1 m1 of 0.120 N NaOH for comp1ete neutra1ization. What is the percentage of H20204 in the materia1? Ea, NmOH = E? H1010. = My, x V}, = 0.19.0? x 00501.6 =o.oosew} a}.th Hkozoq = 90.04 x M2 -.= 115.03: 2.11. _*} 71mm H1010” = 0.003610% X HSOJEB’ = 0.161 cg °/¢> Hall-LO” = O‘lel a: X 100 : 61+?0/o (9.23.005 22. Quinine, a 1ong-used drug for the treatment of ma1aria, is an organic base that may be neutralized with H01. If 0.675 g of quinine requires 42.0 m1 of 0.100 N H01 for neutra1ization, what is its equiva1ent weight? 5%.HOQ = ESLW = A)... x ‘11., = 0.1004 x Oomoi :: c).C)C)*111:>.L3, ’1‘ '4 = 0.6733; :1 [51 £5% C>.C>C>fl 2.CD.L%. :§%; 23. Citric acid, which may be obtained from 1emon juice, has the mo1ecu1ar formu1a 06H 07 . A 0.250 g samp1e of pure citric acid requires 37.2 m1 of 0.105 N NaO for comp1ete neutra1ization. a) What is the equiva1ent weight of citric acid? b) How many acidic hydrogens per mo1ecu1e does citric acid have? ® qu‘NwH ____ 5% (who, r: mxx VJ, = 0.10:3, x 0.03711 -.= 0.003%] *3. E%_upt- 0.614107 ‘1 :: 63.9 i c3.cx>1sil 75 Jfif 0% PM) 0611307 :1 191 ‘a/M 192 a/M _‘ $00 wafiummwk .._____._..____..__.___ .. . 0.04.; Wm M. 24. A method of adjusting the concentration of an aqueous H01 so1ution is to a11ow the so1ution to react with a sma11 amount of magnesium meta1. ad) ,9) How many mi11igrams of magnesium must be added to 500.0 m1 of 1.012 M H01 to reduce the concentration to exact1y 1.000 M H01? IWMHQQW = I‘imxvm = Lori/mic x O.§ooQ,L=O.S‘Oé.Om-£LH0L firm "We... Haw = me‘lm = Lacey-flu; x O.§0001=O.SOOQMHM W Hog Malawian? = (3.8‘OEOM —- 0.5000 M = oooéomflm 2—10 The above caTcuiations indicate that 0.0060 moTe HCT must react with Mg to achieve a 1.000 M CH1 soiution. The coefficients from the baTanced equation yier the moiar reTationship between Mg and H01: W Ha -.~. 0.0060 M HCL y /M MK = ooo'som mlf 1W H01 M Nab, = 0.0030 MM? XlH,$i :- 06 on, M 73m<3 MON 25. A soTution of phosphoric acid was made by dissoTving 10.0 g of H3P04 in 100.0 g of water. The resuiting voiume was 104 m1. Caicuiate for the soiution: a) density b) moTe fraction of soiute and soivent c) moTarity d) moTaTity e) normaTity (triprotic) M. = W - W “3 b - W “WM - [-05 3a m1: AMHSPoq = 10.0%, Yiwmgh; =2 0.101MH3poH 982c3‘3 W4 H10 =- IOOJDX X [M : S1599 WHLO H.013 __ (9.102% _. XHasPOq " O./01m&.+ as:an " O‘Omo SSH? 0M4 )( :: : "‘0 O. 101M + S“.$"i?/m1n O'SYQ‘O ’ .3 mm (:3 W I 3 10.0 .5 HFJPO.g : :JOéimw-LL . z. z; M _ 6 Oneal “I ” 0'0"“ M 0L3 ' __. WW 3 met/ML _, . a n¢4$mKE; KfigM%Nfit D “makfi ~ L01.1§f =1 l CDCL nw exnmafléttra onth x s = 1.9L; .. 1 l :4— if .. .‘1’1N 2—11 26. A 20.0% so1ution of su1furic acid has a density of 1.14 g/m1. Calculate a) the mo1arity b) the norma1ity c) the m01a1ity. W04jikw U 05 .77104ML Q4. L(9CD.JL JDWéL ( LIDCD Xf/CEITwQ) I 3 3 . I ‘1 = ' looxoMXiflfi wamogmm W stoq w}. Loo/(M; : 1.15: x (03% x 0.7.00 = 1735 c5 1H,,qu m H130” w}. Loci/ML : 11%05X1M ~.-. 2.32M 0:th "T2JLLuuf¢kL., fYWAP£QA¢j:6, :: 2.7511/Q?::f:;‘ on, Q 359* p4 ,6} nO(»N-wLQéI%} 2.510122. x l = Him 011. 3.69 N A fi if QW— FMCA\)k/.@01Mifl0gm# /.I‘1 VIC-5% ) lchb/(Q.31M306LWRALLJ6 H1303- . 3 Tit. ~ {kw/14. (5.“! X lo «X X 05600 = 9113.) MICK/MAM H10. 4 ‘ 2 WW 3 1.31M W WW 0.9m.ch “ Q'SHf—Eff‘mlg‘lm ...
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This note was uploaded on 11/16/2011 for the course CHM 2045 taught by Professor Mitchell during the Fall '07 term at University of Florida.

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HW14K-G1-Solution Conc - HW Set 14 - KEY Solution...

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