Chemistry Practice Problem 1 For a given isotope of an clement, the atomic number plus the atomic weight is 148, and their difference is 58. How many protons dues an atom of the isotope contain? (A) (B) 8 (C) 90 (D) 148The atomic number, Z, is equal to the number of pi tons in the nucleus. ‘The atomic weight, A, is appro mately equal to the number of protons and neutrons the nucleus. A+ Z = 148 = P protons +N neutrons + P protor A~Z= 658 = P protons +N neutrons ~ P protor Prom the second equation, 5 = N neutrons From the first equation, 148 = 2(P protons) + 58 45 = P protons) Answer is A. Problem 2 The group of metals that includes lithium, sodium, po- tassium, rubidium, and cesium forms a closely related family known as the (A) rare carth group. (B) halogens. (C) alkali metals (D) alkaline earth metalsLithium, sodium, potassium, rubidium, and cesium oc- cupy the first column of the periodic chart, knewn as Group TA or the alkali metals, Answer is C. Problem 3 ‘Which ofthe flléwing compounds would be font, eon dering the electronegasivities of the elements? electroneyativity K os © 25 1 25 a 30 N 50 ° 35 8 40 (a) co (h) No (©) (b) KCLConsider the differences ctronegativitics for wash compound. compound —clectronegativities difference co C=25 0-35 10 NO N=30 0-35 05 ly 1=25 1=25 0 Kel K=08 Cl=30 22 ‘The difference in electronegativities rinst be greater than 1.7 for the bond to be considered ionic. Only KCL sets this requirement; the other compounds are eo: sidered to have covalent bonds. Answer is D. Problem 4 What is the Lerm for a quantity of a substance to which # chemical formula can be assigned and whose mass is equal to its formula weight? (A) a molecule {B) amole (C)_ an equivalent weight (D) 8 one-normal solutionA mole of an element will have a mass equal to the element's molecular weight. The molecular weight. is generally the same as the formula weight. Answer is B. Problem 5 ~- What is a distinguishing characteristic of the halo- gens? (A) ‘They are phosphorescent. {B) Next to the noble gases, they are the most chemically inactive group. (C) They readily accept an olectron from another atom to form compounds. (D) They have a high electrical conductivity.‘The halogens need one electron to complete an elec- tron shell. They readily accept this electron from other atoms to form compounds. Answer is C. Problem 6 at is the valence (oxidation state) of carbon in sodium carbonate (NagCOg)? (A) -4 (B) -2 (C) 42 (D) +4ince soditun carbonate is a neutral compound, the sum of the oxidation numbers is zero. Oxygen has a valence of ~2, and sodium has a valence of +1. ‘The valence of carbon can be calculated as (2)(41) 42+ (8)(-2) =0 z-4=0 z=+4 Answer is D. Problem 7 Which of the following chemical fornuulas is incorrect? {A) Ca{ON), (B) NagCOs (C) Cac (D) KOHExamine the oxidation numbers of the molecular ele- Imants to check for neutral molecules. compound or oxidation molecule element neutral? CuoH)s Ca ys on NayCOy Na ye 60s cact co 42 20 a 4 KOH K 4" yes on 1 ‘The answer is CaCl Answor is €. Problem 8 Uranium-235 and uranium-238 have the same num- ber of which of the following? (A) neutrons: (B) protons (C) electrons (D) protons and electronsUraniuin-235 and uranium-238 are both isotopes of uranium, Usaniuin has 92 protons and 92 electrons; the isotopes differ in the mmber of neutrons in the nucleus. Another notation for urantuun-235 and uranium-238 is 2431) and ?88U, respectively. Specifying the atomic num- ber (2 = 92) is redundant because, by definition, ura- nium has 92 protons, Answer is D, Problem 9 Which of the following elements is the most clec- tronegative? (A) Br (B) cl () F (D)Electronegativity is the ability of an atom to at- tract an electron, According to periodic trends, elee- tronegativity increases from lefé to 1 to top on the period table. ‘Therefore, fluorine (F) is the most electronegative element, not only among the ‘answer chotees, but among all elements, Answor is C. Problem 10 What is the oxidation number for chromium (Cr) in the compound BaCrO4? (A) +1 (B) +2 (©) +4 (D) +6Since bavium (Ba) has an oxidation number of +2, the chromate ion (CrOg) has an oxidation number of 2. Oxygen always has an oxidation number of ~2, so the eromiuen rust have an oxidation mamber of +6 in order to give the chromate jon a net oxidation number of 2 Answer is D. Problem 11 What are the chemical formulas for the following compounds: aluminum nitrate, magnesium hydroxide, calcium oxide, and cupric carbonate? (A) Al(NOs)3; Mg(OH)2; CaO; CuCOs (B) AlgNOs; Mg(OH); CaO2; CuCOg (C) AINOg; Mg(OH)2; CaO; Cu(COs), (D) AINOs; Mg(QH); Caz03; CuCO,Answer is A, Problem 12 What is the maximum possible positive oxidation aumber for the element Br? (A) +1 (B) +3 (©) +4 (D) +7Bromine has an oxidation number of —1, which means that it normally accepts one electron to com= plete its outer shell of eight electrons. Alternatively, it could give up seven electrons to have a full outer shell. Answer is D. Problem 13 While moving from left to right across the second row of the periodic table (i.e., from Li to Ne), the atomic radii tend to (A) uniformly increase, (B) uniformly decrease. (C) remain the same. (D) first increase, then decrease.Elements with smaller radii are more stable than those with larger radii, Elements increase their stability toward the right side of the periodic table. Answer is B. Problem 14 Which of the following elements has the largest first ionization energy? (A) Ba (barium) (B) Cu (copper) (C) Ne (neon) (D) 8 (sulfur)Ionization energy is the energy required to com- plotely remove an electron from an atom. It is usually ‘expressed in joules or joules per mole. When expressed por unit charge (Le., in J/C, same as volts, V), it known as ionization potential ‘Dhe first ionization energy is the energy required to re- move an eleotron from the outermost shell. Tonization ‘energy decreases as the number of electrons (shells) in- creases, Neon is not only a noble gas, but it has the fewest number of shells. Answer is C. Problem 15 ‘The term “divalent” means (A) an ion’s oxidation number is +2 (B) an ion’s oxidation number is -2 (C) an ion’s oxidation number can be +2 or —2 (D) the ion can have two different oxidation num- bers‘The terms “divalent” and “bivalent” mean the same thing: « valence of 2. Answer is C. Problem 16 ‘Two major types of chemical bonds are observed in chemical bonding: toni and covalent. Which of the following has a bond that is the least ionis in character? (A) Nac (B) CH (©) (®) 40‘The electronegativity difference between two similar atoms is cr. ‘Thorofore, the Hz bond is completely covalent. It has no ionic bond characteristics ‘The answ ©). part2Problem 17 -. What is the percentage (by mass) of hydrogen in glucose (Cglli205)? (A) 6.7% (B) 9.3% (C) 17% (D) 40% The combining Weights of each eloment ace ©: (6 (12-85) = 72 g/m 8:03) (1 $5) = 12 g/mol 2 0: (6)(10 85) = 06 g/met he mou weight of luo is 72% 412 5 496 mot * 2? not + °° jyot = 180 6/mel ‘Thue mass fraction of hydrogen in ghcose ix Answer in A.Problem 18 Vitamin C has the molecular formula of CoHsOo. How many gram-moles are in 23 g of vitamin C? (A) 0.13 mol (B) 0.39 mol (C) 3.08 mol (D) 7.66 mol ‘The approxinnte molecular weight of ClsQp is 9) (2 la) + (1) +0 (16 5) 76 g/mol ‘The number of moles is iW” Tig = 0.13 mol Anawer is A.Problem 19 What is the mass of 0.01 gram-moles of NagSO,? (A) 071g (B) 119g (C) 142g (D) 2.38 g + The combining weights of each element are a (2) (28:09 £5) = 4.98 g/mol a 5: (0 (s2.07 -£5) = 92.07 g/mol 0: (4 (18.00 £5) = 64.00 g/mol he molecular welght of NaySO, i 59m em. 404.0 gm 12.1 gn The mass of NaaSOq fs (0.1 mod (142.1 8 2 (a2 g) Answer is ©.Problem 20 A solution is adjusted from pH 8 to pH 9. ‘The rela- tive concentration of the hydrogen (H"] ion has changed by a factor of what? 1 O) a5 1 ®) a5 5 (D) 10 ‘The definition of pH is pH = — logyo{E*] ww» (5) For pH = 8 (H+]=10"*. For pH = 9, [H+] = 10~°. The change in [H") is by a factor of 10-8 /10~ Answer is B.Problem 21 An unknown quantity of hydrogen gas has a volume of 2.5 L at STP (0°C and 1 atm). What is the mass of hydrogen? (A) 0.073 g (B) 0.19 g (C) 022 ¢ (D) 0.51 ¢ Use the ideal gas laws pV nD yada =O51 08206 atm L nol 73K = in 1116 mo ?)(1.0079 g) 0158/0 uw 4 = (0.16 mop (20188 £,) = 0.235 g (0.22 8) 0.08208, Answer is ©,Problem 22 A compound in gas form has a mass of 0,377 g and occupies 191.6 mL at standard conditions (0°C and 760 mm Hg). What is the formula of the compound? (A) CHa (B) CHs (C) CsMhz (D) CoHe By Avogadro's hypothesis, | gram-mole of any gas oc- ‘eupies 22.4 L. By a simple ratio analysis, the mass of 1 igrannemole of Uhe compound is n= (28h) ware ee Calestave the molecular weights of the compounds listed compound molecular weight Hy 12.011 + (41.0079) = 16.018, Cale (8)(12.011) 4 (8)(1.0079) = 44.006 Gsltig—(6)(12001) + (12)(1.0079) = 72.150 GH, @)(12.001) + (6)(0.0079) = 30.068 Answer is BeProblem 23 ‘An unknown gas with a temperature of 25°C and a pressure of 740 mm Hg is collected in a sampling bag. ‘The volume and mass of the gas are 24.0 L and 34.9 g, respectively. Which chemical formula could represent, the gas? (A) Na (B) Ar (C) Hes (D) HCL + Use the ideal fas law to convert the volume to stan dard conditions. __ (740 msn Hx) (760 mm Hg)(25°C 4 2 =214L Avogadro's hypothesis states that 1 gram-mole of any deal gas occupies 22.4 1, at standard conditions (0° and 760 mm Hg). The molecular weight is _ (4.9 y)(22.4 L) MW) gua = (OM) 2a = 36.5 g/mol‘The molecular weight of HCl is OW) = (1S = 86.5 g/mol +(1) (355 ©) Answer is D. Problem 24 A transportation company specializes in the ship- ment of pressurized gaseous materials. An order is re- ceived for 100 L of a particular gas at STP (0°C and 1 atm). What minimum volume tank is necessary to transport the gas at 25°C and a maximum pressure of 8 atm? (A) 10L (B) 12L (Cc) 4L (D) 16DUse the ideal gas law. _ (1 atmn)(100 L)(25°C + 278.16) = Batm)(OG + 273.18) 3.6L (14 L tank minimum) Answer is C. Problem 25 It is known that ozone (Os) will decompose into oxygen (Oz) at a temperature of 100°C. One mole of ozone is sealed in a container at STP (0°C and 1 atm). What will be the pressure of the container once it is heated to 100°C? (A) 14 kPa (B) 2.1 kPa (C) 37 kPa (D) 210 kPa‘Phe decomposition reaction equation is heat energy + 205 —+ 302 In reaction equations involving ideal gases, the eoefli- clents can be interpreted as the number of molceules, the number of volumes, or the number of moles. In this ‘ease, 3 anol of oxygen are produced from 2 mol of ezone. Use the ideal gas equation of state. pV =nk Since this is a constant-volu my RD PL v (Late) (100°C + 273)(3 mol) ~ (@C +273) mol) 2.05 atm = (2.05 almn)(101.3 kPa) = 207.7 kPa (210 kPa) Answer is D.Problem 26 A sample of an unknown compound is found to be 49.3% carbon, 9.6% hydrogen, 19.2% nitrogen, and 21.9% oxy- gen by weight. What is its molecular formula? (A) OHgNO (B) CiHgNO (C) CyHgN20 (D) C3lNO step 12 Divide the percentage compositions by the atomic weights ofeach element, 49.8 1011 = 4.1046 mal 6 H 9.5248 mol L009 ye 28 set ma 4.007 al nog = 1.3688 mol vom ©:Determine the smallest ratio from step 1. step 8 snvallest ratio = 1.3688 ep 8. Divide all ration by the mollst ratio, 4.1046 o Tyea8 ~* py, 2828 13107 we 1.3688 es step 4 Write the chemical formula using the results from stop 3 CHINO. Answer is D. Problem 27 -. In a laboratory experiment, @ student analyzed a substance with 2.7626 g of lead, 0.00672 g of hydrogen, and 0.8534 g of oxygen, What is the empirical formula for the substance? (A) PbpO4He (B) Pb,O.H (C) Pb,OH2 (D) PbOstep 1: Pind the gravimetric fractions of each ele ment Me = 2.7626 y + 0.00672 4+ 0.8534 = 9.02272 g mary _ 2.7626 g my BG2272 = 0.76258 muy _, 0.00672 g me BODIE g = 0.00185 to = ZO — 0.8534 g ine ~ BORED g. = 0.23557 Divide the gravimetric fractions by the atomic weight of each element. pp, 2.76258 g + = 8.6806 x 10-8 mot aura fH DOO. 1.2365 x 10° mol 10079 Ee AT24 x 10° mol step # Determine the smallest ratio from stop 1 smallest ratio £8855 %€ 107 by inspectionstep 3: Divide all of the ratios from step 1 by the Pb: a 2.005 ve ee o Mem step 4: Write the chemical formula using reaules from step 3. (Recognize that there may he snail errs present in the analysis that will five slight discrepancies.) ‘The formula is PbOyH. Answer is D, Problem 28 What is the gravimetric (i.e., mass) percentage of oxygen in KaCr0,? (A) 33% (B) 42% (C) 57% (D) 66%11. The combining weights are : ‘3 By K: (2) (39 +) = 78 g/mol 9 6 i: (1) (52 S) = 52 ¢/mo Cr: (1) (52 =) 52 g/mol O: (4) (6 =) = 64 g/mol ‘The gravimetric percentage is the same as the mass per- centage. 4 2 1 t re = 0.33, (33%) 78 © 4.52 = +64 mol mol mol Answer is A. Problem 29 ‘The mole is a basic unit of measurement in chemistry. Which of th: following is NOT equal to or the same as 1 mol of the substance indicated? (A) 224 L of nitrogen (Np) gas at STP (B) 6.02 x 10° oxygen (0g) molecules (C) 12g of carbon atoms {D} 16 g of oxygen (03) moleculesOxygen has a molar mass of 16 g/mol. Therefore, 1 mol of Oz has a mass of 32 g. [ tesa) | Part 3Problem 30 During a laboratory experiment at 1.0 atm and 25°C, a student observed that oxygen gas was produced by de- composition of 15 g of sodium chlorate. What was the volume of oxygen? (A) 1271 (B) 3.85 L (C) 517L (D) 6.541 ‘The decomposition reaction is 2NaClOy —+ 2NaCl + 302 ‘The molecular weight of sodiuin chlorate is 22.000 a6 22.990 sol ig + ® (15.000 58 Soi + (19.900 in) = 106.44 g/mol Calculate the moles of O» produced NaCI 106.44 © O44 ol (0.14092 mol NaClOs) 8 mol Ox Tanai NaCIOy = 0.14002 m0] NaCIOy, ) = 921198 motosleal gas law to calculate the volume, WY = nit Fe = 0.08206 atm mol T= 278-4 25K = 200K (0.2188 mot 03) (0.08206 57%) (a8) atm =517L Answer is C. Problem 31 1Ne,O0s reacts with HCI, but not by the stoichiometry implied in the following ‘unbalaneed chemical equation 1NayCO3 + HC] —+ NaCl + Hx0 +0, What is the smallest possible whole-number coefficient for NazCOs in the bale anced equation? (aya 2 4 (0) 5‘The simplest balanced equation is NazOOs + HCI —> 2NaCl + Hz0 + CO2 ‘The smallest whole-number coefficient. for Nays is 1. is ( Problem 32 Which of the following is the result of the reaction given? 5802 + 2KMnOq + 2H,0 — 7 (A) 2MnSO, + K280, + 2H2804 (B) 2MnSO, + K2802 +HSO4 + H20 (C) 2MnSO4 + K2804 + H80, (D) Mn$O4 + 2K,$04 + 2HS04Only the products listed in option (A) would balance the elements on the right and left sides of the equation. The answer is (A\ Problem 33 ‘Which are the oxidizing and reducing agents in the following reaction’? 200, + Kz0r04 —+ 20400 + CrO2Cl, + 2KC1 (A) oxidizing agent: chromium; reducing agent: chlorine (B) oxidizing agent: oxygens reducing agent: chlorine (C) oxidizing ayent: chromium; redueing agent: oxygen (D) There are no oxidizing or reducing agents in this reuction.‘Tho oxidation state of chromium is 6 in each compound, Carbon remains with ¢ 44 oxidation state throughout the reaction, ‘The axi- dation states of both chlorine and oxygen remain the same throughout ‘this reaction. Thus, nothing is exidized or reduced in the reaction, ‘The answer is (D). Problem 34 A voluitric analysis of a gaseons mixture is as follows, CO, 12% 0, 4% N, 82% co What is the percentage of CO on a mass basis? (A) 0.5% (B) 0.8% (©) 1% (D) 2%name vo (%) mole frac. (mo!%) _ mol. wt. _ mass () ee ~ a (cose a om Os 4 oot x No & (Ope ex et 20m a co 2 i 8 8 300 ‘The total mass ofthe mixture is 30.08 kg, ‘Thus, the mass percentage ‘of CO is given as follows. ass % of COm PSE =9.02 (2%) oe ‘Tho answer is (D). Problem 35 "The following equation deseribos the decomposition of potassium chlorate to produce oxygen Has, 2KC10s —+ 2KCI (solid) +302 (gos) Approximately how many grams of KCIOs must be used to produee 4.00 1. of | (x (ges) measured at 7400 tore and 30°C? (A) 110g (19 (©) 10% (©) ugVaaL T= 30°C 4279" = pV sneer ra BE st)(400 L) ) (20) * oh. = 1.87 mol no. of males KO1Os ‘no. of moles of Oy 2 mol/$ mol 2 st x0 a= (228 ne = 1.05 ot ee. £) ( mol Miao, = 301 £5 +365 £5 + (16 5) 4 ml = 128 gfe ne. ame KCIOs = (1.05 mat) (128 8 ‘The answor is (C). = 1296 (1308)Problem 36 12.25 « of pure caleium metal are convertod to 8.13 g of pure CxO, what is the ‘atomic weight of ealcium? ‘The atomie weight of exygen is 16 g/mol. (A) 28 g/mot —(B) 83 g/mol —(C) 37 g/mol (D) 41 g/mol ‘Tho stoichiometric equation is Ca+0—+Ca0 CaO — C20 One mol of oxygen. and 1 mot of calcium are required to make 1 mol ef CaO, 3.13 g~ 2.25 « Mag - 2956 atomic weight of C= Boe ES =A g/mol he answer co}Problem 37 Consider the following reaction at equilibrium, ‘3H (as) + No (gas) =3 2NHs (gos) AH = ~11.0 kcal/mol Which single change in conditions will eae a shift in equilibrium toward an increase in production of NHy? (A) removal of hydrogen gas (B) inerease in temperature (©) increase in volume of the system (D) increase in pressure on the system According to Le Chitelier’s prineiplo, the offects of each change in condition are as follows. removal of hydrogen shifts equilibrium to the reactants increase in temperature shifts equilibruim to the reactants inerease in volume shifts equilibrium to the reuctants increase in pressure _shifts equilibrium to the products The answer is (DProblem 38 Consider the following reaction at equilibrium, 3N2+$HpNH3 AM = -11.0 keal/mol ‘Whnt would be the expected effect on the amount of NH producsd under each of the following conditions? I. raise the temperature TL compress the mixture TIL add additional Hy inerease, Tl: increase, TH: decrease crease, IT: decrease increase, III: decrease According to Le Chételir’s principle, each change has the followings fects. . raise the temperature: shifts equilibrium to the reactantsll, because the reaction is exothermic 1, compress the mixture: shifts equilibrium to the products becaase prod- ucts coutain a simaller number of moles TIL add hydrogen gas: shifts equilibrium to the products because adding addtional reactants will force the formation of more products ‘The answer is (A).Problem 39 Consider the following reaction. MeSOx (solid) + MgO (solid) + 80s (gas) What is the equilibrium constant for the given reaction? (a) w= MAIO] gy y Med 2[MgSO«] [MgO|[S0s] (C) k = [MgO][SOq] (D) & = (80s) Solids have a concentration of 1. Therefore, & = [SOs].Problem 40 Consider the reaction shown. Hy (gas) +z (gas) = 2H (gas) ‘With keq = 25, determine the number of moles of Hy remaining when I mol each of Hy and Ip reach equilibrium in a 1 I, vessel. (A) 1/6 mol (B) 2/7 mol (C) 5/7 mol (D) 5/6 mol Hy (90s) + Ta (gas) = 2A (gas) initial moles 1 1 0 final moles l-« l-« = mol]” i [i wet ~ mol mol tbe = a? ~ @=aj(l—a) =25fat 7 42? = (25)(1~ 224-24) 22? — 502 +25 = 0 = Hoot V(60? = HEH, - (2)(21) = 5/8 mol or 6/7 mol =3 Only the vecond value for # makes sense, because the fret value is ‘genter than the intial amounts of Hs and Jy. ‘Thus, tho remaining, umber of moles of Hat equilibrium is 1 ~ f= 'mol. [the answers | Part 4Problem 41 How is “molality” defined? (A) the mumber of moles of solute in 1000 g of solvent, (B) the number of moles of solute in 1 L of solution (C) the number of gramformula weights of solute per liter (D) the number of gram-equivalent weights of solute in 1 1 of solution Molality is defined as the number of moles of solute per 1000 g of solvent. Option (B) is the definition of molarity, option (C) is the definition of formality, and option (D) is the definition of normality. Problem 42 ‘The solubility of barium sulfate, BaSOg, is 0.0091 g/L at 25°C, The molecular weight of barium sulfate is 233 g/mol. What is the value ofthe solubility product constant ky for BaSO4? (A) 1.52 x 10° mol? /L? (B) 4.24 x 10-8 mol?/L? (©) 8.63 x 10-7 mol? /L? (D) 2.98 x 10-% mol?/L?BaSO4(s) + Ba”* (ag) + SO" (ag} hoy = [Ba?*][S0%7] 7 g) (1mol [BaSO«] = (0.0001 8) (# *) = 3.9 x 1075 mol/L [Ba?+] = [S03-] = [BaSOq] 7 _5 mol\* Kap = (a9% 10 . a) = 1.52 x 107° moP/L? ‘The answer is (. Problem 43 ‘The pH of a 0.001 M HCI solution is (A) 1 (B) 3 (G) 5 (D) 7Molarity (M) is the number of gram-moles of solute per liter of solution. To calculate pH, the ionic conventra- tion of H+ ions in moles per liter is needed. This is equal to the molarity for HCL. L = toes | S008 Answer is B. Problem 44 Which of the following oceurs when table salt (NaCl) is added to continuously heated boiling water? (A) ‘The water continues to boil (B) ‘The water momentarily stops boiling. (C) The water boils even more agitatedly. (D) The temperature of the water decreases but boiling continues uninterruptedSolution ‘The boiling point of a solution is higher than the boiling point of the pure solvent at the same pressure. Addition of the solute will momentarily stop the boiling process until the elevated boiling point temperature is reached. Answer is B. Problem 45 How many milliliters of 1 M NaOH solution will 25 mL of 2 M HeSO, neutralize? (A) 25 ml, (B) 50 mL (C) 75 mL (D) 100 mLBalance the equation. 2NaOH + Hy80, — NagS8O, + 2H,0 2 mol of NaOH neutralize 1 mol H28O,, ‘The number of moles of H_SO, is iSO. = ( a) (0.025 b) = 0.05 ‘The number of moles of NaOH needed is mol NaOH mol 1250, Mx.on = (0.05 mol HySO,) (2 = 0.10 ‘The volume of NaOH needed is (0.10 mot Naow) { —-— | = 0.10 1, NaOH mol Tt (100 mL NaOH) Auswer is D,Problem 46 _. What is the normality of each of the following solu- tions? 1. 500 ml, of 0.25 M HzSOq I, 41.7 g of K,Cr,0; in 600 ml. of solution (ion- izes to Cr**) TIL 0.135 gram-equivalents of HO, in 400 mL of solution (A) 2.N; 5.2 Nj 1.66 N (B) 6.25 Nj 0.7 N; 0.338 N 42 N; 0.338 N 28 N; 0.56 N Normality is the number of gram equivalents of s0- lute per liter of solution. 1, 500 mb of 0.25 M H,SO4 N = M x A oxidation mumber HgSOq + HpO —+ 2H* + $O;~> + H20 A oxidation number = 2 NV = (0.25 M)(2) =0.5NIl. 41.7 g of K2Cr207 in 600 ml, of solution KyCrQ7 + H20 — 2Cr** + K,08" +120, A oxidation number = 6 MW of K2Cr07 = (2 1) + (2)(52.0) + (7)(16) = 294.2 g/mol = 49.033 g/mol va (tite ( 49.033 a 0 iol TIL. 0.135 equivalents of HgSOq in 400 ml. of solu- tion _ 0.135 equivalents o4L 0.3875 N_ (0. Answer is C. N y)Problem 46 A wastewater treatment plant uses chlorine gas as areactant. A tank js filled with 800 m of 20°C water, and chlorine is added at a dosage of 125 g per cubic meter of water. (Assume all of the chlorine dissolves and none initially reacts chemically.) Henry’s law constant for the chlorine is 15.2 atm. If the atmospheric pressure is 1.0 atm, what is the theoretical partial pressure of the chlorine gas at the tank surface immediately after the gas is added? (A) 3.2% 107% atm (B) 4.8 x 10~* atm (C) 0.039 atm (D) 0.11 atm « thdifiags oP Shoring in the tank is (02s S) (600m) 100 kg, ‘The molecular weight of chlorine gas ie “)W, (2) (365 85) <1 g/mol The number of moles of chlorine is m__ 100000 ¢ eT ol = 1408 mol ‘The mans of water in the tank is (500 (1000 3) = 0000 keThe number of moles of watar in the tank is m0 = SW 5 (800000 ke) (1000 4) 18 = 4.44 x 107 mol ‘Tho mole fraction is, Ton + mH 1408 mol = Tae mol + 4,44 107 tol 17 x 10-F From Henry’s law, the partial pressure is proportional to the mole fraction, (15.2 atm)(3.a7 x 10-4) = 48x 1074 atm, or is BeProblem 47 How many liters of 2M solution (i.¢., a molarity of 2) can be produced from 184 g of ethyl alcohol (CHsCH. on)? (A) 1.5L (B) 20L (C) 2.51 (D) 5.0L, ‘The moleularvelgt of ety lool i (2) (12 585) +06) (0,8) +00 (16 585) = 46 g/mol ‘The number of moles is na, I84e = MW" jg & 4 mal ‘The molarity is the number of motes per liter, Since the molarity is M = n/V, Answer is B.Problem 48 ‘ohn poured 100 lof 40M NaCl, 100 mLofS0N KOH, a5 ml of 020 HC soins nether ad then sade tet lmetabe 10 L ais inte froin 01M x 10mL-+02M x 5.0m = 00200 100-0 a fa]