Rem em ber: On NAPLEX, your answer m ay be expressed in a different set of units other than theones you used to work the problem . For exam ple, a problem that gives an answer of 100 m g m ay be expressed as 0.1 gram . Sim ilarly, 0.5 m g could be expressed as 500 m cg or 0.0005 gram s. Use the inform ation provided for questions 1 through 3. You receive an order for 500 m g of am inophylline in norm al saline in a total volume of 250 m l. The patient weighs 132 pounds. The am inophylline is to be adm inistered at a dose of 0.3 m g per kg per hour. Aminophylline Injection is supplied in vials with 25 mg per ml. The IV set you will use delivers 60 drops per m l. 1. How m any m illigram s of theophylline will the patient receive each hour? 2. How m any hours will the 250 m l last if adm inistered at the correct dose? 3. W hat will be the flow rate, in drops per minute, to administer the dose ordered? 4. A dose of 120 m g of gentam icin is administered IV. After 10 m inutes, a blood sam ple is assayed for gentamicin content. The result is 8 mcg per m l. W hat is the volume of distribution for gentamicin? Use the prescription shown for questions 5 through 8. You have tablets that contain 0.25 m g of reserpine per tablet. You can m ix ground up reserpine tablets in cherry syrup to m ake a liquid that will be stable for at least 3 months. The patient is a child that weighs 22 pounds. Reserpine Liquid 0.1 m g per m l Disp: 60 m l Sig: 0.01 m g per kg per dose one dose twice a day 5. How m any reserpine tablets will be needed to com pound the prescription? 6. W hat volum e (m l) will be in each dose? 7. How long will this prescription last at the dose ordered? 8. If reserpine tablets cost $20.00 per 100 tablets, what will be the cost of the tablets used? Questions 9 and 10. You need to prepare 30 gram s of 0.25% triam cinolone in Cold Cream. Triam cinolone is available in a suspension of 40 mg per m l that has a specific gravity of 1.0. 9. How m any ml of the triam cinolone solution will be needed? 10. How m any grams of the Cold Cream will be needed? Question 11 thru 13. You are to mix this IV using the m aterials shown. You must ADD the volumes of the dextrose and NaCl to the 1000 ml of Sterile W ater. 11. How many ml of the Sodium Chloride Injection, 23.4% , will you add to the sterile water? 12. How m any m l of the Dextrose 70% in W ater Injection will you add to the sterile water? Sterile W ater for Injection Dextrose Sodium chloride Adm inister over 24 hours 1000 ml 10 % 60 mEq Sterile W ater for Injection, 1000 ml Sod Chloride Inj, 23.4% (MW = 58.5) Dextrose 70% in W ater Injection 13. After all of the ingredients are m ixed, what will be the flow rate in milliliters per hour? Page 1 of 27 Provide the correct conversion factor for each instance below: 14. 1 grain = m illigrams 15. 1 avoirdupois pound = 16. 1 fluid ounce = 17. 1 US gallon = 18. 1 US pint = gram s milliliters fluid ounces m illiliters 19. Digoxin Injection is supplied in am pules of 500 m cg per 2 m l. W hat quantity, in m L, m ust a nurse administer to provide a dose of 0.2 mg? 20. Gentamicin Injection is supplied in a concentration of 80 m g per 2 m l vial. How m any m illiliters are needed to give a single dose of 4 mg per kg to a patient weighing 165 pounds? 21. You have Sodium Phenobarbital Injection, 13% . How m any m l will be needed to obtain 2 grains? 22. Concentrated Acetic Acid is supplied at a concentration of 90% weight/weight and has a specific gravity of 1.11. W hat volum e is needed to obtain enough acetic acid to prepare 60 m ls of Acetic Acid Solution, 5% (e.g., vinegar)? Use the prescription to the left for questions 23 and 24. Aspirin 300 m g Caffeine 30 m g Codeine 15 m g DTD 15 caps SIG: i cap q 4 hrs PRN pain 23. How many milligram s of codeine will be needed to com pound the prescription? 24. W hat will be the total final weight, in grains, of ingredients for one capsule? 25. A patient is to prepare one liter of a 1:5000 potassium perm anganate (KM nO 4) solution as a foot soak. You have a stock solution of KMnO 4, 5% . How m any m illiliters of the stock solution m ust be used to prepare one liter of the foot soak? Inform ation For Problem s 26 and 27. Your patient is receiving lithium carbonate (Li 2CO 3) capsules but needs to change to a liquid form (lithium citrate; Li3C 6H 5O 7). Molecular w eights: lithium = 7, carbon = 12, hydrogen = 1, oxygen = 16 26. If the patient’s dose of lithium carbonate was 300 mg three tim es a day, how many m illimoles of lithium did the patient receive in a day? 27. Lithium citrate syrup is available as 300 mg of lithium citrate per 5 ml of syrup. How many m illiliters of syrup will be required per day to provide the dose calculated in No. 26? 28. You have a vial that contains 5 gm s of drug and the chart to the right for m ixing the drug for use. Nurses want to m ix the vial so that it com es out with 200 m g per m l and ask you how m uch diluent they should add to m ake that concentration. Diluent Added Concentration Resulting 9.6 m l 500 m g per m l 19.6 m l 250 m g per m l 49.6 m l 100 m g per m l Page 2 of 27 How many milliequivalents of sodium chloride did you add to the bag? Molecular weights sodium = 23. A physician orders one liter of 3% sodium chloride injection for a sodium depleted patient. In order to obtain the sodium chloride am ount you calculated in No.4% . W hat is the least am ount you can weigh on this balance if you need to work with an accurary of 2% ? Page 3 of 27 . How m any will you need for the RX? 39. 23. W hat quantity m ust a nurse administer to provide a dose of 0. Pediatric Lanoxin Injection is supplied in am pules of 100 mcg per m l.5 32. How many milliliters of the 23. If the codeine injection has 60 mg per ml. use the RX in the box to the left.5 31.0 m g DTD: caps no. You have available 1 percent hydrocortisone cream and cold cream that can be m ixed with the hydrocortisone. per liter. chlorine = 35. You receive an order for 30 grams of 0. 2.4% NaCl to m ake the 3% NaCl solution. How m uch sodium chloride will you need to add to m ake the final 30 m l volum e isotonic? E value for tetracycline HCl = 0. 37.5 33.5 m Eq per ml as your source for NaCl.4% NaCl solution will you ADD TO the one liter bag of 0. how m any m illiosm oles of sodium chloride. How many gram s of the cold cream will be needed to fill the order? 30.5 m g 300. How m any m l will you use? MW Na = 23. are in the final product? Molecular weights sodium = 23. 32. You have 0. You can make Robitussin-AC by adding codeine injection to Robitussin. XX Sig: i or ii caps q 4-6 hrs prn pain Dr Lotta Scripts The inform ation provided in the prescription shown to the left is for questions 37 and 38.12 35. how m any milliliters of the injection will you ADD TO a four fluid ounce bottle of Robitussin to produce the sam e am ount of codeine as in Robitussin-AC? Tetracycline HCl NaCl to isotonicity 2 per cent qs 30 m l SIG: 2 drops into OS QID For questions 34 and 35.9% NaCl to prepare the 3% sodium chloride injection? Molecular weights sodium = 23. After you finished m ixing the 3% sodium chloride. chlorine = 35. Cl = 35.1 percent hydrocortisone cream for use on an infant. You have Hyoscine HBr tablet triturates (similar to nitroglycerine tablets) that contain 1/150th grain per tablet.5 36. chlorine = 35. you calculated the num ber of m l of 23.04 m g? Hyoscine HBr Oxycodone Acetam inophen 0. 34. You have tetracycline HCl powder and sodium chloride crystals.3 m g 7. in 50 m l vials.9% sodium chloride injection in one liter bags and sodium chloride injection.29. W hat will be the total am ount of Hyoscine HBr used to fill the prescription? 38. you will need to use Sodium Chloride Injection. A balance has a sensitivity of 5 milligram s. You are out of Robitussin-AC. a cough syrup that contains 10 mg of codeine per 5 ml. In question num ber 30. 01% of hydrochloric acid in a form ula for 1000 m l of a topical liquid. How many m illiosmoles are present in a single 10 ml vial? Ca = 40. C = 12. 50% 1000 ml Amino Acids. Mr. H = 1 51. Am inophylline Injection is supplied as 25 m g per m l. If am inophylline is 80% theophylline. Cl = 35.4 calories (or kilocalories or kcals) per gram and Am ino Acids provide 4 calories per gram . Johns weighs 198 lbs and is to receive a dose of 15 mg per kg.Morphine Sulfate 7. A form ula calls for using 0.4% solution. O = 16. O = 16 50. 10% 400 ml Electrolytes / Vitamins 200 ml Sterile W ater 800 ml Inform ation for questions 53 and 54 is contained in the box to the left. W hat volum e of Am inophylline Injection will the pharm acist need to obtain the dose in Num ber 43 above? 45. H = 1. Calcium Chloride Dihydrate Injection is supplied as a 10% solution in water. A patient is to receive IV theophylline 24 m g every hour. How many m illiliters will be needed for the dose? 49. How many milligram s will be consum ed in total? 46. How m any m illiequialents of sodium are in a single vial of 50 ml? Na = 23. how long will a 500 ml bag run? 52. This form ulation is for 24 hours of a total parenteral nutrition (TPN) form ulation. If each m orphine tablet weighed a total of 100 m g. W hat will be the cost of the 60 m g of drug you need for a prescription? 43. You have a stock drug solution that contains 10% active drug.) Page 4 of 27 .05% active drug. (Fat provides 9 calories per gram . Morphine Sulfate is available in 15 m g tablets. I have 10 ml of a solution that contains one gram per 5 m l. The electrolytes and vitam ins do not provide any calories. How m any m l of water m ust I add to the 10 ml to m ake a solution that is 1:5000 in strength? Total Parenteral Nutrition Formula Dextrose Injection.5. Dextrose provides 3.5 m g Acetam inophen 325 m g DTD: XX caps Sig: one cap q 4 hrs prn pain Use the prescription shown for questions 40 and 41.20 an avoirdupois ounce. How m any m l of hydrochloric acid are needed to prepare the 1000 m l? 48. what would be the weight of m aterial in one capsule? 42. how m uch aminophylline must the pharm acist use to prepare a 24 hr supply? 44. Am ikacin Injection is supplied as 500 m g in 2 m l. If the dose is one-half (½) teaspoonful four tim es a day for ten days. If the dose is 2 m g per m inute. Sodium Bicarbonate Injection is an 8. 40. A drug product has 0. HCl has a specific gravity of 1.2 and is supplied as a 40% weight/weight solution. A drug product costs $62. Lidocaine Injection for cardiac use is a 4% solution. How many tablets would provide the am ount of m orphine needed for this prescription? 41. W hat volum e of this stock solution will you use to prepare 120 ml of a 1:1000 solution? 47. to adm inister the form ula above over 24 hours? 55. III only C. W hat will be the absolute neutrophil count for a patient with the W BC result shown above? 63. How m any tablets will the patient need per dose? (NOTE: Bioavailability will be the sam e for both. How m any vials will you need to obtain the needed am ount of indom ethacin? Use the white blood cell differential count shown to answer questions 61. I only B. will the patient receive in 24 hours? 54. A patient has been receiving oral Aminophylline Tablets. The IV set will deliver 10 drops per m l. II and III only 62. 59 and 60.16 Indom ethacin 0. Boric Acid can be used as a preservative in ophthalm ic solutions. W hat will be the absolute granulocyte count for a patient with the W BC result shown above? Page 5 of 27 . Total W BC Count 3 14.52. 62 and 63.05% Boric Acid qs W ater qs ad 15 m l 59. fungal? I. The doctor wants to change giving oral theophylline. This patient is most likely to have what type of infection – bacterial. How many drops will you need for a 30 m l bottle of eye drops? 56. Fungal A. in drops per m inute. How m any calories. W hat quantity of boric acid will be needed to m ake the eye drop shown isotonic? E values are boric acid = 0. If the patient has been receiving 500 mg of aminophylline.5% boric acid. viral. Dimethyl Sulfoxide Liquid is supplied as a 50% weight / weight solution with a specific gravity of 1. It can also be used to replace NaCl when m aking a product isotonic.000/m m Bands Segs Basophils Eosinophils Lymphocytes M onocytes 5 65 1 2 24 3 61. Viral III. Indom ethacin is available in an injection of 1 m g powder for solution per vial. The proper concentration of benzalkonium to prevent bacterial contam ination in an eye drop is 1:750. Bacterial II. W hat will be the flow rate. II and III only E. You have boric acid in a saturated solution that is 5.40. I. W hat volum e of this solution will you need to use to obtain the needed am ount of boric acid? 60. 58. W hat is the percent weight / volum e of liquid dim ethyl sulfoxide? The prescription to the right has inform ation for questions 58. and theophylline is available in tablets of 100 m g. indomethacin = 0. from all sources. Aminophylline is 80% theophylline.) 57.53. I and II only D. You have a stock solution of 17% benzalkonium chloride and a dropper that delivers 50 drops per m l. 000 from your pharmacy. How m any m Eq’s of sodium will be added as a result of using sodium phosphate? Na = 23. A physician has requested a 20 m L vial of Lidocaine 1% with Epinephrine 1:200. How m uch of the Epinephrine Injection m ust you add to the vial of Lidocaine Injection to prepare the needed product? 72. For adults. S = 32.4 m illigram s of deliverable drug in 200 puffs.3 m g per Kg per week given in equally divided daily doses. Nutropin-AQ Injection is supplied at a strength of 10 mg per 2 m L. Magnesium sulfate is available as a 50% solution of m agnesium sulfate heptahydrate (7 waters of hydration). On a m ercy mission to a Third W orld country you have been asked to calculate the correct amounts of ferric gluconate injection (Ferrlecit) to adm inister to severely anem ic patients. 0 = 16 67. Ms Thom as thinks this is a very great overdose and wants you to explain the dose to her and tell her how m uch calcium the recom mended dose would provide. The shelf spot for that product is em pty but you do have Lidocaine 1% Injection and Epinephrine Amps. How many calories (or kilocalories. Ferrlecit contains 12. Sodium Phosphate is being added as Na 3PO 4. what volum e of drug will be given each dose? Norm al Hgb (15 gm /dl) minus Patient’s Hgb (gm /dl) = Hgb deficiency Hgb deficiency (x) Patient’s weight (in kg) = m illigram s of elemental iron required Give 5% of the total dose calculated every other day 74. W hat volum e of this injection is needed for this form ula? Mg = 24. You are given the following form ula for such calculations. H = 1. the m axim um daily dose from a Beclovent Inhaler is 840 m icrogram s.5 mg elem ental iron per m L. She is confused. A Beclovent Inhaler contains 8. Nutropin-AQ) is dosed at a rate of 0.) The first patient weighs 110 pounds and has a hem oglobin of 8 gm s/dl.Use the TPN form ula shown for problem s 64 to 69. 1% . How many ml of 23. W hat quantity of Pepcid Injection will be used? 68. 64. Page 6 of 27 . to prepare the product. however. Ms Thomas wants to take one gram of elem ental calcium every day and thinks calcium citrate is the best way to do so. by instructions on the container she is purchasing that say each tablet contains 560 m g of Calcium Citrate Tetrahydrate (20% calcium ) and she should take three tablets 3 tim es a day to obtain 1 gram of elem ental calcium daily. You have dextrose available as a 70% solution. Human Growth Hormone (somatropin.5 mg Pepcid 10 mg Regular Insulin 20 units Sterile W ater qs ad 960 ml Flow Rate = 80 m l per hour 69. W hat quantity will be needed to provide the required amount? 66. kcals) will the dextrose provide per 24 hours? 65. The pharm acist will be using Pepcid Injection. (NOTE: this is an old form ula that is no longer used. Cl = 35. 40 m g per 4 m l vial.5 Ingredient Quantity Dextrose 200 grams HepatAm ine 60 grams Sodium Chloride 50 mEq Potassium Chloride 40 mEq Sodium Acetate 20 mEq Magnesium Sulfate 10 mEq Sodium Phosphate 9 m Mol Potassium Acetate 15 mEq Calcium Chloride 2 mEq M ultivitam ins-12 5 ml Trace Elem ents-5 1 mL Vitam in K-1 0. W hat is the m axim um number of Beclovent puffs a patient may use in one day? 71. O = 16 70. P = 31.4% Sodium Chloride Injection will be needed to prepare the TPN? Na = 23. W hat will be the daily dose for a patient who weighs 55 pounds? 73. in milligram s. H = 1. W hat will be the total weight of ingredients. At the end of 8 hours. W hat was the initial concentration of potassium chloride in the IV fluid? K = 39.6 m l. How m any m Eq of calcium are present in one liter of Lactated Ringer’s Injection? Molecular weights: Ca = 40. 79. such as 1:1000) 86. How much Zephiran concentrate will you mix with enough water to make one gallon of the 1:750 solution you need? 77. The concentration in the diluted solution will be (express as N:XXXX. O = 16 85. Lactated Ringer’s Injection contains 20 m g of calcium chloride dihydrate per 100 m l of IV fluid. m illiosm oles and m illiequivalents. Cl = 35. Neupogen Injection is supplied at a concentration of 0. H = 1. You need a 1:5000 solution of benzalkonium chloride as a preservative.5. You need a 1:750 solution of Zephiran Chloride to disinfect the pharm acy counters. Cl = 35. and sim ilarities. O = 16 83. Cl = 35. 83 and 84 use the sam e inform ation to dem onstrate the differences. Cl = 36 78.5. 82. Your patient (weighs 132 pounds) and is to receive a dose of 5 mcg/kg/day. between m illim oles. How m uch water m ust you add to the 80 m L to reduce the concentration to 3 percent? Use the inform ation in the box provided to answer questions 79 through 81.48 m g/1. You have 20 ml of a 1:200 solution in stock. You dilute 1 m l of Fungizone Injection (50 m g per 10 m l) to one liter.75. Lactated Ringer’s Injection contains 20 m g of calcium chloride dihydrate per 100 m l of IV fluid. How m any grams of acetam inophen will you need for the 20 capsules? Atropine Sulfate 0. will you need to use to com pound the prescription? 80. Approximately how m uch atropine sulfate. IN GRAINS.4 mg Morphine Sulfate 5 mg Acetaminophen 325 mg DTD: 20 caps SIG: 1-2 caps q 4 hrs prn Dr Anodyne 81. the patient has received 30 m Eq of potassium chloride. to prepare the entire quantity? Questions 82. W hat volum e of Neupogen Injection is required for this dose? Page 7 of 27 . H = 1. O = 16 84. Lactated Ringer’s Injection contains 20 m g of calcium chloride dihydrate per 100 m l of IV fluid. You told your assistant to m ake a 3 percent m orphine solution but something got misunderstood and the result is 80 m L of a 20 percent solution. An IV solution containing potassium chloride is being adm inistered at a rate of 30 drops per m inute using IV tubing that delivers 15 drops per mL. How m any milliOsm oles of calcium chloride are present in one liter of Lactated Ringer’s Injection? Molecular weights: Ca = 40. How m uch of the 1:5000 solution can you prepare by using all 20 m l of the 1:200 solution? 76. Zephiran is available as a 17% concentrated solution.5. How m any m illiMoles of calcium are present in one liter of Lactated Ringer’s Injection? M olecular weights: Ca = 40. 2 and 5. W hat pH would represent the m iddle of this range? 93. Ca = 40. Cl = 35.75 buffer. 94.45% Sodium Chloride Injection and 40 m Eq of potassium chloride in a total volum e of 1000 m L through an IV set that delivers 15 drops per m L. A table of pKa values gives 4. Use the Henderson-Hasselbach Equation to calculate the relative molar ratios of sodium acetate and acetic acid required to prepare a pH 4. or above. per 1000 m L. Dextrose = 180 89. H = 1. Cl = 35.75 as the pKa for acetic acid.5 88. K = 39. A patient is receiving Dextrose 5% in 0. You have an IV set that delivers 12 drops per mL. 92. How m uch water m ust you use to change the final concentration of this container to 150 mg per 5 mL? Inform ation for questions 88 through 91. Na = 23. Determ ine the actual quantities.87. W hat is the total osmolarity of the IV fluid being adm inistered? Express your answer as m illiosmoles. A form ulation for an oral liquid of sum atriptan succinate (Im itrex – for m igraine headaches) indicates that the final product should have a pH between 4. Na = 23. The IV fluid has been running at a rate of 12 drops per minute for 15 hours. How m any grams of KCl have been administered over the 15 hour period? 90. H = 1. Page 8 of 27 . How m any mLs of this injection must you add to the bag of IV fluid to m ake the desired product? 96. At what flow rate (drops per minute) will the IV fluid be adm inistered to give the desired dose? Inform ation for questions 97 through 99. for a buffer that uses one m ole of sodium acetate (NaC 2H 3O 2) and the corresponding am ount of acetic acid (C 2H 4O 2). Molec W eights: Dextrose = 180. A physician wants to put 10 m Eq of calcium into a 500 m l bag of Norm al Saline Solution and adm inister the calcium at a rate of 0. The parents of a child cannot read or speak English and a pediatrician knows the m other is going to give the child one teaspoonful of m edication per dose no m atter what dose is ordered.5.5 m Eq per hour. The pediatrician wants you to add enough water to make Biaxin Oral Suspension to a concentration of 150 m g per 5 m L. Molecular W eights: Na = 23. How m any millim oles of KCl have been adm inistered over the 15 hour period? 91. The m icrobiology lab reports that the infecting agent will require a higher than norm al blood level to be effective. You have Calcium Chloride Dihydrate.5. You have a container of Biaxin powder for oral suspension that says that addition of 55 mL of water will result in a final volum e of 100 m L at a concentration of 250 m g per 5 m L. Cl = 35. The attending physician asks you to determine exactly how often each dose of the antibiotic should be adm inistered to keep the blood level at.3. 10% Injection in 10 m L vials. C = 12. How many mEq of potassium chloride have been adm inistered so far? K = 39. O = 16 Inform ation for questions 95 and 96. in gram s. O = 16 95. Mr Shirrah is very ill and needs antibiotic therapy for an infection. the determ ined minim um inhibitory concentration (MIC). rounded to the nearest whole num ber. Am ino Acids Injection. 1:5000 concentration.5 Inform ation for questions 101 through 104. IV bolus dose:0. You have vials of am photericin B.) 102. 5% in W ater C. W hat will be the dose. 105. If the IV set delivers 15 drops per mL. what will be the flow rate for the dose ordered? Inform ation for questions 105. You are told to prepare a double strength Dopamine IV drip because a patient is requiring large doses to obtain the drug’s effects. Answer the following questions: 101. Fox will receive a total of 80 mEq of potassium over the next 12 hours. The usual Dopamine drip concentration is 400 m g in 250 m l. contain 50 mg of amphotericin B in a total volume of 10 m L. Dextrose 5% in W ater B. that. A recent lab report indicates that Mr. Sterile W ater E. Peak level: 12 m cg/m L. Dopam ine Injection is available as 40 m g per mL in 10 m L vials. How often should a dose of the drug be adm inistered in order to m aintain the drug level at. that will be adm inistered at a rate of 80 m L per hour. W hat is the volum e of distribution of this drug? 98. Sodium Chloride Injection. 106 and 107. so that Mr. You are to prepare a 2 liter bag of am photericin B irrigation solution.5 hours. What will be the flow rate. Charles Fox has an IV hanging that has 40 mEq of potassium chloride in 1000 m L of Dextrose 5% in Half Normal Saline. 14.75 mcg/ml Therapeutic MIC level: 3 mcg/ml 97. 100. How m any m L of the 40 m g per m L Dopam ine Injection should be added to a 500 m g bag of IV fluid to prepare the double concentration? (Ignore volum e of the dopam ine injection. The physician ask you to slow the IV flow rate to 40 m L per hour and also add enough potassium chloride injection. Fox’s serum potassium level is only 3. 50 m g. 97.9% . To what IV fluid should the dopam ine be added? A. Roberta Snagg has a recurring m ycotic infection of the bladder. the therapeutic level? Use the half-life calculated in No. in ml per hour of dopamine. Based on first order pharmacokinetics. when properly reconstituted. The patient weighs 176 pounds and needs the pressor dose level of dopam ine (10 m cg/kg/m in). will the patient be receiving? 104. to administer the am photericin B at the desired dose? Page 9 of 27 . Trough level at 12 hrs: 0. in drops per minute. Your irrigation set delivers 12 drops per mL. Sodium Bicarbonate Injection. K = 39. Cl = 35. or above. Shirrah for a drug elim inated 100% by the kidney. 3% D.5 m Eq/L and his physician wants to increase the potassium dose.5 gram . what is the half life of this drug? 99.You receive the following pharm acokinetic data on Mr. How m uch am photericin B m ust be mixed in the 2 liter bag to provide the desired concentration of irrigation solution? 106. Ms. Mr Shirrah has normal renal function. The IV has been running at a rate of 80 m L per hour for the past 6. Her physician wants to use a bladder irrigation of am photericin B to be continuously run through the bladder for at least 24 hours. 10% 103. 6 mg/dL and has held at that sam e level for 5 days. 109 and 110. A patient (weight. Argatroban E. 10 m L Sterile W ater qs ad 1000 mls Flow Rate: 1 mL per kg per hour 112. Cl = 35.2% M VI-12. W hat will be the new flow rate in drops per m inute? 110. The concentration of heparin in the IV drip is 10. 0. The patient weighs 85 kilogram s. The form ulation is shown in the box to the right. Cl = 35. A physician has found a new form ula for an IV fluid to be adm inistered after surgery that can be described as “TPN Lite”. will be needed to prepare one liter of the form ula? Dextrose. 2 m Eq/m l. 4 m Eq per mL. How m any mL of water will be required to prepare one liter of the form ula? 116.5 115. The first patient to receive this IV fluid had GI tract surgery and needs to be NPO for 5 days.107. Mr. Mr. A possible drug for this purpose would be A.5 114. will be needed to prepare one liter of the form ula? Na = 23. will be needed to prepare one liter of the formula? K = 39. 700 m g/m L. Swan-Ganz catheter B. Once the heparin drip was stopped. Central catheter E. Don Smith is a 35 year old male who has been diagnosed with AIDS. Lepirudin Inform ation for questions 111 through 117 is included in the box to the right. currently running at 15 units per kilogram per hour. 176 pounds) in cardiology is receiving a heparin drip. Foley catheter D.000 units per 100 m L and the IV set delivers 15 drops per m L. W arfarin B. Sm ith’s physician wants to use Epivir and knows the drug m ust have its dose adjusted based on a patient’s renal function. W hat will be the new dosage in units per kilogram per hour? 109. 4% Sod Chloride. How m any calories from the dextrose and the am ino acids will the patient receive in 24 hours? 117. 0. How many m Ls of Am ino Acids Injection. How many m Ls of Potassium Chloride Injection. the physician wanted to change to a subcutaneous agent for further anticoagulation. Calculate Mr.75% Pot Chloride. W hat will be the flow rate to adm inister the solution at the dose ordered for this patient? 118. Page 10 of 27 . W hat type of catheter will be inserted into the patient’s bladder and used for adm inistration of the am photericin B bladder irrigation? A. How many mLs of Dextrose Injection. 8 inches tall and weighs 180 pounds. Sm ith’s creatinine clearance. Robinson catheter Inform ation for questions 108. How m any m Ls of Sodium Chloride Injection. Daltaparin C. The IV set used to adm inister this solution delivers 12 drops per m L. PICC catheter C. Use the form ula provided to answer the following questions: 111. Mr. Sm ith is 5 feet. 10% will be needed to prepare one liter of the formula? 113. Answer the following questions: 108. Sm ith’s serum creatinine is 2. Alteplase D. 15% Am ino Acids. The m ost recent partial throm boplastin tim e (PTT) indicates that the patient is being under dosed and that the heparin rate should be increased by 20% according to the hospital’s weight based heparin protocol. The ethylenediamine is the other 20% of the total m olecular weight. what will be the actual volume adm inistered to Mr Smith? And Now – The Answers 1. There are two m olecules of theophylline per molecule of ethylene diam ine in am inophylline. 2.77 hrs (x) 60 m ins/hr = 1666 m ins 15000 drops (div by) 1666 mins = 9 drops per m inute ALTERNATIVELY: 250 ml (div by) 27. If the physician prescribes Epivir Oral Solution. 10 m g per ml.2 lbs/kg = 60 kg 60 kg (x) 0. Mr Sm ith is unable to take his Epivir and will have a NG tube inserted to make oral drug administration m ore practical.3 m g Aminophylline/kg = 18 m g/hr Question calls for theophylline dose per hour 18 m g Am in (x) 0.8 = 14. W hat is the appropriate dose for Mr. Page 11 of 27 . McClure? Creatine Clearance Initial Dose M aintenance Dose < 5 ml/m in 50 m g 25 m g once daily 5 – 14 ml/m in 150 m g 50 m g once daily 15 – 29 ml/m in 150 m g 100 m g once daily 30 – 49 ml/m in 150 m g 150 m g once daily 120.4 m g Theo NAPLEX will expect you to know that Aminophylline (MW = 420) is theophylline ethylenediamine and that the am ount of theophylline (MW = 180) in am inophylline is 80% . 500 m g Am ino (div by) 18 mg/hr = 27. The dosing literature for Epivir has the following inform ation concerning use in renal failure.119.77 hrs 3. 132 lbs (div by) 2.77 hrs = 9 ml per hour 9 m l/hr (x) 60 drops/m l = 540 drops per hour (div by) 60 m ins/hr = 9 drops/m in NOTE: with a pediatric IV set (delivers 60 drops per m l) m l/hour and drops/m in are the same number. 250 m l fluid (x) 60 drops/ml = 15000 drops 27. 60 ml (x) 0.1 mg/m l = 6 m g reserpine needed 6. $ 20. then there will be 1 m l per dose. Vd = Dose Co 8 m cg/m l = 8 m g/liter Vd = 5. then 125 m l/hr = 8 hr.W hen doing flow rates you wind up expressing your answer in m l per hour. Always m ultiply volum e (x) specific gravity to get weight. 22 lbs (div by) 2.25% = 2. 100 ml/hr = 10 hr.875 m l (x) Sp Grav of 1. or keep vein open rate) at 42 m l per hour. 4.875 m l 10.2 mg per day = 30 days ALTERNATIVELY: 60 ml (div by) 1 ml/dose = 60 doses (div by) 2 doses/day = 30 days 8. or drops per minute.0 = 1. m l per m inute. 15 or 20 drops per m l (for adult IV sets) and 60 drops per ml (for pediatric IV sets).5 m g/gram (x) 30 gram s = 75 m g needed (div by) 40 m g/m l = 1.80 100 tabs 1. Since a drop rate im plies that the nurse is going to actually count the drops falling in the drop cham ber.25 mg/tab 10 kg (x) 0. If one liter (1000 ml) of fluid is to be administered. Page 12 of 27 . and 42 m l/hr = 24 hr. The m ost com m on flow rates for large volum e IV fluids range from 75 to 125 ml per hour but smaller and larger rates are possible to correct volum e deficits (rarely greater than 250 ml per hour) or to sim ply keep an IV line open and flowing (KVO.01 m g/kg/dose = 0. 7. 12.2 mg needed per day 6 m g in Rx (div by) 0. Rearrange for other values.1 mg per ml and dose is 0. IV sets generally deliver 10. ask this sim ple question – how high can you count in one minute? Few people would say they could count to 100 in one m inute.1 mg/dose (x) 2 doses per day = 0. 84 m l/hr = 12 hr. 0.2 lbs/kg = 10 kg 120 m g = 15 liters 8 mg/L 6 mg = 24 tabs needed 0.875 gram s of weight for triam cinolone injection 30 gram s total (m inus) 1. $X = 24 tabs 9.125 gram s of cold cream needed NOTE: This problem can be changed by changing the specific gravity of the drug being added. your drops per m inute rate must by less than 100 to be realistic and 60 is a m ore likely num ber.00 = $ 4. NOTE: W ith a pediatric IV set (60 drops per m l) flow rate in ml per hour = drops per m inute.1 mg.875 grams = 28. 0. 75 ml/hr = 14 hr. Therefore.1 mg per dose Since liquid is 0. 17 15.2 mg = 200 m icrogram s 500 m cg = 0.00 1184. Best solved by using alligation 1000.17 m l (div by) 24 hrs = 49.05 (= 5% ) = 3 gram s needed.5 m g/m Eq = 3510 m g needed 23. % weight/weight (x) specific gravity = % weight/volum e 2 m l = 7. so 58.17 ml ml ml ml 70% X ml = 10 parts 10 parts 1015 m l 60 parts = 169. 0. MW of NaCl = 58. 1 fluid ounce = 29.5 m g (div by) 1 = 58.100% % = gram s per 100 m l.4 % = 234 m g per m l 3510 m g (div by) 234 m g/m l = 15 m l required Another way to work No.9 % wt/vol -or.5 m g/m Eq = 4 m Eq per m l 60 m Eq (div by) 4 m Eq/m l = 15 m l required 12. so you need 1 m l 90% wt/wt (x) 1.1 gram per m l 60 m l (x) 0.17 ml 10% 0% 60 parts = 1015 ml (1015 ml includes the NaCl added) 70 parts (will be total volum e after all additions = 1184. 8 pints) 480 m ls – 16 fluid ounces (x) 30 m l/ounce) – is often used Use either one 75 kg (x) 4 m g/kg = 300 m g needed X m l = 2 m l = 0. but not both. I US pint = 473 mls 19.11 (SpGr) = 99.4% Total Volum e 1184. if concentration is 1 gm /m l. 13. then need 3 m l Page 13 of 27 .11.17 ml) of sterile water of dextrose 70% of NaCl 23. thus 58. 11 234 m g 58.34 m l/hr W ould really run at 49 or 50 m l per hour. 1 US gallon = 128 fluid ounces 18. 60 m g only in some circum stances 15. 1 avoirdupois pound (one we all know) = 454 gram s 16.8 m g exactly 65 m g is usable.5 ml needed 80 m g 13% = 130 m g per m l. 1 grain = 65 m g (x) 2 = 130 m g needed 22.5. NAPLEX answers are multiple choice and would include 49 or 50. 14.2 lbs/kg = 75 kg (exact is 453.5 mg = 1 m illim ole Na + is monovalent (one charge). 4 quarts.5 m g = 1 m illiequivalent 60 m Eq (x) 58.00 169. 1 grain = 64.6 gram s) 30 m l routinely used (also 3785 m l. so 100% = 100 gram s per 100 m l -or.8 ml needed 200 m cg 500 m cg X ml = 300 m g 21. 165 pounds (div by) 2.5 milligram s 20.57 milliliters exactly 17. 4 m l (the powder volum e) = 24. DTD m eans the formula is for one capsule 15 m g/cap (x) 15 caps = 225 m g needed (MFT was once used to mean the form ula was for 15 caps but is now routinely used to simply indicate to “mix and make” capsules. 5 grams = 5000 m g 5000 m g (div by) 200 m g/m l = 25 m l of total volum e 25 m l (m inus) 0. Similarly. and 5 gram s of drug. In this case 10 m l TID would be a good dose.23. 1: 5000 m eans 1 gram in 5000 m l 1 gram = 1000 m g.) BUT.4 m l to the volum e. Therefore. At a concentration of 500 m g per m l.1 parts 30 grams 1 part = 3 grams of 1% hydrocortisone 30 gm s (m inus) 3 gm s = 27 grams of cold cream needed Page 14 of 27 . 74 m g of Li Carb = 2 m illim oles of lithium 300 m g (x) 3 doses/day = 900 m g per day 27.4 m l to the final volume.) 28.) MPJE Law Note on No. 300 m g + 30 mg + 15 mg = 345 mg per capsule 345 m g (div by) 65 mg per grain = 5.4 m l less than the final volume. cherry syrup or som e other diluent with no active ingredients. so the powder contributed 0.3 m Moles 900 mg 74 mg per day Thus.9 parts 1 part = 30 gram s X grams = 0. 2 tablets. 210 mg of Li citrate = 3 millim oles of lithium = 1701 m g of Li citrate needed daily X ml = 1701 mg 5 m l = 28. 26.3 grains per capsule 25.6 m l of diluent to add to m ake 200 m g/ml 29. there would be 20 m l of volum e and at 100 m g/m l there would be 50 ml of volum e. there would be a total volum e of 10 ml.3 mMoles 3 mMoles X m Moles = 2 m Moles = 24.1 parts 0. at 250 m g/m l. then the product would rem ain a C-II because codeine only converts to a CIII or C-V when mixed with other active ingredients that lim it abuse of the codeine. In each case the am ount of diluent added was 0. But only 9. 23: You probably would use C-II codeine to prepare the product. but we are only m aking one liter (1000 ml) MW of Li (two) X mg 1000 m l 5% = 50 m g/m l MW of Li 200 m g for 1000 m l 200 m g = 4 m l needed 50 m g/m l carbonate = 74 Thus. but after mixing the codeine with other drugs you would now have a C-III product (equivalent to Tylenol No.1% 0% 0.6 m l of diluent was added.35 m l 300 mg per day (NAPLEX could ask for a specific dose. thus we would have 1000 mg in 5000 ml. If you simply dissolve codeine in water. = 1000 m g = 5000 m l (three) citrate = 210 X mg = 210 m g 24. the powder will contribute 0. 24. Can be done by alligation 1% 0. 5 m Eq/m l = 146.29.2 parts 0. 0% 120 m l = 4.5 and monovalent.5 mg per mMole NaCl One charge.8 parts codeine injection 5.4 parts to 1000 ml 3% 20.5 = 58.025.1 parts 0.. thus 58. 9000 m g from the Norm al Saline plus 24.078.5 m g = 1 m Eq 24. (cont) 30.9 ml to add 2.5 mg/mEq = 411.9% NaCl = 9 mg/ml (x) 30 m l = 270 m g NaCl to m ake 30 m l Norm al Saline Saline 270 m g (m inus) 72 mg from tetracycline = 198 m g NaCl needed 35.2% solution 60 m g per ml = 6% solution 6% 0. 33.6 mg NaCl added 23 + 35. The problem could be changed to ask for the total num ber of m illiosm oles in the BAG .87 milliosm oles in the bag. Another Method that can be used Alligation again 0. 10 mg per 5 m l = 2 m g per 1 m l 2 m g/m l = 0.9 ml = 24.6 m Eq of NaCl 32.2 parts 34. 000 m g (or 30 gram s) per liter MW of NaCl = 58. 100 m cg = 0.4% 2.25 m g per ml = 1.6 mg (div by) 58. Cl = 35. X m l = 1000 m l = 102.25 m g/ml 198 m g (from No. MW NaCl = 58..) 2% = 20 m g per ml (x) 30 ml = 600 m g needed (x) 0.04 m g = 40 mcg per dose Can use either one X ml = 40 m cg 1 ml = 0. 078 mg in the bag.5) NaCl — > Na + and Cl – (two particles per m illim ole).5 m g per m Eq (x) 2..1% = 1 m g per gram (x) 30 gram s = 30 m g hydrocortisone needed 1% Hydrocortisone = 10 mg hydrocortisone per gram 30 mg needed (div by) 10 mg hydrocortisone per gm of cream = 3 gm cream 30 gm s (m inus) 3 gm s = 27 gram s of cold cream needed 23.5 m g = 1.8 parts = 120 ml (This method allows for the increased final volum e of adding the codeine.12 = 72 m g of NaCl equivalency Normal = 0. 34) (div by) 146. com es out to be 1130.1 parts 20. 000 m g = 2 m illiosomoles 58. so divide by one = 58.64 m illiosm oles NOTE that the question asked for milliosm oles per LITER.9% 31.14 m l of 5.078.4 parts = 1000 m l 23. so 58.35 m l needed 36.4 m l 100 m cg Page 15 of 27 .1 m g per m l 0.078 mg added to the bag = 33. 3% NaCl = 30 mg per ml (x) 1000 mls = 30.5 m g = 2 m illiosm oles X milliosm oles 30.2% 4 fluid ounces = 120 m l X ml = 0.4% = 234 m g per ml (x) 102.5 mg per millim ole (Na = 23. 39. as such. 1 avoirdupois ounce = 28.8 = 720 m g of aminophylline needed 44.350 m g 576 m g (div by) 0. This sam e concept is the usual reason for drug products where 60 m g = 1 grain.5 m g/m l = 1.8 m l 25 m g 0. X ml = 720 mg 1 m l = 28. 100% 2% = 50 (Accuracy Factor) 50 (x) 5 mg sensitivity = 250 m g least am ount that can be weighed with 2% accuracy 40.25 mg per dose (x) 4 doses/day (x) 10 days = 50 m g 46. you can get the answer they want. so you will need 120 m g 10% = 100 m g per ml in stock solution 120 m g (div by) 100 m g/m l = 1. 45.20 = $ 0. NAPLEX would likely set up the answer to be based on the 64. 1:1000 m eans 1 gram in 1000 m l -or. 6 m g (div by) 60 mg per grain = 1/10th of a grain X tabs = 1 tab 1/10th grain 1/150th grain = (6 m g was determined in No. you would get a dose of 20 mg for the acetam inophen and that is just too low. DTD = 20 caps (x) 7.5 m g per m l one teaspoonful = 5 m l.37. the grain to m g equivalent is reduced to 60 mg per grain to provide a safety factor when adm inistering the drug. above) 1/10 (x) 1 1/150 Invert and multiply 1 (X) 10 150 = 15 tabs 1 NOTE: Hyoscine HBr is a potent alkaloid and. with NAPLEX being m ultiple choice. 24 m g/hr (x) 24 hrs = 576 mg theophylline 1000 m g 6500 m g 7500 m g (div by) 20 = 375 mg per capsule $ X = $ 62. just go by the dose per capsule.8 mg grain. In this case.5 ml/dose (x) 0.5 mg/cap = 150 m g needed (div by) 15 m g/tab = 10 tabs needed 41. 300 m g is the usual dose of acetam inophen.5 m l 2. Again. If you divided that by 20.350 m g 43. 38.1 m g per 1 m l 120 m l are to be prepared.3 mg/cap (x) 20 caps = 6 m g needed NOTE: W hen unsure about is the form ula for one capsule or for several capsules. DTD m eans formula is for one capsule 0.2 ml needed Page 16 of 27 .1000 m g in 1000 m l -or.35 gram s = 28. 10 tabs (x) 100 m g/tab = 1000 m g for the m orphine 325 m g (x) 20 caps = 6500 m g for the acetam inophen 42.13 (13 cents) 60 m g 28. so one-half teaspoonful = 2.05% = 0. 37. 84 mg = 1 m Eq) 50 m l therefore = 50 mEq 51. thus there is 1 mEq per m l (84 mg/m l. You m ust add 9.1 mg/m l (x) 1000 ml = 100 m g HCl needed 40% wt/wt (x) 1. 1700 + 160 = 1860 cals per day 1000 ml + 400 m l + 200 m l + 800 m l = 2400 m l per day 2400 m l/day (div by) 24 hours = 100 m l per hour (x) 10 drops/ml = 1000 drops/hr 1000 drops/hr (div by) 60 mins/hr = 16. 1:5000 means one gm in 5000 ml You have 10 ml with 1 gm per 5 ml = 2 gm thus.7 (or 17) drops per m in Page 17 of 27 . 1:5000 would m ean 2 gm in 10.000 m l 53. 198 lbs (div by) 2.000 m g (div by) 2 m g/m in = 10.990 m l of water Dextrose 50% m eans 50 gm in 100 m l or 500 gm in 1000 m l Amino Acids 10% m eans 10 gm in 100 m l or 40 gm in 400 m l 500 gm dextrose (x) 3. 1350 m g (div by) 250 m g/m l = 5.4% is to make the concentration 1 m Eq per m l for easy dosing.2 lbs/kg = 90 kg (x) 15 m g/kg = 1350 m g needed 500 m g in 2 m l = 250 m g/m l 49.21 m l HCl needed 48. CaCl 2 * 2 H 2O = MW of 147 Dissociates into Ca + Cl + Cl or 3 m illiosm oles/m illimole (W ater does not count as particles) 10% = 100 m g per ml or 1000 mg per 10 m l vial (147 m g = one millim ole) X mOsm s 1000 m g 50.4 cals per gm = 1700 cals 40 gm am ino acids (x) 4 cals per gm = 160 cals 54.4 m l = 3 m Osm s 147 m g = 20.000 m ins (div by) 60 m ins/hr = 166. (NOTE: The reason sodium bicarbonate injection is 8.7 hrs 52.2 sp grav = 48% wt/vol 48% = 480 m g per m l 100 m g (div by) 480 m g/m l = 0.4 m illiosm oles NaHCO 3 = MW of 84. thus 84 mg = 1 m illiequivalent (sodium is m onovalent) 8.000 mg in the bag 20.01% = 0.) 4% = 40 m g per ml (x) 500 ml = 20. 0.47.4% = 84 m g per ml. 5 m g of indom ethacin needed 7. 70% = 70 gm per 100 m l X ml 200 gm = 100 m l 70 gm = 285.5 mg of indom ethacin needed 61. 5.5 vials but you must use 8 vials to have 7.73 = 10.3 mg (div by) 55 m g/m l = 4.7 m l Page 18 of 27 .9% or 9 mg/ml (x) 15 m l = 135 m g NaCl needed if no drug present 135 m g (m inus) 1.4 cals/gm = 680 cals in 12 hrs (x) 2 = 1360 calories in 24 hours 65.76 (or 12) drops needed 500 m g am inophylline (x) 0.2 mg = 133.52. = 50 drops 170 m g = 11.52 mg NaCl = 257. 7.2 m g NaCl Norm Saline = 0.8 (for 80% theophylline) = 400 m g theophylline 400 mg theophylline (div by) 100 mg per tab = 4 tabs 57. Thus 14. Granulocytes are bands.2 mg Thus. 1:750 = 1 gm in 750 ml -or.16 (E-value) = 1.68 m l needed 60. 7. thus 5 + 65 = 70 1 m g per vial so need equal to 7.3 mg of boric acid needed 59. 1 m g of boric acid = 0.40 Sp Gr = 70% wt/vol 58. thus. segs.7 = 9800 neutrophils cells 63. Bacterial – due to the increased num ber of neutrophils.05% = 0.52 m g of NaCl for isotonicity X m g Boric Acid 133. Bands and Segs are neutrophils. 62.5 mg (x) 0. m illiMETER) (x) 0.55. 0.8 mg NaCl needed after allowing for indom ethacin Boric Acid E-value is 0.5% = 55 mg per ml 257.200 64. 50% wt/wt (x) 1.000 cells per cubic m illim eter (yes.1000 m g in 750 m l X mg 30 m l = 1000 m g 750 m l = 40 m g for the 30 m l 17% = 170 m g per ml and 170 mg in 50 drops X drops 40 m g 56.5 m g available This m eans that 70% of the white blood cells are neutrophils W BC count = 14.000 (x) 0.5 m g indom ethacin replaces 1.5 mg per ml (x) 15 ml = 7. 200 gm (x) 3. basophils and eosinophils (5 + 65 + 1 + 2 = 73) so 73% of white blood cells are granulocytes.8 mg NaCl = 1 m g Boric Acid 0. 1 mg 73.54 m l per dose Page 19 of 27 .5 m g elemental iron per m l = 1. 55 pounds 2.1 m g = 1 mg 200 m l = 1 ml 10 m g = = 20 m l (x) 1 mg 200 m l 0.4 mg = 8.1 m g Epinephrine Amps. MW of NaCl = 58.5 mg per week 7. thus 246 m g = 2 m Eq of Mg sulfate 246 m g (div by) 2 = 123 m g per mEq 50% = 500 m g per ml 123 m g/m Eq (x) 10 m Eq = 1230 m g needed 1230 m g (div by) 500 m g/m l = 2.2 lbs/kg = 25 kg of patient (x) 0.1 m g (x) 1 m l 10 m g = 0.25 m g iron per dose 19.1 mg per day (approx) X ml = 1. 27 mMoles = 27 mEq of sodium added with the Na phosphate 70.4% = 234 m g per ml 69.22 ml per dose per day 10 m g Ideal Hgb (15) minus Patient Hgb (8) = 7 gram Hgb deficit (x) 55 kg = 385 m g total iron needed 385 m g iron (x) 0. then use 1 ml of the diluted solution.1000 m g per 200. MgSO 4 * 7 H 2O = MW of 246 Mg is divalent.5 mg/week (div by) 7 days/week = 1.5 m g = 1 m Eq 58.400 mcg (div by) 200 puffs = 42 m cg per puff 840 mcg lim it per day (div by) 42 mcg per puff = 20 puffs per day 71. 8.66. 1:200.05 (or 5% ) = 19. 2 m l = 0.5 m g/m Eq (x) 50 mEq = 2925 m g needed 23.000 m l -or. dilute to 100 m l with sterile water. 40 mg per 4 m l = 10 mg per ml 68. 72.000 m l -or. 164 m g = 3 m illim oles of Na per m illim ole Na phosphate 9 mMoles of Na phosphate (x) 3 = 27 mMoles of sodium 1 m Mole of Na = 1 mEq of Na because Na is m onovalent Thus. thus 58.46 m l needed 67. 1% = 10 m g per m l = 0.01 m l How best to obtain such a sm all volum e? Take 1 m l of Epinephrine.3 mg per kg = 7.1 m g per 200 ml X mg 20 m l X ml 0.5 m l needed MW of Na phosphate = 164 So.000 = 1 gram per 200.5 10 m g (div by) 10 m g/m l = 1 m l needed Na is monovalent. 2925 m g (div by) 234 m g/m l = 12.25 mg per dose (div by) 12. 33 m l 79.25 gram s) that have been adm inistered over 8 hours IV set is 15 drops per ml. running at 30 drops per m inute = 2 m l per m inute 8 hours (x) 60 minutes per hour = 480 m inutes 480 m ins (x) 2 m l/m in = 960 m l infused No 77 continues – To change to percent 78.74.0467 gm 17 gram s (17% ) m ake the 1:750 solution 77.234 grams per 100 ml -or.33 m l total volume (m inus) 80 m l = 453. but also done with m athematics ALLIGATION METHOD 20 3 = 80 ml 3 0 MATHEMATICS METHOD 80 m l (x) 0. 1:750 = 1 gram in 750 m l X ml 100 m g = X mg 20 m l = 1000 m g 200 m l = 100 mg in the 20 ml 5000 m l = 500 m l would be produced 1000 m g X gram s = 1 gram = 5.5 gram s 81.33 ml 100 m l 3 gm = 533.2 (for 20% ) = 112 m g of elemental calcium per tablet 1000 mg of elemental calcium per day 112 m g of elem ental calcium per tablet = 8. so 75 m g = one m Eq 30 mEq (x) 75 mg per mEq = 2. 560 m g Ca Citrate Tetrahydrate (x) 0. KCl = 39 + 36 = 75 for the molecular weight.0467 grams per gallon of 1:750 3785 m l (1 gal) 750 m l X ml = 100 m l = 29.0. 0.4 mg (x) 20 caps = 8 mg total 8 m g (div by) 64.500 m g (div by) 1000 m g/gram = 6.20 (= 20% ) = 16 gram s of m orphine X ml = 16 gm s 17 = 453.250 m g (2.1000 m g in 200 m l 1:5000 = one gram in 5000 ml -or. 6500 m g + M orphine 100 m g (5 m g/cap (x) 20 caps) = 6608 mg Page 20 of 27 .1000 m g in 5000 m l 76.93 (or nine) tablets per day The explanation is that the tablets are not pure calcium and most of the tablet is not calcium but the citrate and the water. it is a m onovalent salt.25 grams 960 m l = 0.33 m l total volume 533.33 m l 20 = 533. 8 m g + Acetam inophen.69 m l ~ 1 ounce would be added to one gallon of water to 5.8 m g = 1/8th grain 80. Atropine. X grams 100 m l = 2.234% Best done by alligation. 1:200 = one gram in 200 ml -or. 75. 325 m g (x) 20 caps = 6. 67 ml (additional am ount) = 121.67 m L (final volum e) minus 45 m L (powder volum e) = 121. 200 m g (div by) 147 m g per millim ole = 1.67 m L will be total final volume 166. 3 particles = 4 particles per 200 m g = 4 m illiosm oles 147 m g MW = 147. m ust also m ultiply the 1000 ml by 200 Final expression of concentration is 1:200.67 m l will be total final volume 166.67 mL of water to add Page 21 of 27 . 20 m g per 100 m l = 200 m g per 1000 m l (liter) = 2 m Eq 147 m g = 2.6 ml X ml 300 m cg 250 mg per 5 m l = 50 m g per m l 50 m g per m l (x) 100 m l = 5000 m g of drug in the bottle X mL = 5 mL 5000 mg 150 m g X mL = = 1.2 lbs per kg = 60 kg of patient (x) 5 mcg/kg/day = 300 mcg/day 0.6 ml 480 m cg 5 m L (x) 5000 m g 150 m g = 1 m l per day = 166. 1000 mg (div by) 5 m g = 200.000 86.82.72 m Eq per liter 50 m g in 10 ml = 5 mg in 1 ml diluted to 1000 m l.48 mg = 480 m cg per 1. Molecular weight CaCl 2 + 2 H 2O = 147 One m illim ole = 147 m g 83.67 m l of additional volum e needed 55 m l (original am ount to add) plus 66. calcium is divalent. since you multiplied the 5 m g by 200. hence 147 m g = 2 m Eq of calcium (and calcium chloride) m Eq 200 m g 85. so 5 m g in 1000 m l is the actual concentration m ust raise the 5 m g to 1000 m g (1 gram) to express ratio concentration. W eight = 132 pounds (div by) 2.36 m illim oles per liter CaCl 2 = Ca ++ plus Cl – plus Cl – = 3 particles per m illim ole (147 m g) = 3 m illiosm oles X particles = 200 m g 84.67 ml is amount to add ALTERNATE METHOD: 100 m L (volum e of product) minus 55 m L (vol of water added) = 45 m L (vol of drug powder) 250 mg per 5 m l = 50 m g per m l X mL = 5000 mg 5 mL 150 m g X mL = 50 m g per m l (x) 100 m l = 5000 m g of drug in the bottle 5 m L (x) 5000 m g 150 m g = 166.67 m l (new volum e) minus 100 m l (original volum e) = 66. 87. one liter of half Norm al Saline (0. since Na+ is m onovalent.5 mg per m Eq (x) 28.6 m g (divide by 1000 m g/gram ) = 2.88.6 tim es hypertonic 92.145.8 millim oles (from Que 90) (x) 2 milliosm oles per m illim ole = 57.145. is not an electrolyte. 5.8 mL per m in (x) 60 m in per hr (x) 15 hr = 720 m L in 15 hr 2980 mg 1000 mL = 2. X mL 12 drops = 1 mL 15 drops = 0. 50 gm = 50.9% NaCl) = 154 m Eq of sodium chloride.1456 gram s 90.5 m olecular weight of KCl.5 m g = 1 m Eq 74.5 mg = 28.75 would be choosing the middle ground. 1 mL 15 drops = = 0.6 m illiosm oles for KCl (B) Dextrose is 5% or 5 gm per 100 m l (x) 10 = 50 gm in 1000 m l.000 m g Dextrose.8 m L per m inute 0.8 m Eq KCl 39 + 35.3 minus 4.000 mg = 277. Page 22 of 27 . 39 + 35. one can determine that acetate buffers (sodium acetate plus acetic acid) are com monly used to achieve such a pH. Further.2 + 0. so 74.8 m Eq = 2. so 1 m illim ole (180 m g) = 1 m illiosm ole 50. thus.2 = 1. m onovalent salt so 74.6 m g = 1 m illim ole 74.75 Selecting a pH of 4. MW = 180.55 4.8 m illimoles (A) Since KCl — > two particles.6 m g millim oles 2145.5 = 74.6 (KCl) + 277.8 mL per m inute (x) 60 minutes per hour (x) 15 hours = 720 m L infused in 15 hours m Eq KCl 720 m L 89.5 = 74.1 divided by 2 = 0.78 (dextrose) + 154 (NaCl) = 489.45% ) = 77 m Eq of sodium chloride (154 div by 2). = 40 mEq KCl 1000 m L = 28.5 molecular weight of KCl. Thus. so 1 millim ole (or m illiequivalent) of NaCl = 2 m illiosm oles 77 millim oles of NaCl (x) 2 milliosm oles per m illim ole = 154 m illiosm oles for NaCl TOTAL for the Answer: 57. this solution is about 1. Using any of several chemical reference texts. 1 millim ole = 1 milliquivalent NaCl — > Na+ and Cl–.55 = 4.38 m illiosm oles NOTE: Osm olarity of blood is 308.1 1. then 1 m illim ole or milliequivalent = 2 m illiosm oles 28.5 mg per m Eq (x) 40 m Eq = 2980 m g in the 1000 m L X mL 12 drops = m illigram s 720 m L 91.78 m illiosm oles for the dextrose 180 m g/m m ole (C) One liter of Norm al Saline (0.5 m g = 1 m illim ole (and 1 m illiequivalent) 74. 93.75 + 0 (0 = log of 1). Dose Conc Orig (peak) Vd = Trough level at 12 hrs: 0. In No. Peak level: 12 m cg/m L.5 + 4 (4 H) + 32 (2 O) = 147 mg for MW Since Calcium is divalent. thus the m olar concentrations of (BASE) Sodium Acetate and (ACID) Acetic Acid should be the sam e (e. Vol Dist = 0.8 mEq of KCl have been given 40 m Eq (m inus) 20.g. Ca Chloride Dihydrate = CaCl 2 * 2 H 2O. Shirrah: IV bolus dose: 0. so flow rate is 25 m l per hour 25 m l/hour (x) 12 drops/m l = 300 drops per hour (div by) 60 m ins/hr = 5 drops per m inute Pharm acokinetic data on M r. pH = pKa + log Conc Base Conc Acid 4.5 mcg/m l Four half lives 0.5 hr = 520 ml adm inistered (480 m l rem ain) 500 m g = 41. 60 gram s = one m ole of acetic acid These quantities would be dissolved in sufficient water to make one liter to prepare the buffer solution. Since one half live is 3 hours.75 m cg/mL Therapeutic drug level: 3 m cg/m L 97. 95.67 liters 12 m g/liter Since it took four half lives to fall from peak to trough.2 mEq that rem ain in the bag Page 23 of 27 . then 147 mg of Ca Chloride dihydrate contains 2 m Eq of calcium X mg of Ca Chloride 10 m Eq 10% = 100 m g per mL 96.5 + 35. we found it takes two half lives for the drug level to decline to the MIC of 3 m cg per ml. then 3 hrs (x) 2 half lives = 6 hours as tim e between doses Peak at tim e O = 12 m cg/m l One half life 6 m cg/ml Two half lives 3 mcg/m l Three half lives 1.8 m Eq = 19. so 40 + 35. and 12 hours passed between peak and trough. Cannot directly calculate because needed form ula requires use of logarithm s.5 gram . For Sodium Acetate: (Na = 23 x 1) + (C = 12 x 2) + (H = 1 x 3) + (O = 16 x 2) = 23 + 24 + 3 + 32 = 82 Thus. 80 m l/hr (x) 6. X = 10 m Eq (x) 147 m g 2 m Eq = 735 mg is 10 mEq 735 m g = 7.75 m cg/m l (12 hr later) 100. 99. = 147 mg of Ca Chloride 2 m Eq so. 82 gram s = one m ole of sodium acetate For Acetic Acid: (C = 12 x 2) + (H = 1 x 4) + (O = 16 x 2) = 24 + 4 + 32 = 60 Thus.35 m l needed 100 m g/m l 10 mEq in 500 ml = 1 mEq per 50 m l = 0.75 = 4.5 gram = 500 m g = 12 m cg/m l = 12 m g/Liter 98. 98.5 mEq per 25 m l. then 12 hrs divided by 4 half lives gives 3 hours per half live: t½ = 3 hrs 480 m l = 12 hours of fluid 40 m l/hr available 40 m Eq KCl in 1000 m l = 4 m Eq per 100 m l X m Eq 520 m l = 4 mEq 100 ml = 20. one m ole of each agent) 94.. So do by hand m ethod. Foley catheter – YES. 10% NO.8 m Eq needed to add to the bag K = 39 + Cl = 35.9% = 149 m g per m l (This is sam e as 2 mEq/m l) 101. 80 m l/hour (x) 12 drops per ml = 960 drops per hour 1 gram 5000 m l = 0. PICC means peripherally inserted central catheter. Dextrose 5% in W ater YES. 5% in W ater NO. 30 ml/hr (x) 15 drops/m l = 450 drops per hour (div by) 60 m in/hr = 7.5 = 74. goes into a vein and term inates just outside the heart Page 24 of 27 .2 mEq of KCl rem aining in the bag 100 m l 80 m Eq (m inus) 19. D. Sodium Chloride Injection. To what IV fluid should the dopam ine be added? A. X mg 500 ml = 400 m g 250 m l 4529. could result in hemolysis of RBCs due to low osmolari ty E. 1:5000 means 1 gram in 5000 ml X grams 2000 m l = 106.6 m g needed 14.6 m g (div by) 149 m g per m l = 30.No 100 continued: Could also do for KCl still in the bag X m Eq 480 m l = 4 m Eq = 19. Swan-Ganz catheter – NO.2 mEq = 60. 176 pounds (div by) 2.8 m Eq = 4529. Am ino Acids Injection.2 lbs per kg = 80 kg 80 kg (x) 10 m cg/kg/m in = 800 m cg per min 800 mcg/min (x) 60 min/hr = 48. Head 3 is used to inflate the ballon that keeps the catheter in the bladder. Central catheter – NO. Sodium Bicarbonate Injection. PICC catheter – NO. basic (hi pH) solution incompatible C. norm al saline okay D.4 m l = 800 mg dopam ine needed (div by) 40 m g per ml = 20 m l needed 102. W hat type of catheter will be inserted into the patient’s bladder and used for adm inistration of the am photericin B bladder irrigation? A. preferred diluent B.4 grams (or 400 mg) 960 drops/hr = 16 drops / minute 60 m ins/hr 107. used for TPN form ulations and u su ally diluted 103. 3% NO.5 m g per m Eq (x) 60.000 mcg (= 48 m g) per hour X ml 48 m g = 500 ml = 30 ml per hour 800 m g 104. Sterile W ater NO. inserted thru right side of the heart into the pulm onary artery B. indwelling urinary bladder catheter that would have 3 heads. goes into a vein and terminates just outside the heart C.5 drops per m inute (would actually run at either 7 or 8 drops per m inute 105. Head 1 is the outflow point for the urine. Head 2 is the inflow point for the irrigation solution. IV only 111. so need 400 ml 4 0 6 10 = 1000 m l 113. 10 m l Sterile W ater qs ad 1000 ml Flow Rate: 1 m l per kg per hr 15.15 (= 15% ) = 150 gram s of dextrose needed X ml 150 gm 112. Alteplase NO. IV only E. Once the heparin drip was stopped. the physician wanted to change to a subcutaneous agent for further anticoagulation. 4% Sod Chloride. 0. (Continued from above) E. W hat will be the new flow rate in drops per minute? 10.2 (= 20% increase) = 18 units per kg per hr 109.2 lb/kg = 80 kg 18 units/kg/hr (x) 80 kg = 1440 units/hr 1440 units/hr = 14. W arfarin NO. Lepirudin NO. A possible drug for this purpose would be A. 108. Robinson catheter – NO.75% Pot Chloride. 1000 m l (x) 0. 15% Am ino Acids. 15 units per kg per hr (x) 1. Cl = 35.5 gm (7500 m g) per 1000 m l 58. 110.000 gm/m l = 214. this is a “straight” catheter that is inserted into the bladder to drain urine but is not intended to remain in the bladder and also does not have a second channel for use for inflation of the balloon or as an irrigation line. Argatroban NO.75% ) = 7. 700 m g/m l = 70% so can do by alligation 70 15 X m l = 1000 m l = 214.107. 10 = 100 m l 70 gm = 4 Dextrose.5 mg/mEq (x) 4 mEq per ml = 234 m g/ml Na = 23.6 drops per m inute (NAPLEX would increase dose to get bigger and rounded off num ber. oral and IV only B.000 units (div by) 100 ml = 100 units per m l 176 lbs (div by) 2.0075 (= 0.2% MVI-12.5 = 58. subQ route only C.3 m l 15 parts 70 parts 15 0 55 70 = 1000 m l Could also do by mathematical method: 1000 m l (x) 0. 0. then 4 parts = 400 m l.4 ml per hour (x) 15 drops/ml = 216 drops per hour (div by) 60 min/hr 100 units per ml = 3. IV only D. Daltaparin YES.5 mg per one milliequivalent since Sodium is m onovalent 7500 m g (div by) 234 m g/m l = 32 m l needed Page 25 of 27 .3 m l 70 gm If 10 parts = 1000 m l. 4 kg [(140 – 35) x 68.3 x 8 inches) = 68.8 cals in 24 hours 117.4 cals per 24 hrs 2040 m l (x) 0. 0. Then see which answer is provided as one of the choices.2 = 38.75% Pot Chloride. The IV set used to adm inister this solution delivers 12 drops per m L.5 mg/mEq (x) 2 mEq/m l = 149 m g per m l 2000 m g (div by) 149 m g per ml = 13. 0. 1000 m l (x) 0.3654 m l/m in Page 26 of 27 . Cl = 35. The usual form of the equation is given below: Clcr (m L/m in) = [(140 – age) x weight (kg)] [72 x serum Cr (m g/dL)] x 0. Since the m ath problem s on NAPLEX are multiple choice.002 ( = 0.2% ) = 2 gm (2000 m g) per 1000 m l K = 39.85 (for wom en) The weight used should be the ideal body weight (IBW ) for the patient involved.4 kg}] 72 x 2.6 m g/dL = 7182 187.4 cals/gm = 1040. The first patient to receive this IV fluid had GI tract surgery and needs to be NPO for 5 days. NAPLEX has not consistently dem anded use of the ideal body weight.4 + 326. The Cockcroft-Gault equation is com m only used to estim ate the clearance of creatinine in the kidney. If the serum creatinine is stable. The patient weighs 85 kilogram s. 15% Am ino Acids.6 gm s am ino acids (x) 4 cals/gm = 326. then choose the ideal body weight answer.5 mg per one milliequivalent since Potassium is m onovalent 74.2% M VI-12.3 m l Dextrose.480 drops in 24 hours or 1440 m inutes (24 hr x 60 min/hr) 24.04 (4% am ino acids) = 81.7 m l 330.7 m l Total vol Tot additives Water to add 1000 m l 669.4 m l needed 115. If both answers are provided. 10 m l Sterile W ater qs ad 1000 ml Flow Rate: 1 m l per kg per hr 116.5 = 74.4 cals per 24 hr Total calories in 24 hours = 1040.4 = 1366. How m any m L of water will be required to prepare one liter of the form ula? Dextrose Amino Acids Sod Chloride Potassium M VI-12 Total additives 214.3 m l 400 m l 32 m l 13.114. BUT. How m any calories from the dextrose and the am ino acids will the patient receive in 24 hours? 1 m l (x) 85 kg (x) 24 hrs = 2040 m l in 24 hours 2040 m l (x) 0.480 drops (div by) 1440 minutes = 17 drops per m inute 118.4 m l 10 m l 669. For Don Smith: Ideal Body Weight Clcr (ml/m in) = IBW in kg = 50 + (2. 4% Sod Chloride. you should work the problem using the actual body weight given and the ideal body weight. the creatinine clearance is a good approxim ation of the glom erular filtration rate. W hat will be the flow rate to adm inister the solution at the dose ordered for this patient? 2040 m ls (x) 12 drops per m l = 24.15 (15% dextrose) = 306 gm s of dextrose (x) 3. 3 (x) inches over 5 ft) Fem ales 1-18 yrs and > 5 ft tall Fem ales > 18 yrs) and > 5 ft tall IBW in kg = 42. One m ethod to determ ine is the NG tube is in the lung is to put the end of the tube extending from the patient’s nose into a cup of water and check for bubbles. If the bubbles occur. A common set of form ulas is as shown below: Males 1-18 yrs and > 5 ft tall Males > 18 yrs and > 5 ft tall IBW in kg = 39 + (2. then 150 mg once daily for a creatinine clearance between 30 and 49 m l/m in (read this from the chart. Page 27 of 27 . no calculations required). Care must be exercised to avoid placing the NG tube in the lung. also N/G) is inserted through the nose into the stom ach.2 + (2.65 1000 2 119. they represent air coming out of the lungs and the tube needs to be repositioned.27 (x) inches over 5 ft) IBW in kg = 50 + (2. 120.5 + (2.3 (x) inches over 5 ft) Children.27 (x) inches over 5 ft) IBW in kg = 45.Several m ethods of calculating ideal body weight from actual body weight and height exist. 150 m g dose (div by) 10 mg per ml = 15 m l per dose A NG (nasograstic tube. 150 mg once. 1-18 yrs and < 5 ft tall IBW in kg = (height in cm) (x) 1.