CHEM 207 REVIEW EXAMS QUESTIONS AND ANSWERS

CHEM 207 REVIEW EXAMS QUESTIONS AND ANSWERS Practice questions for the exam with solutions. For you to do well on the exam, please work on the questions before looking at the solutions. If you find an error, let me know so I can let everyone know. REMEMBER THE EXAM IS ON Monday, Sept 28th on Top Hat. The exam will have 20 questions and only one hour. Studying the lecture ppts and lecture questions would also be helpful. Lecture 1A questions: Density and stat analysis, Q and T tables, % Tolerance tables are in the back. Density question 1. Density of water was the first experiment we did in the lab. Which statements are false. a. Density of any substance is define as mass per unit of volume. b. The unit used in the lab for density was ml/g. c. Density of water can increase with increasing atmospheric pressure. d. Density of any substance is constant at a constant pressure and temperature. e. When doing the density experiment, the sig figs were dictated by the precision of the balance and the 25 ml transfer pipette (25.00ml). Answer: b is false: density units were g/ml Sig fig question: 2. How many sig figs are in 0.00870 a. 6 b. 5 c. 4 d. 3 e. 2 Answer: d – 3 sig fig—you must recognize that leading zeros are not significant but trailing zeros are. 3. What statement is true? a. Accuracy is a measure of how reproducible the result is. The precision of an experiment is a measure of how close the result comes to the true value. b. In research, you can always find the true value of the data. c. Fluctuations in observations, which yield results, can change the true value of the experiment. d. Accuracy of the experiment is a measure of how close the result of the experiment comes to the true value. The precision of an experiment is a measure of how reproducible the results are. e. None of the above Answer: D- need to know the difference between precision and accuracy 4. A student performs an experiment four times and gets the following results: 18.1, 16.9, 18.4, and 17.5. All of these data are acceptable according to the Q-test. It turns out that the true value for this unknown is 19.0 What is the % RELATIVE ERROR for these results? a. 0.0684% b. 6.84% c. 5.37% d. 0.0537 % e. none of the above Solution: When you have data points and a true value , need to know how to calculate absolute error and % relative error. I true value – exp. valueI = absolute error I true value – exp. valueI /true value x 100 = % rel error Don’t mix up with absolute / % rel error when talking about measuring instruments (remember we use the tolerance table which is in the back.) Exp value = average of data points = 18.1 + 16.9 +18.4 +17.5 /4 = 17.7 19.0 -17.7/19.0 x 100 = 6.84 % answer B 5. A student performs a titration experiment four times and gets the following results for the %KHP in the unknown sample: 66.70%, 66.83%, 67.24%, and 66.65%. It turns out that the actual %KHP in the unknown is 65.54%. Which of these results should be rejected (90% confidence level)? A) 67.24% B) 66.65% C) 67.24% and 66.83% D) 66.65% and 66.70% E) none of these results should be reject Solution: This is a Q test question. You should know the formula of how to check for an outlier. Actual value (65.54) is there so you can identify the outlier easier. Put data in ascending order. 66.65, 66.70, 66.83, 67.24 I 67.24-66.83 I/I 67.24-66.65 I = 0.41/0.59 = 0.69 if Q exp Q crit you can reject, Q crit is 0.76 at 90% so can’t reject -----Q table will be on the test—E 6. A student has 4 data points with the average being 11.9 and has the standard deviation of 0.75. All of these data are acceptable according to the Q-test. What is the confidence limit at a 95% confidence level? A) 11.9 +/- 1.20 B) 11.9 +/- 1.04 C) 11.9 +/- 1.19 D) 11.9 +/- 4.46 E) 11.9 +/- 0.521 Solution: Need to know the formula for calculating the confidence limit which is ts/square root of N Need to look up t in t table in back --- 4 data pts with 95% conf. level is 3.18 So 0.75 x 3.18/2 = 1.19 so answer is C Lecture 1B questions: Glassware 1. Determine the grams to weigh out to make a 0.0500 M XY stock solution in a volume of 500ml. The molecular weight of XY is 329.2 g/mol. A) 0.0075 g XY B) 8.23 g XY C) 32.92 g XY D) 8230 g XY E) none of the above Solution: Basic molarity problem: convert ml to l so 500ml/1000ml = 0.500 L Then 0.500L x 0.0500 Mol/ L x 329.2 g/mol = 8.23 g (notice all units cross out except grams) so answer is B 2. Determine the molarity of XY stock solution with 0.6784 g of XY in 40ml of water. The molecular weight of XY is 329.2 a. 0.05152 M b. 0. M c. 0.M d. 0. M Solution: Another basic molarity problem: convert gram and volume to molarity 0.6784 g x 1mol/329.2 g= 0. mol of XY then molarity is mol/liter so convert 40ml to liters 40ml x 1 liter/1000ml = 0.04000L so mol/liter = 0. mol XY/0.04000L = 0.05152M so A is the correct answer. 3. What is the dilution factor if you took 20ml of stock solution and dilution to 500ml? a. 30 b. 25 c. 20 d. 15 e. 10 Solution: Dilution factor is total volume/ aliquot volume so 500ml/ 20ml = 25. DF is 25 and answer is B. 4. A student transfers 3.0 mL of a 6.5 x 10-3 M NaOH stock solution into a 100 mL volumetric flask and fills it to the mark with deionized water. What is the molarity of the new solution? (Formula weight of NaOH = 40.00 g/mole) A) 2.0 x 10-5M B) 2.0 x 10-4M C) 0.22 M D) 0.44 M E) none of the above Solution: Just a dilution problem MV=MV so 3.0ml x 6.5 x 10^-3M = 100ml x ? ? =0. but w/ 2 sig fig answer is B 5. What is the % RELATIVE ERROR when a 80 mL volume is measured with a 100 mL buret? A) 0.08% B) 0.2% C) 0.8% D) 0.1% E) none of the above Solution: Now we are talking about rel error with measuring instruments and using the tolerance table (tolerance is the same as rel error) everything on that table is a %. Relative error is always is % and the delivery volume is taken in to account. Look up 100 ml buret tolerance on table----0.08% Take 0.08 X 100 (max vol)/80(delivery vol) = 0.1% answer is D Remember if we are talking absolute error, it is never a % (so divide by 100) and delivery volume is not taken in account (always max volume). 6. What instrument do you need to deliver 23.5 ml +/- 0.03ml? (Note that there may be more than one correct answer.) A) 25 mL graduated cylinder B) 25 mL buret C) 25 mL measuring pipet D) 25 mL transfer pipet E) none of the above Solution: deliver 23.5 ml +/- 0.03ml , not a percent so we know we are talking absolute error. Need to look at the % tolerance table that is given…. A) 25 mL graduated cylinder 1.5/100 x 25 = 0.38 ml B) 25 mL buret 0.12/100 x 25 = 0.030ml correct C) 25 mL measuring pipet 0.40/100 x 25 = 0.10 ml D) 25 mL transfer pipet can’t deliver 23.5 ml with a transfer pipet E) none of the above Lecture 2A: Calibration Curve 1. Which of the following term does NOT appear in Beer’s Law? A) concentration (c) B) absorbance (A) C) path length (b) D) wavelength (λ) E) molar absorptivity (ε) Solution: Beer’s law is A= ebc need to know formula. Answer is D. 2. A 1.0 x 10-3M solution of Congo Red has a % transmittance of 76.6 when measured in a 1.0-cm cuvette at a wavelength of 500 nm. What is the ABSORBANCE of this solution? a. 0.001 b. 0.142 c. 0.848 d. 1.88 e. 0.116 Answer: convert % transmittance to absorbance- the rest of the other info is not important. 76.6% convert to 76.6%/100= 0.766. Need to know A= -log (T) formula so A= -log 0.766= 0.116 so answer is E. 3 . In Experiment 3, the absorbance of four standard Congo Red solutions were measured to obtain a calibration curve that gave the following best fit line: y = 0.1812x +.0154 250 mL of a Congo Red unknown was obtained from the TA. Then 5.0 mL of this solution was removed and diluted to 100 mL in a volumetric flask. The absorbance of this 100 mL solution was measured to be 0.67. What is the concentration of the Congo Red in the initial 250 mL volumetric flask? (Don’t worry about the units.) A) 0.14 B) 2.8 C) 3.6 D) 72 E) none of the above Line equation problem will be on the test. Remember absorbance is always the Y and concentration is X. Also this has dilution factor too. First calculate the concentration using the absorbance value given in the problem. y = 0.1812x +.0154 So 0.67= 0.1812x +0.0154 or (0.67-.0154)/.1812= x or x = 3.6 But the 5ml aliquot was diluted in 100ml. Dilution factor = V final/ Valiquot = 100/5 = 20 so your sample was diluted by 20 so the original concentration of sample is 3.6 x 20 = 72 Lecture 2B: Acids and Bases 1. What is the definition of an acid and a base? Solution: Acid is a proton donor. Base is proton acceptor. 2. What makes a strong acid a. One which donates hydrogen ions and is virtually 100% ionized in solution b. One which doesn’t ionize fully when it is dissolved in water. c. One which is concentrated in solution. d. One which accepts hydrogen ions and is virtually 100% ionized in solution. e. None of the above. Solution: A. Know the definitions of acids and bases along with the differences of strong vs weak. 3. What is the pH of a 0.25 M H2SO4 solution? a. 1.7 b. 0.60 c. 0.30 d. 3.3 e. Can’t calculate with the information given. Solution: remember that Sulfuric acid is a strong diprotic acid so when it fully ionizes it will form 2 hydronium ions from one sulfuric acid molecule. [H3O+] = 0.25M x2= 0.50 M. pH = -log [0.50M] = 0.30. So answer is C. 4. What is the pH of a 0.25 M HCl solution? a. 1.7 b. 0.60 c. 1.2 d. 3.3 e. Can’t calculate with the information given. Solution: remember that hydrochloric acid is a strong monoprotic acid so when it fully ionizes it will form 1 hydronium ion from one hydrochloric acid molecule. [H3O+] = 0.25M , pH = -log [0.25M] = 0.60. So answer is B. 5. What is the pH of 0.0031 M NaOH a. 2.5 b. 5.0 c. 9.5 d. 11.5 Solution: Remember the NaOH is a strong base. 0.0031 M NaOH = [OH-] pOH = -log [ 0.0031] = 2.5 . From 14= pOH + pH, we can calculate pH of the base. 14= 2.5 + pH pH= 11.5 6. The pH of a test solution is found to be 1.0. What is the [H3O+]? a. .010 b. 0.10 c. 1.0 d. 10 e. 100 Solution: [H+] = 10 –pH so to calculate the molarity of the hydronium ion use the base of 10 to the power of the negative pH. 10 -1 = 0.1 M . Answer is B. Make sure you know how to do this on your calculator Lectuce 3A: titrations 1. What has to be true for titration to occur? a. Must have an indicator. b. Titrant and analyte must react. c. Titrant and analyte must be liquid. d. Titrant is added until equivalence point occurs and it is accurately measured. e. All of the above. Solution: answer is E. Make sure you know what the purpose of titration is and what is needed for a titration. 2. A student pipets 24.00 mL of 0.1256M HCl solution into a Erlenmeyer flask. How many mL of 0.0984M NaOH will be required to neutralize the HCl or reach equivalence point? a. 18.80 ml of NaOH b. 0.2966 ml of NaOH c. 30.63 ml of NaOH d. 3.014 ml of NaOH e. 2.362 ml of NaOH Solution: 1 to 1 molar ratio- I’m assuming you know that HCl and NaOH will react. HCl + NaOH H2O +NaCl (neutralization reaction) when it is 1 to 1 ratio and both are liquids, we can use MV=MV formula (0.1256M)(24.00ml)= (0.0984M) (V) V= 30.63 ml of NaOH (c ) 3. Which statement is false? a. The equilibrium constant, Ka is used to compare weak acids. b. Weak acid is not fully ionized but as you titrate with NaOH, more of the weak acid will dissociate as the hydronium ions are being consumed to reach a new equilibrium point. c. The equilibrium constant, Ka is a constant ratio from the concentration of ionized acid over un-ionized acid. d. The larger the Ka, the weaker the acid. e. Potassium hydrogen phthalate is a weak organic acid which donates one hydrogen ion. Solution: Ka = [H3O+] [ A-] / [HA] , the more the acid dissociates the stronger the acid and the larger the Ka. Answer is D which is False, the larger the Ka, the stronger the acid 4. A solution of NaOH is prepared and is standardized against 0.3050 g of KHP. The endpoint was reached by addition of 15ml of NaOH. What is the molarity of NaOH? (Formula weights are 204.2 KHP and 39.99 g NaOH/mol) a. 0.9342 M b. 0. M c. 0.09958 M d. 0.5085 M e. None of the above Solution: Need to know that KHP and NaOH is a 1 to 1 ratio. 0.3050 g KHP x 1 mol KHP/ 204.2 g = 0. mol KHP x 1mol NaOH/1 mol KHP = 0. mol NaOH 0. mol NaOH/ 0.015L= 0.09958 M NaOH (Answer is C) 5. A student weighs out 0.6945 g of KHP unknown. The phenolphthalein endpoint (colorless to pink) was reached after the addition of 17.30 mL of the 0.1458 M KOH solution. What is the % KHP in the unknown? (Formula weights: KOH, 56.11 g/m; KHP, 204.23 g/m) A) 25.76% B) 51.51% C) 20.38% D) 74.17% E) none of the above Solution: % of KHP problem: need to know molar ratio of KHP to KOH is 1 to 1 Find out moles KOH then convert to moles of KHP M (V) = moles So 0.1458 M or mol/L x 0.01730 L = 0. mol KOH x 1mol KHP/1 mol KOH = . mol of KHP Convert mole of KHP to grams using the formula wt of KHP so we can get % unknown . mol of KHP x 204.23 g/1 mol KHP = 0.5151 g of actually KHP Actually KHP/g of unk x 100 = % of KHP so 0.5151 g KHP/0.6945 g unk x 100 = 74.17% t Table Number of Value of t for desired trials confidence N 80% 90% 95% 99% 2 3.08 6.31 12.7 63.7 3 1.89 2.92 4.30 9.92 4 1.64 2.35 3.18 5.84 5 1.53 2.13 2.78 4.60 6 1.48 2.02 2.57 4.03 7 1.44 1.94 2.45 3.71 8 1.42 1.90 2.36 3.50 9 1.40 1.86 2.31 3.36 10 1.38 1.83 2.26 3.25 % TOLERANCE OF VOLUMETRIC APPARATUS VOLUME (ML) GRADUATED CYLINDER MEASURING PIPET* BURET* VOLUMETRIC FLASK TRANSFER PIPET 0.1 0.1 5.0 0.2 4.0 0.5 4.0 1 2.0 1.0 0.60 2 1.0 0.80 0.35 3 0.40 0.37 4 0.30 5 2.0 0.80 0.20 6 0.17 7 0.15 8 0.25 10 1.0 0.60 0.20 0.20 0.20 15 0.15 20 0.12 25 1.5 0.40 0.12 0.12 0.10 30 0.12 40 0.10 50 0.80 0.10 0.10 0.080 75 0.080 100 0.70 0.08 0.080 0.060 200 250 0.60 0.060 500 0.55 0.040 1000 0.50 0.035 2000 0.50 0.030 *NOTE: The tolerance given on the graduated equipment in the first three columns is % of full capacity.