87. A solution contains 5% of anhydrous dextrose in water for injection.How many milliosmoles per liter are represented by this concentration? (MW of dextrose is 180 g). Question could be read: What is the osmolarity of D5W? 5 g of dextrose are found in 100 ml of water a. 255.50 mOsmol/L. x g of dextrose are found in 1000 ml (L) of water. Weight ( g / L ) xSpeciesx1000 b. 278.00 mOsmol/L.* Osmolarity = MolecularWeight ( g ) c. 287.00 mOsmol/L. 50 x1 x1000 d. 301.30 mOsmol/L. = 277.8 mOsmol/L = 180 e. 310.00 mOsmol/L. 88. A solution contains 156 mg of K+ ions per 100 ml. How many milliosmoles are represented in a liter of the solution? (MW of K+ = 39 g) 156 mg of K+ are found in 100 ml of water a. 30 mOsmol/L. x g of K+ are found in 1000 ml (L) of water, x = 1560 mg = 1.56 g b. 35 mOsmol/L. Weight ( g / L ) xSpeciesx1000 c. 40 mOsmol/L.* Osmolarity = MolecularWeight ( g ) d. 45 mOsmol/L. 1.56 x1 x1000 = 40 mOsmol/L = 39 e. 50 mOsmol/L. 89. A solution contains 10 mg% of Ca++ ions. How many milliosmoles are represented in 1 liter of the solution? (MW of Ca++ = 40 g). 10 mg of Ca++ are found in 100 ml of water a. 0.5 mOsmol/L. x g of Ca++ are found in 1000 ml (L) of water, x = 100 mg = 0.1 g b. 1.0 mOsmol/L. Weight ( g / L ) xSpeciesx1000 Osmolarity = c. 1.5 mOsmol/L. MolecularWeight ( g ) d. 2.0 mOsmol/L. 0.1 x1 x1000 = 2.5 mOsmol/L = 40 e. 2.5 mOsmol/L.* 90. How many milliosmoles are represented in a liter of an 0.9% sodium chloride solution? (MW of Nacl = 58.5, species = 2). 0.9 mg of Nacl is found in 100 ml of water a. 308 mOsmol/L.* x g of Nacl are found in 1000 ml (L) of water, x = 9 g b. 358 mOsmol/L. Weight ( g / L ) xSpeciesx1000 Osmolarity = MolecularWeight ( g ) c. 399 mOsmol/L. d. 413 mOsmol/L. 9 x 2 x1000 = 307.7 mOsmol/L = 58.5 e. 429 mOsmol/L. 91. A ready-to-use enteral nutritional solution has an osmolarity of 470 mOsm/L. How many ml of purified water are needed to adjust 8 fluid ounces of the enteral solution to an osmolarity (280 mOsm/L)? a. 120 ml. b. 160 ml.* c. 240 ml. d. 300 ml. e. 400 ml. One of the most convenient methods of solving this problem is using the equation: (Q1) (C1) = (Q2) (C2) Page 24 of 24 d. 3000 days.* 77. 78. Approximately 50% of the drug is excreted unchanged in the urine. If the normal dosage scheduled for Dicloxacillin is 125mg every 6 hours. A patient with renal function 20% of the normal should receive? a. 25mg q6h. b. 31.25mg q6h. c. 75mg q6h.* d. 62.5mg q6h. 79. 80. What is the minimum quantity that can be weighed on a balance with a sensitivity requirement of 6 mg if an error of 5% is permissible? a. 100 mg. b. 120 mg.* c. 140 mg. d. 160 mg. Sensitivity Requirement (SR) = Minimum Qty to be weighed x Permissible error % SR = (x) mg x 5% 6 mg = (x) mg x 5/100 (x) mg = 100 x 6/5 = 120 mg 81. 82. If the pka of Aspirin is 6.4, what fraction of the drug would be ionized at pH 7.4? a. 60% Charge: 10-1=0.1 100 b. 75% ch arg e ( PH − pKa ) % of ionized drug = Acid = -1 10-2=0.01 1+10 c. 80% 10-3-0.001 Base = 1 100 100 100 100 100 = 1+ ( 0.1) = 1.1 = = = d. 90%* −1 −1(1) −1( 7.4 −6.4 ) = 10-4=0.0001 1+10 1+10 1+10 e. 100% 83. If half life elimination of a drug is 2 hours, what fraction of the original dose of the drug will remain in the body after 4 hours? a. 80% b. 50% c. 25%* d. 12.5% e. 7.5% 84. A graph (log Cp versus time) of Cp = CO e-kt will give: (Cp = concentration of drug, Dose = e-kt) a. A straight line with a positive slope. b. A straight line with a negative slope.* c. Does not exist. d. A circle. e. Negative slope. 85. A pharmacist adds one gram of calcium chloride (CaCl2. 2H2O) to a 500 ml bottle of water. How many mEq. Of chloride are present in each ml of solution? (Ca = 40; MW CaCl2 = 111; H2O = 18). a. 0.014 b. 0.027 c. 0.036 d. 0.041 e. 13.60* 86. Suppose 12 suppositories, each containing 300 mg aspirin, are required. Given the density factor of aspirin is 1.1, what is the amount of Cocoa butter required for the preparation? a. 20.40 g. Total amount of aspirin = 12 x 300 = 3,600 mg = 3.6 g. b. 20.04 g. TotalAmountOfAspirin 3.6 Amount of aspirin replacing Cocoa butter = DensityFactorofAspirin = 1.1 = 3.27 c. 20.54 g d. 20.73 g.* Total amount of Cocoa Butter = 12 x 2 = 24 g. Amount of Cocoa Butter Needed = 24 – 3.27 = 20.73 g e. 18.22 g. Page 23 of 24 20.15 hr-1. c. Page 22 of 24 .weight. 120 ml/min. b. 25%. 660 mg. By applying the method of Cockroft and Gault ClCR = (140 − age. 300 mg. b. b.45% experience the same side effect. c. 100-60=40% 74. 75. 200 mg. 150. 72.mg / dl ) (140 − 20)(72) ClCR = = 120 ml/min 72(1) 68. 500. b.000 unit of penicillin G is equal to: (1 unit = 0.e. the minimum number of patients that would have to receive the new drug for 3 years to statistically demonstrate the prevention of one episode of this side effect in at least one patient is: a. 50%. 10%. In an adequately powered. 200. 76.e. In patients who receive a newly discovered drug. Divide (15%/25%)x100= 60 then take 60 off 100. in. 70.* c. 5000 days. 140 ml/min. The amount of the drug will provide an adequate therapeutic level. d. d. Based on these results. A drug degrades at the rate of 1mg in 60 day from 100mg. c.0 mg/dl? a. The relative risk reduction achieved with the new drug over the study period is: a. 2000*. 40%*. 15%.6 mcg) a. only 0. c. 69.66.45%)x100=0.0. the total amount of drug in each tablet is: a. What is the Creatinine clearance (ClCR) of a 20 year old man weighing 72 kg and has a Serum Creatinine Concentration (CCR)=1. 560 mg. a specific serious side effect (i. If the elimination rate constant of this drug is 0. only 15% obtain the same clinical benefit.in. 6000 days. reduction in leukocytes) with conventional therapy is seen in 0. b. then divide 100/0. e. 73. 460 mg. 600 mg.in. randomized controlled trial conducted over 3 years.5% of the study sample. prevention of a serious cardiovascular event) with a new drug is achieved in 25% of the study sample. 80 ml/min. In adequately powered. d. In the patients who receive a placebo. The drug in the slow release core must sustain therapeutic level for 12 hours.kg ) 72(Ccr. 15. the desired clinical outcome (i. what is the t1/2 of the drug? a.* c. years)(body.5=2000. randomized controlled trial conducted over 2 years. The press coating of a tablet contains 200 mg of a drug for immediate release.05. Subtract (0. 67. d. 360 mg.5%-0.* d. 100 ml/min. 4000 days. b. 400 mg. e. 71. A total parentral nutrition (TPN) order requires 500 ml of D30W. 150 x100 40 g 150 Then x = = 375ml = 40 100ml x Page 21 of 24 . 0. 0. b. e.6 x100 = 0. Sensitivity Requirement (SR) = (Minimum weighable amount) x (acceptable error) Minimum weighable amount = (6)/2% = 300 mg. 100 mg/dL. 400 ml.000ml 100ml then x = o. c.61. 0. b. Because 1000 µg = 1 mg and 100 ml = 1 dL. 300 ml.000. 64. 100 µg = 0. 120 mg. 0. 667 ml. The upper therapeutic drug concentration for valproic acid is considered to be 100 µg / ml .1 mg/dL. The concentration of sodium fluoride (NaF) in a community’s drinking water is 0.0006%. Using data based on the patient’s measurements. allows the estimation of the patient’s BSA.6 ppm concentration indicates 0.6 g of sodium fluoride present in 1. How many ml of D40W should be used if the D30W is not available? a. d. c. 500 ml of D30W will contain 150 g of dextrose. c. The intercept on the middle line. 65. the grams present in 100 ml will be: 0.00006%. 300 mg. c. Which of the following measurements must be known in order to use this nomogram? a. The o. a.006%. a line is drawn between the two outside parallel lines.000. 1 mg/dL. b. Express this value in terms of mg/dL. e. 375 ml. Express this concentration as a percentage. a. Height and weight. b. which is calibrated in square meters of body surface area.000. The left line is calibrated with height measurements in both centimeters and inches. D40W contains 40 g of dextrose per 100 ml.00006 g 1. e. c.6 x = 1. whereas the right line lists weights in kilograms and pounds. 10 mg/dL. 200 mg. 180 mg.6% Sodium fluoride is a solid chemical. The nomogram in the USP consists of three parallel vertical lines.1 mg/ml = 10 mg/dL 63. 0. Age and weight. Height and Creatinine clearance. Weight and sex. d. Therefore.6 ppm.06% e. What is the minimum amount of a potent drug that may be weighted on a prescription balance with a sensitivity requirement of 6 mg if at least 98% accuracy is required? a.000 62. Age and height. 1000 mg/dL. 6 mg. 0. d.000 ml of solution. The USP contains nomograms for estimating body surface area (BSA) for both children and adults. d. 125 ml. b. e. d. Page 20 of 24 .6)/1 = 78. c. Grams. e. Use the equation of: (Q1) (C1) = (Q2) (C2) (4 ml) (1/2000) = (x ml) (1/1000) then x = 2 ml. c. Therefore. 5 ml. Microgram. 0. How many ml of adrenaline chloride solution (0.15 µg . 0.05 mg of digoxin per ml.1%) may be used to prepare the solution. e.5 µg . Five thousand (5000) nanogram equals 5: a. None of the above. c. d. e. 1 ml. 2 ml. 0. b.04 ml. 1. b. d.0015 µg .5860 mg = (x) (74. Lanoxin pediatric elixir contains 0. O.002 ml. b.15 mg of digoxin contained in 3 ml of the elixir would be equivalent to 150 µg of drug. Centigrams. 0. How many micrograms are there in 3 ml of the elixir? a. 59. d. 60.015 µg . Kilogram. 0. Milligram. a.5 mg 58. 1 mg = 1000 µg . a. when comparing salts of a drug. because the 300 mg weight is based on a chemical formula containing 7 waters of hydration. The valence of iron has no significance in this type of problem because only one atom of iron is present in each molecule of ferrous sulfate. 110 mg.6 mg 1mEq x = 74. b. 1 equivalent weight of KCL = 74. None of the above. . 78. 0. How much elemental iron is present in every 300 mg of ferrous sulfate (FeSO4 u 7H2O)? (Atomic weights: iron = 55. b. This concentration expressed in terms of mg/dL will be: a. 0.004 = 1. 150 mEq. Weight qualities expressed in molar amounts allow a more realistic evaluation of the actual number of drug molecules present.96 mg 56.9. S = 32. 30 mg.wt. e.54. d.7 mEq.6mg 5860mg then x = 5860 x1 = 78. d. 14 mg. c. 1.6 or the problem may be solved by using the equation: mg of chemical = (mEq) (mol. 12.54 mg/dL. 164 mg. e. 150 mg. a.5 mEq. . b.32 g . How many milliequevalents (mEq) of KCL are present? (Mol. This is incorrect.27 kg Dose in mg = 0. b. The body weight will be = 16/2. A patient’s serum cholesterol value is reported as 4 mM/L. O = 16.5mEq 74. 120 mg.27 = 13.6) a.004 moles.08 x 24 x 7. 20 mEq.9 x 300)/278 = 60.86 g of potassium chloride (KCl). c. H = 1.This is the amount of anhydrous ferrous sulfate present in each 300 mg.154 mg/dL. 60 mg. The formula weight of ferrous sulfate is (55. A 250 ml infusion bottle contains 5. the mM/L concentration is converted by realizing that 1 mole of cholesterol weighs 386 g and 4 mmoles equals 0.08 mg/kg/hr.This result assumes that the ferrous sulfate is anhydrous with a molecular weight of 152.6 g 1 milliequivalent (mEq) = 74. 596 mg/dL. Wt. d. The infusion rate of theophylline established for a neonate is 0.004mole = 386 g x then x= 386 x0.54 g 1 x = 1. 1mole 0. How many mg of drug are needed for one daily bottle if the body weight is 16 lb. c. 1540 mg/dL. The amount of iron present in 300 mg of the chemical will be: 300 mg → 278 x → 55. c.Ditto. 154 mg/dL.This result would be obtained if the correct answer was either doubled or halved to reflect the +2 valence of iron. e. . 8 mg. In this problem. d.9 x = (55.9+32+(4x16)+(7x2)+(7x16) = 278.)/(valence) Page 19 of 24 . e. An increasing number of laboratory test values and drug doses are being reported in terms of millimoles (mM).2 = 7. The question asks for iron (Fe) only.54 g in 1 L or 1450 mf/L and 154 mg/dL 57. 30 mg. Iron has valences of +2 and +3). for example. 55.58 mg. KCL = 74. 4 g/oz. or 100 mg. 240 mg. 160 mg. c. 100 Page 18 of 24 . Two ounces (avoir.8 g. 60 mg. 100. How many jars can be filled? a.) consist of 28. e.52.8 = 80 jars. Because the normal Creatinine clearance rate is 100 to 120 ml/min 40ml / min 100ml / min = x 240mg then x = 40 x 240 = 96 . 73. Thus. 53. 100 mg. d. The normal maintenance dose would be = 120 lb x 2 mg/lb = 240 mg. 88. c. b. 83.). d. e. 4540 g divided by 56. What maintenance dose should be administrated if the normal maintenance dose is 2 mg/dl of body weight? a. A pharmacist repackages 10 lb of an ointment into jars to be labeled 2 oz (avoir. Ten pounds contain 454 g x 10 = 4540 g. or 56. The estimated Creatinine clearance rate for a 120 lb patient is 40 ml/min. 80. 120 mg. b. 2% Qs 100 ml What percentage of patients had a reduction between 5 and 15 mm? a. a. whereas 2 SDs will include approximately 97% to 98%. a. 1. How many ml of glycerin would be needed to prepare 1 lb of an ointment containing 5% w/w glycerin? (The density of glycerin is 1. Step 2: Multiply the amount of chemical by its “E” value: 1200 mg x 0. c. Therefore.45 g. e. e.4 ml. A density or SG of 1. d. It shows the dispersion of numbers around the mean (average value).31 g.5 ml. A radiopharmacist prepares a solution of 99mTC (40 mCi/ml) at 6:00 am. c. 51.5 ml. 5.37 g. 50% d. approximately one-half of the original strength has decayed. 24 ml. 0.25 g. e. 22. Because 1 lb of the ointment contains 5% w/w glycerin.31) Rx qs 0. c. On SD will include approximately 67% to 70% of all values.0 ml.9% = 900 mg. b.2 ml. If the solution is intended for administration at 12:00 pm at a dose of 20 mCi. 70%. Step 1: 50.2 ml.25 g/ml) a.7/1. or 1200 mg.2 g. d. b. 1. b. Because the time interval between preparation and administration is 6 h.53 g. e. How much sodium chloride is needed to adjust the following prescription to Isotonicity? (E value for sodium thiosulfate is 0.90 g. Step 3: Determine amount of NaCl needed as if no other chemical was present: 100 ml x 0. Blood pressure measurements were made 1 week on five patients with the following averages: Patient BP 1 140/70 2 160/84 3 180/88 4 190/90 5 150/70 Sodium thiosulfate Sodium chloride Purified water 1. Step 4: Subtract contribution by chemical (step 2) from the amount of NaCl (step 3): 900 mg – 372 mg = 528 mg (amount of NaCl needed to render the solution isotonic).25 indicates that 1 ml of the liquid weighs 1. 28. 1.48. c.7 g of glycerin. 2. then v = w/d = 22.31 = 372 mg (equivalent amount of NaCl). 1 ml of the solution now assaying at 20 mCi/ml is needed. all of the patients listed had a systolic blood pressure reduction of 10 mm with a Standard Deviation (SD) of 5 mm. how many ml of the original solution are needed? The half-life of the radioisotope is 6 h.0 ml. 0. 0. b.2% = 1. 18. A standard deviation is calculated mathematically for experimental data. and the half-life of the radiopharmaceutical is 6 h. After one month of therapy. Density = w/v. 454 g x 5% w/w = 22. 20%.25 = 18.7 ml. 0. 90%. 40%. 49. d. 0. Determine amount of sodium thiosulfate in the prescription: 100 ml x 1.0 ml.2 ml Page 17 of 24 . 15 mCi. 2160 mg. convert the weight in pounds to kilograms: (1 kg = 2.55 x 2 = 159. Calculate the dose of a drug to be administered to a patient if the dosing regimen is listed as 2 mg/kg/day. b.5 mCi/ml Page 16 of 24 . 12 mg.2 lb) 1kg 2. 770 mg. The adult intravenous (IV) dose of zidovudine is 2 mg/kg q4h six times daily. e. In this example.45. 2. 350 mg.2 47. d. First. 160 mg.1 mg 2. 140 mg. b. 7. d.8 x 2 mg x 6 doses = 981.0 mCi. the halflife of 6 h allows a quick comparison of the amount of radioactivity remaining. 10 mCi. d. 980 mg.8kg 2.2 lb 1kg 2. How many mg will a 180 lb patient receive daily? a.2lb = x 180lb then x = 1x180 = 81. Because 1 kg = 2.2 Second. e. What concentration of the original 99mTC solution described below will remain 24 h after its original preparation? Solution of 99mTC (40 mCi/ml) a. determine the total daily dose = 81.2lb = x 175lb then x = 1x175 = 79. 650 mg.5 mCi. e. b.8 mg 46. 164 mg. c. a. c. 78 mg. The patient weighs 175 lb.55kg Then total daily dose = 79.5 mCi. c. Original Activity 40 mCi/ml After 6 h 20 mCi/ml After 12 h 10 mCi/ml After 18 h 5 mCi/ml After 24 h 2. 5. The loss in first-order kinetics is a constant fraction of the immediate past concentration. 1/2 gallon of 200 proof (100%) alcohol is a proof gallon. Two gallons of 25% alcohol is 50% proof. 25 gallons of 70% alcohol contain how many proof gallons? Pr oofGallons = 25 x70 = 35 50 Page 15 of 24 . 43. The wine gallon is the common unit of volume measure. Convert 212oF to Co? 9C = 5F – 160 = (5 x 212) – 160 = 900 ∴ C = 100oC ---------------------------------------------------------------------------------------------------------------------42. Convert 100oC to Fo? 9C = 5F – 160 5F = (9 x 100) + 160 = 900 + 160 F = 1060/5 ∴F = 212oF "PROOF SPIRITS" Proof Gallon = Wine gallon x Proof Strength One proof gallon is defined as the gallon of 100 proof (50%) ethyl alcohol. Therefore. Any quantity of alcohol containing the equivalent of 1 gallon of 50% alcohol is a proof gallon.5 50 50 44. 55 gallons of 45% alcohol contains how many proof gallons? 9C Pr oofGallon = GallonxStrength 55 x 45 = = 49.160 40. Convert 32oF to Centigrade? = 5F – 160 = (5 x 32) – 160 =0 ∴C = 0oC ---------------------------------------------------------------------------------------------------------------------41."TEMPERATURE CONVERSION" All problems of conversion of Fahrenheit (F) to Centigrade (C) temperature may be solved by the following formula: 9Co = 5F . contains (x) ∴ 5000 ml 100ml ∴x = 100 x 250 = 50% 5. In these cases. Page 14 of 24 . consider the volume of the diluent as having 0% concentration of the drug. 200mg of 10% + 50gm of 20% + 100gm of 5%? 200 50 100 350 X X X 10% 20% 5% = = = 20 10 5 35 00 00 00 00 contains 35 00 absolute alcohol. you will run into a problem where the addition of a diluent or solvent is contained. contains (x) ∴ 350ml 100ml ∴x = 100 x35 = 10% 350 NB: Occasionally. 000ml absolute alcohol. mixture of solids possessing different % strengths. What is the % v/v of alcohol in the following mixture. 1 Liter 60%. 1000 ml of 70%? 1000 3000 1000 5000 X X X 60% 40% 70% = = = 60 120 70 250 000ml absolute alcohol."ALLIGATION" It is an arithmetic method of solving problems that involve the mixing of solutions. 3000ml of 40%.000 absolute alcohol. ointments. 000 contains 250.000 39. 38. What is the final % of Zinc Oxide ointment made by mixing ZnO ointment of the following strengths. Alligation medically used to calculate the % strength of a mix made by mixing two or more components of a given % strength. 000ml absolute alcohol. 36.0002 = 0.25 = 1.25% Ephedrine SO4 Rose Water ad 10ml How many ml of 1:50 stock solution of ephedrine SO4 are necessary for dispensing? 1:50 = 2% ∴ 0.5% = (x) x 0.25% x 10 = 2% x (x) ∴x = 10 x0. How many ml of 1:5000 Potassium Permanganate solution can be made from 50ml of 0.02% 50 x0.02 37.02% 5000 ∴ 50 x 0.25ml 2 Page 13 of 24 .5 ∴x = = 1250ml 0.5%? 1 = 0. You receive the following prescription: 0. 5 parts then 1. 0.20 g. 1. b. the final weight of the cream will be greater when hydrocortisone powder is added.02 x x = 0. Because the amount of 0.90 g. If 20ml of a 1:200 w/v solution are diluted to 500ml.5% = (x) x 500 20 x0. A floor nurse requests a 50 ml minibottle to contain heparin injection 100 u/ml. d.92 g 98 or.53 g.32. b.2 + 0.000 = 0.5 x60 = 0. How many grams of pure hydrocortisone powder must be mixed with 60 g of 0. 5 ml. what will be its concentration? ∴ 65% x 100 = (x) x 85 65 x100 ∴x = = 76. 1. The number of ml of heparin injection 10.5 ml 33.1 ml.000 = 1ml x then x = 100 x 50 = 5000 u then x = 5.5% hydrocortisone cream if one wishes to prepare a 2.5 ∴x = = 0. let x = weight of 100% HC powder (x g) (100%) + (60 g) (0. by algebra. what is the final concentration? ∴ 20 x 0. 2.5% hydrocortisone cream is exactly 60 g. 1 ml.000/10.5%) = (60 g + x g) (2%) x + 0. e. 100u x = 1ml 50ml 10.000u 5. 0.0% w/w preparation? a.5% 60 g x = 98 parts 1. d.92 g. 0.30 g.3 = 1.5 ml.5 ml. 0. If a 65% w/v sugar solution is evaporated to 85% of its original volume.5 Parts of 100% 2% 0.5% 98 Parts of 0.02% 500 34.000 u/ml needed for this order will be: a. the problem may be solved by allegation alternate method or by simple algebra.92 g 35. Therefore. c. 100% 1:200 = 5% 1.47% 85 Suppose the volume is 100ml Page 12 of 24 . c. e. 0. reduce them to the lowest possible common denominator (. add 17 ml of Sterile Water for Injection to obtain 500 units per ml. the final volume of solution is: 500u 10. If 500ml of a 15% v/v solution is diluted to 1500ml.000 units: to reconstitute.)اﻟﻤﻘﺎم 30. 20 ml. Therefore. are reconstituted. c.Whenever the problem deals with proportional parts. 17 ml.000 15% = 500 x% 500 x15 ∴x = = 5% 1500 or Strength x Volume = Strength x Volume 15(%) x 500 = (x) x 1500 ( x) = 15 x500 = 5% 15000 31. When some drug powders. Page 11 of 24 . e.When ratio strengths are given in the problem. 10 ml. the resulting volume will be 10 ml.000u = 1ml x 1000u 10.000u = 1ml x then x = 20 ml.5 ml."DILUTION & CONCENTRATION OF SOLUTIONS" There are two (2) important rules to follow: 1. b. especially bulky antibiotics. In this example. the total volume of solution that must be prepared will be: then x = 10 ml. When a concentration of 1000 u/ml is desired. 7 ml.” How many ml of SWFI must a pharmacist add if a 1000 u/ml concentration is needed by the nurse? a. the pharmacist must add 7 ml of SWFI to the vial. convert to percentage % strength before beginning your calculations. 8. the volume occupied by the bulk powder once it has dissolved must be considered. The previously listed volume means that the volume occupied by the bulk powder must have been 20 – 17 ml = 3 ml. d. what is the resulting % strength of the final solution? 15. A vial of lyophilized drug is labeled “10. 2. When the powder has dissolved. c. 10000 Page 10 of 24 . 10 mg.5 x 40)/(2) = 30 mg/dL or 300 mg/L 29.5 mEq of calcium per 100 ml. 50 mg. 30 mg/L. 30 mg. A solution contains 1. 1 mEq equals 40 mg divided by 2 = 2. 150 mg/L. 500 mg.5. 300 mg/L.)/(valence) = (1. e.05 gm . b. 600 mg/L Because the valence of calcium is +2.mg. Therefore. 20 mg. Wt. 1 x = 10000 500 ∴x = 1x500 = 0. b. e. and that is equal to 50mg. How much of Gentian Violet is used to make 500ml of 1:10000 solution? a. If using the equation: Mg of chemical = (mEq) (mol.24 5.25 28. Express the solution’s strength of cacium in terms of mg/L (The atomic weight of calcium is 40) a.5 mEq = 30 mg. 1. 60 mg/L.25 g 1 = 1046ml x Then x = 1046 x1 = 199. d. d. c. 02% as ratio strength? 0. Express concentration of this solution as a ratio strength? Multiply by 1000: 2mg/ml = 2000mg/1000ml then change mg's into grams = 2gm/1000ml then… = 1gm/500ml or 0. d.086 grains 27. 1/10.25% w/v sodium hypochlorite.43grain = 0. One hundred ml of Clorox contains 5. 1/180. 1/200.25 of sodium hypochlorite."RATIO SOLUTIONS" For solids in liquid Gram/1000ml of mixture. The ratio strength will be: Page 9 of 24 .002 25.05 1 = = 500 100 2000 (=0.25gm ∴ ∴ The ratio is 1:2000 0. What is the ratio strength of the solution made by dissolving 125mg 2 tablets of mercury bichloride in 500ml of a solution? (Solution) Total mgs = 125 x 2 = 250mg. c. For solids in solids Grams/1000grams of mixture or Grains/1000grains of mixture 22. 1/9.2gm x 15.43 = 3.25 0. 1/100. a. A pharmacist dilutes 100 ml of Clorox with 1 quart of water. e. Express 1:4000 as a % concentration? 1 x = 4000 100 100 x1 Therefore x= = 0.2 gm x= 2500 Then change grams into grains according to 1gm=15. Express the concentration of sodium hypochlorite in the final solution as w/v ratio.02 1( part ) = 100 x( parts ) 2 1 = 10000 5000 = 1:5000 23. The final dilution will be 100 + 946 ml of water for a total of 1046 ml.002 gm 1ml = 1gm xml then x= 1 = 500 0. is equal to 0. Express 0. Which.025% 4000 24. How many grains of Potassium permanganate could be used to make 500ml of 1:2500 solution? 1 gm in (x) gm in 2500ml 500ml 1gm` ( x) gm = 2500ml 500ml 1x500 = 0. Commercial Clorox contains 5.05%) 26. A prescription calls for a solution of drug equal to 2 mg/ml. b. For liquids in liquids Ml/1000ml of mixture. 18 ---------------------------------------------------------------------------------------------------------------------21.16 0. then the change in pH is going to be: pH = 4. 9. What is the slope if.24 0.76.94 – 4. For the curve equation where y is function of x. y = 12x – 3x2. d.= = 4.176 = 4. c. Page 8 of 24 . b.76 = 0. y = 9? a.76 4.76 + + Log 0. 6. x = 1.94 And since pH before addition of NaOH was 4. 7. 3. 18.4 g.04 0.04M of NaOH converts 0.2 0. 0.04M of acetic acid to 0.76 4.7782 + (-4. Calculate the change in pH upon adding 0.2 g.5 g. 363.2218 6 x 10-4 = 0. A hospital clinic requests 2lb of 2% hydrocortisone ointment.(-3.2218) = + 3. 0. 0. calculate H+ concentration of the solution? PH Log H = = = -Log H .76 + Log Page 7 of 24 . pH = -Log 6 x 10-4 = .8 d.2 + 0.16 b.2 g.18 d.76 The addition of 0.0)) = .19 PH = Pka + Log salt acid 0.76 + + Log Log 1 = 4. e. We have first to convert 2lb into grams = 2 x 454 = 908gm 908gm Xgm = X 2% = Xgm X 5% 908x 2% = 363.(Log 6 + Log 10-4) = . 27.1 7.04 = 4. 545 g. d.17. 3.18 c.11. PH of a certain solution is 11.2gm 5% 18. 32. b.(0.2 − 0.218 b.2M concentration of sodium acetate.76 at 25oC? a.17 c.2218 19.2 = = 4. 322. Pka value of acetic acid is 4. consequently the concentration of acetic acid is decreased and the concentration of sodium acetate is increased by equal amount according to the following equation: PH = Pka + Log Salt + base Salt − base 0.0006. 0. 3221.2M concentration of sodium acetate and acetic acid.1.94 x 10-12 20. Find the pH of a solution which has H+ concentration of 6 x 10-4 a. 45.04 mole of NaOH to a liter of a buffer solution of 0. How many grams of 5% hydrocortisone ointment would be diluted with white petrolatum to prepare this order? a. c. c. One tablespoonful (tbsp) delivers 15 ml of liquid. 600 ml. c. 400 ml. 5ml Page 6 of 24 . 10 capsules. 160 ml.3 120 15. 20 capsules.20 ml → x mg 200mg x = 120ml 20ml then x = 20 x 200 = 33. 16.100ml of solution Number of capsules = 100mlx1cap = 20 Capsules. 800 ml. 15 capsules. 250mg (One capsule) -------------------. b. d. How many capsules of 250mg of a drug should be used to prepare 100ml solution which has a concentration of 250mg/5ml? a. the patient is receiving four doses per day for 10 days. 15 ml x 4 x 10 days = 600 ml total. Therefore.5ml of solution ?? -------------------. In this prescription. b. e. The directions intended for the patient on a prescription read “1 tbsp ac & hs for 10 days” what is the minimum volume the pharmacist should dispense? a. 200 ml. 5 capsules. d. Ampicillin is mixed in a 5% dextrose bag. d. 10mg% c.4 11. b.67capx30ml = 5 Capsules.25 mg.8 h. Solution of a substance is 1:10000. 5. c. The symbol’s original meaning as a drachm (weight) or fluidrachm (volume) quantities is archaic and should not be used. 6. the symbol “i” is used to represent a 1 teaspoon dose. The mean half-life is: a. p. e.0 h. the patient in this prescription is receiving four daily doses for a total of 20 ml. what is the percentage w/v in mg? a.25 mg.s. 6.i. Prepare 30ml solution of 2. 10 Capsules. 5.67 Capsules 150mg → → 100ml 30ml 16.5% Clindamycin using 150mg Clindamycin capsules. 19 mg. b. Mean half-life= (3+9+6+5+4)/5 = 5.0 h. 7 Capsules. d. 100mg% 12. 5. 100ml 14.0 h. 3 Capsules. b.s. 500 mmole.67 Capsules Y Y = 16. 15mg% d. The degradation rate of ampicillin is 12mh/hr.4 h. c. 120 ml Sig: 3 I t. 5 Capsules. 33 mg. Because a standard teaspoon is considered to be 5 ml.c. d. c. 10. How many capsules are needed? a. and 4 hours.9. 4. 5. e. 25 mmole.d. 2. In today’s health practice. a. A 10ml vial of concentrated stock solution for electrolyte replacement therapy contains 55g of calcium chloride. What is T90 of the ampoule? 13. d. c. 6. 8. How many mg of codeine phosphate are being consumed daily by a patient taking the following prescription as described? Rx Codeine phosphate 200 mg Dimetapp Elixir q.5g in 100ml is equal to 2500mg in 100ml 150 mg 2500 mg X = → → 1 Capsule X 2500mgx1capsule = 16. 9. 100 mmole. 120 ml → 200 mg codeine phosphate Page 5 of 24 . b. & h. 50 mmole. The amount of calcium in each ml is: a. 5mg% b. 25 mg. Five subjects given a single IV dose of a drug have the following elimination half lives: 3. d.5 Parts of 100% 2% 0. How many grams of pure hydrocortisone powder must be mixed with 60 g of 0. c. This problem can be solved by the allegation alternate or simple parts method. 600 ml.7.92 g Page 4 of 24 .5 parts then 1. c. How many ml of 2% iodine solution must be mixed with a 7% iodine solution to obtain 1 L of 5% strength? a. 1. 300 ml. 1. by algebra.5% hydrocortisone cream is exactly 60 g. 2 parts 5 parts = x 1000ml 2 x1000 x= = 400ml of 2% iodine solution. 100% 1. b. 0.5% 60 g x = 98 parts 1. the problem may be solved by allegation alternate method or by simple algebra. d. 500 ml.0% w/w preparation? a. b.90 g. 0. 7% 3 Parts of 7% 5% 2% 2 Parts of 2% Thus. 400 ml.92 g 98 or.92 g.53 g. let x = weight of 100% HC powder (x g) (100%) + (60 g) (0.5%) = (60 g + x g) (2%) x + 0. 5 8. the final solution will contain 2 parts of 2% iodine for every 3 parts of 7% iodine.5 x60 = 0. 200 ml.2 + 0.20 g.5% hydrocortisone cream if one wishes to prepare a 2. e. Therefore. the final weight of the cream will be greater when hydrocortisone powder is added. 0.5% 98 Parts of 0. Because the amount of 0. e.3 = 1.30 g.02 x x = 0. How should he mix to make the 10% product? 50% 5 Parts of 50% 10% 5% 40 Parts of 5% 20% 5 Parts of 20% 10% 5% 10 Parts of 5% The total = 5 parts of 50% + 5 parts of 20% + 50 part of 5% 6. He has some 50%. In what proportion must the following strengths be mixed to obtain a 10% mixture. and 3%? 20% 15% 10% 5% 3% 5 Parts of 5% 10 Parts of 3% 7 Parts of 20% 5 Parts of 15% Proportions are 7 parts of 20% + 5 parts of 15% + 5 parts of 5% + 10 parts of 3% Page 3 of 24 . or more different strengths required to prepare another requested strength. 15%. 5%. 5. 20%.NB. four. This system can be used to determine the relative concentration of three. and 5% in stock. 20%. A pharmacist wishes to prepare a 10% ointment of drug. = In what proportion should 15% Boric acid solution be mixed with white petrolatum to produce 2% boric acid ointment? 15% 2 Parts of Boric acid 2% 0% 13 Part of petrolatum The final proportion is 2:13 Page 2 of 24 . 68 High alcohol 4.78% 22 Parts 32% 10% 46 Parts Low alcohol = 473x 46 = 320ml. 68 473x 22 = 153ml. 3 and Tween 80 is 15) Span 80 4. 1. how much low alcohol elixir (10%) and high alcohol elixir (78%) must be mixed to prepare 1 pint (473ml) of the requested elixir? Page 1 of 24 . A formula for a cosmetic cream requires 5g of an emulsifying blend consisting of Span 80 and Tween 80. Part of low concentration ingredient needed.5 parts of span 80 10.5 = 2. Desired concentration Low concentration ingredient. 10.2 Tween 80 = = 2. 25 Part The final proportions are 20:25 or (4:5 95%:50%) A physician requested that an elixir containing 32% alcohol be prepared.5 2.3 4.14 g.5 Tween 80 Span 80 = 15 6.5 5x6. how many grams of each emulsifying agent should be used in preparing the emulsion? (HLB of span 80 is = 4. If the required HLB (Hydrophilic Lipophilic Balance) is 10. Which proportion of 95% alcohol and 50% alcohol should be used to make a solution of 70% alcohol? 95% 20 Part 70% 50% 3.2 parts of tween 80 5x 4."PHARMACEUTICAL CALCULATIONS" "ALLIGATION ALTERNATE" To find out the relative amounts of solute or other substances of different strengths.95 g 10. High Concentration ingredient Part of high concentration ingredient needed. you must use the X to make a mixture of a required strength.5.