PEBC-Calculation Questions

Page 24 of 24

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?

a. 255.50 mOsmol/L. b. 278.00 mOsmol/L.* c. 287.00 mOsmol/L. d. 301.30 mOsmol/L. 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)

a. 30 mOsmol/L. b. 35 mOsmol/L. c. 40 mOsmol/L.* d. 45 mOsmol/L. 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). a. 0.5 mOsmol/L. b. 1.0 mOsmol/L. c. 1.5 mOsmol/L. d. 2.0 mOsmol/L. 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). a. 308 mOsmol/L.* b. 358 mOsmol/L. c. 399 mOsmol/L. d. 413 mOsmol/L. 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)

5 g of dextrose are found in 100 ml of water x g of dextrose are found in 1000 ml (L) of water.

Osmolarity = )(1000)/(

geightMolecularWxSpeciesxLgWeight

= 1801000150 xx = 277.8 mOsmol/L

156 mg of K+ are found in 100 ml of water x g of K+ are found in 1000 ml (L) of water, x = 1560 mg = 1.56 g



= 391000156.1 xx = 40 mOsmol/L

10 mg of Ca++ are found in 100 ml of water x g of Ca++ are found in 1000 ml (L) of water, x = 100 mg = 0.1 g



= 40100011.0 xx = 2.5 mOsmol/L

0.9 mg of Nacl is found in 100 ml of water x g of Nacl are found in 1000 ml (L) of water, x = 9 g



= 5.58100029 xx = 307.7 mOsmol/L

Page 23 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% b. 75% c. 80% d. 90%* 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. b. 20.04 g. c. 20.54 g d. 20.73 g.* e. 18.22 g.

% of ionized drug = )(arg101100

pKaPHech −+

= )4.64.7(1101100

−−+= )1(1101

100−+

= 1101100

−+=1.1

100)1.0(1

100 =+=

Charge: Acid = -1 Base = 1

10-1=0.1 10-2=0.01

10-3-0.001 10-4=0.0001

Total amount of aspirin = 12 x 300 = 3,600 mg = 3.6 g.

Amount of aspirin replacing Cocoa butter = intorofAspirDensityFactOfAspirinTotalAmoun

= 1.16.3 = 3.27

Total amount of Cocoa Butter = 12 x 2 = 24 g. Amount of Cocoa Butter Needed = 24 – 3.27 = 20.73 g

Page 22 of 24

66. 67. What is the Creatinine clearance (ClCR) of a 20 year old man weighing 72 kg and has a Serum Creatinine

Concentration (CCR)=1.0 mg/dl? a. 140 ml/min. b. 120 ml/min.* c. 100 ml/min. d. 80 ml/min.

By applying the method of Cockroft and Gault

ClCR = )/..(72

)...)(,,140(dlmginCcr

kginweightbodyyearsinage−

ClCR = )1(72

)72)(20140( − = 120 ml/min

68. 69. 500,000 unit of penicillin G is equal to: (1 unit = 0.6 mcg)

a. 200 mg. b. 300 mg.* c. 400 mg. d. 600 mg.

70. 71. The press coating of a tablet contains 200 mg of a drug for immediate release. The amount of the drug will

provide an adequate therapeutic level. The drug in the slow release core must sustain therapeutic level for 12 hours. If the elimination rate constant of this drug is 0.15 hr-1, the total amount of drug in each tablet is:

a. 360 mg. b. 460 mg. c. 560 mg.* d. 660 mg.

72. 73. In an adequately powered, randomized controlled trial conducted over 2 years, the desired clinical outcome (i.e.

prevention of a serious cardiovascular event) with a new drug is achieved in 25% of the study sample. In the patients who receive a placebo, only 15% obtain the same clinical benefit. The relative risk reduction achieved with the new drug over the study period is:

a. 10%. b. 15%. c. 25%. d. 40%*. e. 50%.

Divide (15%/25%)x100= 60 then take 60 off 100, 100-60=40%

74. In adequately powered, randomized controlled trial conducted over 3 years, a specific serious side effect (i.e. reduction in leukocytes) with conventional therapy is seen in 0.5% of the study sample. In patients who receive a newly discovered drug, only 0.45% experience the same side effect. Based on these results, 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. 15. b. 20. c. 150. d. 200. e. 2000*.

Subtract (0.5%-0.45%)x100=0.05, then divide 100/0.0.5=2000. 75. 76. A drug degrades at the rate of 1mg in 60 day from 100mg, what is the t1/2 of the drug?

a. 6000 days. b. 5000 days. c. 4000 days.

Page 21 of 24

61. The concentration of sodium fluoride (NaF) in a community’s drinking water is 0.6 ppm. Express this concentration as a percentage.

a. 0.00006%. b. 0.0006%. c. 0.006%. d. 0.06% e. 0.6%

Sodium fluoride is a solid chemical. The o.6 ppm concentration indicates 0.6 g of sodium fluoride present in 1,000,000 ml of solution. Therefore, the grams present in 100 ml will be:

mlx

ml 100000,000,16.0

= then gxox 00006.0000,000,1

1006.==

62. The upper therapeutic drug concentration for valproic acid is considered to be 100 mlg /µ . Express this value in terms of mg/dL.

a. 0.1 mg/dL. b. 1 mg/dL. c. 10 mg/dL. d. 100 mg/dL. e. 1000 mg/dL.

Because 1000 gµ = 1 mg and 100 ml = 1 dL. 100 gµ = 0.1 mg/ml = 10 mg/dL

63. The USP contains nomograms for estimating body surface area (BSA) for both children and adults. Which of the following measurements must be known in order to use this nomogram?

a. Age and height. b. Age and weight. c. Height and Creatinine clearance. d. Height and weight. e. Weight and sex.

The nomogram in the USP consists of three parallel vertical lines. The left line is calibrated with height measurements in both centimeters and inches, whereas the right line lists weights in kilograms and pounds. Using data based on the patient’s measurements, a line is drawn between the two outside parallel lines. The intercept on the middle line, which is calibrated in square meters of body surface area, allows the estimation of the patient’s BSA.

64. 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. 6 mg. b. 120 mg. c. 180 mg. d. 200 mg. e. 300 mg.

Sensitivity Requirement (SR) = (Minimum weighable amount) x (acceptable error) Minimum weighable amount = (6)/2% = 300 mg.

65. A total parentral nutrition (TPN) order requires 500 ml of D30W. How many ml of D40W should be used if the

D30W is not available? a. 125 ml. b. 300 ml. c. 375 ml. d. 400 ml. e. 667 ml.

500 ml of D30W will contain 150 g of dextrose. D40W contains 40 g of dextrose per 100 ml.

xmlg 150

10040

= Then mlxx 37540

100150==

Page 20 of 24

5860 mg = (x) (74.6)/1 = 78.5 mg

58. Lanoxin pediatric elixir contains 0.05 mg of digoxin per ml. How many micrograms are there in 3 ml of the elixir?

a. O.15 gµ . b. 0.015 gµ . c. 1.5 gµ . d. 0.0015 gµ . e. None of the above.

1 mg = 1000 gµ . Therefore, 0.15 mg of digoxin contained in 3 ml of the elixir would be equivalent to 150 gµ of drug.

59. How many ml of adrenaline chloride solution (0.1%) may be used to prepare the solution. a. 0.002 ml. b. 0.04 ml. c. 1 ml. d. 2 ml. e. 5 ml.

Use the equation of: (Q1) (C1) = (Q2) (C2) (4 ml) (1/2000) = (x ml) (1/1000) then x = 2 ml.

60. Five thousand (5000) nanogram equals 5: a. Centigrams. b. Grams. c. Kilogram. d. Microgram. e. Milligram.

Page 19 of 24

54. How much elemental iron is present in every 300 mg of ferrous sulfate (FeSO4 u 7H2O)? (Atomic weights: iron = 55.9; S = 32; O = 16; H = 1. Iron has valences of +2 and +3).


The formula weight of ferrous sulfate is (55.9+32+(4x16)+(7x2)+(7x16) = 278. The amount of iron present in 300 mg of the chemical will be: 300 mg → 278 x → 55.9 x = (55.9 x 300)/278 = 60.32 g - This result would be obtained if the correct answer was either doubled or halved to reflect the +2 valence of iron. 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. - Ditto. - This result assumes that the ferrous sulfate is anhydrous with a molecular weight of 152. This is incorrect, because the 300 mg weight is based on a chemical formula containing 7 waters of hydration. - This is the amount of anhydrous ferrous sulfate present in each 300 mg. The question asks for iron (Fe) only.

55. The infusion rate of theophylline established for a neonate is 0.08 mg/kg/hr. How many mg of drug are needed

for one daily bottle if the body weight is 16 lb. a. 0.58 mg. b. 14 mg. c. 30 mg. d. 150 mg. e. 8 mg.

The body weight will be = 16/2.2 = 7.27 kg Dose in mg = 0.08 x 24 x 7.27 = 13.96 mg

56. A patient’s serum cholesterol value is reported as 4 mM/L. This concentration expressed in terms of mg/dL will

be: a. 0.154 mg/dL. b. 1.54 mg/dL. c. 154 mg/dL. d. 596 mg/dL. e. 1540 mg/dL.

An increasing number of laboratory test values and drug doses are being reported in terms of millimoles (mM). Weight qualities expressed in molar amounts allow a more realistic evaluation of the actual number of drug molecules present, for example, when comparing salts of a drug. In this problem, the mM/L concentration is converted by realizing that 1 mole of cholesterol weighs 386 g and 4 mmoles equals 0.004 moles.

xmole

gmole 004.0

3861

= then gxx 54.11

004.0386==

x = 1.54 g in 1 L or 1450 mf/L and 154 mg/dL

57. A 250 ml infusion bottle contains 5.86 g of potassium chloride (KCl). How many milliequevalents (mEq) of KCL are present? (Mol. Wt. KCL = 74.6)

a. 12.7 mEq. b. 20 mEq. c. 78.5 mEq. d. 150 mEq. e. None of the above.

1 equivalent weight of KCL = 74.6 g 1 milliequivalent (mEq) = 74.6 mg

mgx

mgmEq

58606.741

= then mEqxx 5.786.74

15860==

or the problem may be solved by using the equation: mg of chemical = (mEq) (mol.wt.)/(valence)

Page 18 of 24

52. A pharmacist repackages 10 lb of an ointment into jars to be labeled 2 oz (avoir.). How many jars can be filled? a. 73. b. 80. c. 83. d. 88. e. 100.

Ten pounds contain 454 g x 10 = 4540 g. Two ounces (avoir.) consist of 28.4 g/oz, or 56.8 g. Thus, 4540 g divided by 56.8 = 80 jars.

53. The estimated Creatinine clearance rate for a 120 lb patient is 40 ml/min. What maintenance dose should be administrated if the normal maintenance dose is 2 mg/dl of body weight?


The normal maintenance dose would be = 120 lb x 2 mg/lb = 240 mg. Because the normal Creatinine clearance rate is 100 to 120 ml/min

mgml

xml

240min/100min/40

= then 96100

24040==

xx , or 100 mg.

Page 17 of 24

48. A radiopharmacist prepares a solution of 99mTC (40 mCi/ml) at 6:00 am. If the solution is intended for administration at 12:00 pm at a dose of 20 mCi, how many ml of the original solution are needed? The half-life of the radioisotope is 6 h.

a. 0.5 ml. b. 1.0 ml. c. 1.5 ml. d. 2.0 ml. e. 5.0 ml.

Because the time interval between preparation and administration is 6 h, and the half-life of the radiopharmaceutical is 6 h, approximately one-half of the original strength has decayed. Therefore, 1 ml of the solution now assaying at 20 mCi/ml is needed.

49. How much sodium chloride is needed to adjust the following prescription to Isotonicity? (E value for sodium thiosulfate is 0.31)

Rx Sodium thiosulfate 1.2% Sodium chloride Qs Purified water qs 100 ml

a. 0.37 g. b. 0.45 g. c. 0.53 g. d. 0.31 g. e. 0.90 g.

Step 1: Determine amount of sodium thiosulfate in the prescription: 100 ml x 1.2% = 1.2 g, or 1200 mg. Step 2: Multiply the amount of chemical by its “E” value: 1200 mg x 0.31 = 372 mg (equivalent amount of NaCl). Step 3: Determine amount of NaCl needed as if no other chemical was present: 100 ml x 0.9% = 900 mg. 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).

50. After one month of therapy, all of the patients listed had a systolic blood pressure reduction of 10 mm with a Standard Deviation (SD) of 5 mm. Blood pressure measurements were made 1 week on five patients with the following averages:

Patient 1 2 3 4 5

BP 140/70 160/84 180/88 190/90 150/70 What percentage of patients had a reduction between 5 and 15 mm?

a. 20%. b. 40%. c. 50% d. 70%. e. 90%.

A standard deviation is calculated mathematically for experimental data. It shows the dispersion of numbers around the mean (average value). On SD will include approximately 67% to 70% of all values, whereas 2 SDs will include approximately 97% to 98%.

51. 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 1.25 g/ml)

a. 1.2 ml. b. 18.2 ml. c. 22.7 ml. d. 24 ml. e. 28.4 ml.

A density or SG of 1.25 indicates that 1 ml of the liquid weighs 1.25 g. Because 1 lb of the ointment contains 5% w/w glycerin, 454 g x 5% w/w = 22.7 g of glycerin. Density = w/v, then v = w/d = 22.7/1.25 = 18.2 ml

Page 16 of 24

45. The adult intravenous (IV) dose of zidovudine is 2 mg/kg q4h six times daily. How many mg will a 180 lb patient receive daily?


First, convert the weight in pounds to kilograms: (1 kg = 2.2 lb)

lblb

xkg

1802.21

= then kgxx 8.812.2

1801==

Second, determine the total daily dose = 81.8 x 2 mg x 6 doses = 981.8 mg

46. Calculate the dose of a drug to be administered to a patient if the dosing regimen is listed as 2 mg/kg/day. The patient weighs 175 lb.


Because 1 kg = 2.2 lb

lblb

xkg

1752.21

= then kgxx 55.792.2

1751== Then total daily dose = 79.55 x 2 = 159.1 mg

47. What concentration of the original 99mTC solution described below will remain 24 h after its original preparation?

Solution of 99mTC (40 mCi/ml) a. 15 mCi. b. 10 mCi. c. 7.5 mCi. d. 5.0 mCi. e. 2.5 mCi.

The loss in first-order kinetics is a constant fraction of the immediate past concentration. In this example, the half-life of 6 h allows a quick comparison of the amount of radioactivity remaining.

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 mCi/ml

Page 15 of 24

"TEMPERATURE CONVERSION" All problems of conversion of Fahrenheit (F) to Centigrade (C) temperature may be solved by the following formula:

40. Convert 32oF to Centigrade? 9C = 5F – 160

= (5 x 32) – 160 = 0

∴C = 0oC ----------------------------------------------------------------------------------------------------------------------

41. 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. Any quantity of alcohol containing the equivalent of 1 gallon of 50% alcohol is a proof gallon. Therefore, 1/2 gallon of 200 proof (100%) alcohol is a proof gallon. Two gallons of 25% alcohol is 50% proof. The wine gallon is the common unit of volume measure.

43. 55 gallons of 45% alcohol contains how many proof gallons?

5.4950

455550

Pr ===xengthGallonxStroofGallon

44. 25 gallons of 70% alcohol contain how many proof gallons?

3550

7025Pr ==xoofGallons

9Co = 5F - 160

Page 14 of 24

"ALLIGATION" It is an arithmetic method of solving problems that involve the mixing of solutions, ointments, mixture of solids possessing different % strengths. Alligation medically used to calculate the % strength of a mix made by mixing two or more components of a given % strength.

38. What is the % v/v of alcohol in the following mixture, 1 Liter 60%, 3000ml of 40%, 1000 ml of 70%?

1000 X 60% = 60 000ml absolute alcohol. 3000 X 40% = 120 000ml absolute alcohol. 1000 X 70% = 70 000ml absolute alcohol. 5000 250 000

∴5000 ml contains 250,000 absolute alcohol.

100ml contains (x)

%50000,5

250100==∴

xx

39. What is the final % of Zinc Oxide ointment made by mixing ZnO ointment of the following strengths, 200mg of

10% + 50gm of 20% + 100gm of 5%?

200 X 10% = 20 00 50 X 20% = 10 00 100 X 5% = 5 00 350 35 00

∴350ml contains 35 00 absolute alcohol.

100ml contains (x)

%10350

35100==∴

xx

NB: Occasionally, you will run into a problem where the addition of a diluent or solvent is contained. In these cases, consider the volume of the diluent as having 0% concentration of the drug.

Page 13 of 24

36. How many ml of 1:5000 Potassium Permanganate solution can be made from 50ml of 0.5%?

50001

= 0.0002 = 0.02%

∴50 x 0.5% = (x) x 0.02%

mlxx 125002.0

5.050==∴

37. You receive the following prescription:

Ephedrine SO4 0.25% Rose Water ad 10ml

How many ml of 1:50 stock solution of ephedrine SO4 are necessary for dispensing? 1:50 = 2% ∴0.25% x 10 = 2% x (x)

mlxx 25.12

25.010==∴

Page 12 of 24

32. A floor nurse requests a 50 ml minibottle to contain heparin injection 100 u/ml. The number of ml of heparin injection 10,000 u/ml needed for this order will be:

a. 0.1 ml. b. 0.5 ml. c. 1 ml. d. 2.5 ml. e. 5 ml.

mlx

mlu

501100

= then x = 100 x 50 = 5000 u

xmlu 000,5

1000,10

= then x = 5,000/10,000 = 0.5 ml

33. If 20ml of a 1:200 w/v solution are diluted to 500ml, what is the final concentration?

1:200 = 5% ∴20 x 0.5% = (x) x 500

%02.0500

5.020==∴

xx

34. How many grams of pure hydrocortisone powder must be mixed with 60 g of 0.5% hydrocortisone cream if one

wishes to prepare a 2.0% w/w preparation? a. 0.90 g. b. 0.92 g. c. 0.30 g. d. 1.20 g. e. 1.53 g.

Because the amount of 0.5% hydrocortisone cream is exactly 60 g, the final weight of the cream will be greater when hydrocortisone powder is added. Therefore, the problem may be solved by allegation alternate method or by simple algebra.

partsx

partsg

5.19860

= then gx 92.098

605.1=

or, by algebra, let x = weight of 100% HC powder (x g) (100%) + (60 g) (0.5%) = (60 g + x g) (2%) x + 0.3 = 1.2 + 0.02 x x = 0.92 g

35. If a 65% w/v sugar solution is evaporated to 85% of its original volume, what will be its concentration?

Suppose the volume is 100ml ∴65% x 100 = (x) x 85

%47.768510065

==∴xx

100%

2%

1.5 Parts of 100%

0.5% 98 Parts of 0.5%

Page 11 of 24

"DILUTION & CONCENTRATION OF SOLUTIONS" There are two (2) important rules to follow: 1- When ratio strengths are given in the problem, convert to percentage % strength before beginning your

calculations. 2- Whenever the problem deals with proportional parts, reduce them to the lowest possible common

denominator (المقام).

30. If 500ml of a 15% v/v solution is diluted to 1500ml, what is the resulting % strength of the final solution?

%%15

500000,15

x=

%51500

15500==∴

xx

or Strength x Volume = Strength x Volume 15(%) x 500 = (x) x 1500

%515000

50015)( ==xx

31. A vial of lyophilized drug is labeled “10,000 units: to reconstitute, add 17 ml of Sterile Water for Injection to

obtain 500 units per ml.” How many ml of SWFI must a pharmacist add if a 1000 u/ml concentration is needed by the nurse?

a. 7 ml. b. 8.5 ml. c. 10 ml. d. 17 ml. e. 20 ml.

When some drug powders, especially bulky antibiotics, are reconstituted, the volume occupied by the bulk powder once it has dissolved must be considered. In this example, the final volume of solution is:

xu

mlu 000,10

1500

= then x = 20 ml.

The previously listed volume means that the volume occupied by the bulk powder must have been 20 – 17 ml = 3 ml. When a concentration of 1000 u/ml is desired, the total volume of solution that must be prepared will be:

xu

mlu 000,10

11000

= then x = 10 ml.

Therefore, the pharmacist must add 7 ml of SWFI to the vial. When the powder has dissolved, the resulting volume will be 10 ml.

Page 10 of 24

xmlg 1

104625.5

= Then 24.19925.5

11046==

xx

28. A solution contains 1.5 mEq of calcium per 100 ml. Express the solution’s strength of cacium in terms of mg/L (The atomic weight of calcium is 40)

a. 30 mg/L. b. 60 mg/L. c. 150 mg/L. d. 300 mg/L. e. 600 mg/L

Because the valence of calcium is +2, 1 mEq equals 40 mg divided by 2 = 2- mg. Therefore, 1.5 mEq = 30 mg. If using the equation: Mg of chemical = (mEq) (mol. Wt.)/(valence) = (1.5 x 40)/(2) = 30 mg/dL or 300 mg/L

29. How much of Gentian Violet is used to make 500ml of 1:10000 solution? a. 10 mg. b. 20 mg. c. 30 mg. d. 50 mg. e. 500 mg.

500100001 x

=

gmxx 05.010000

5001==∴ , and that is equal to 50mg.

Page 9 of 24

"RATIO SOLUTIONS" For solids in liquid Gram/1000ml of mixture.

For liquids in liquids Ml/1000ml of mixture. For solids in solids Grams/1000grams of mixture or

Grains/1000grains of mixture 22. Express 0.02% as ratio strength?

)()(1

10002.0

partsxpart

=

50001

100002

=

= 1:5000

23. Express 1:4000 as a % concentration?

10040001 x

=

Therefore x= %025.04000

1100=

x

24. A prescription calls for a solution of drug equal to 2 mg/ml. 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

xmlml

gmgm 1

1002.0

= then 500002.01

==x

25. 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. Which, is equal to 0.25gm

20001

10005.0

50025.0

==∴ (=0.05%)

∴The ratio is 1:2000

26. How many grains of Potassium permanganate could be used to make 500ml of 1:2500 solution?

1 gm in 2500ml (x) gm in 500ml

mlgmx

mlgm

500)(

2500`1

=

2.02500

5001==

xx gm

Then change grams into grains according to 1gm=15.43grain = 0.2gm x 15.43 = 3.086 grains

27. A pharmacist dilutes 100 ml of Clorox with 1 quart of water. Express the concentration of sodium hypochlorite in the final solution as w/v ratio. Commercial Clorox contains 5.25% w/v sodium hypochlorite.

a. 1/9. b. 1/10. c. 1/100. d. 1/180. e. 1/200.

One hundred ml of Clorox contains 5.25 of sodium hypochlorite. The final dilution will be 100 + 946 ml of water for a total of 1046 ml. The ratio strength will be:

Page 8 of 24

= 4.76 + Log 16.024.0

= 4.76 + 0.176

= 4.94 And since pH before addition of NaOH was 4.76, then the change in pH is going to be:

pH = 4.94 – 4.76 = 0.18 ----------------------------------------------------------------------------------------------------------------------

21. For the curve equation where y is function of x, y = 12x – 3x2. What is the slope if; x = 1, y = 9? a. 3. b. 6. c. 7. d. 9.

Page 7 of 24

17. A hospital clinic requests 2lb of 2% hydrocortisone ointment. How many grams of 5% hydrocortisone ointment would be diluted with white petrolatum to prepare this order?

a. 18.2 g. b. 27.5 g. c. 45.4 g. d. 363.2 g. e. 545 g.

We have first to convert 2lb into grams = 2 x 454 = 908gm

908gm X 2% = Xgm X 5%

Xgm = %5

%2908x = 363.2gm

18. Find the pH of a solution which has H+ concentration of 6 x 10-4

a. 32.218 b. 322.18 c. 3221.8 d. 3.2218

6 x 10-4 = 0.0006, pH = -Log 6 x 10-4 = - (Log 6 + Log 10-4) = - (0.7782 + (-4.0)) = - (-3.2218) = + 3.2218

19. PH of a certain solution is 11.1; calculate H+ concentration of the solution?

PH = -Log H

Log H = - 11.1 = 7.94 x 10-12

20. Calculate the change in pH upon adding 0.04 mole of NaOH to a liter of a buffer solution of 0.2M concentration

of sodium acetate and acetic acid. Pka value of acetic acid is 4.76 at 25oC? a. 0.16 b. 0.17 c. 0.18 d. 0.19

PH = Pka + Log acidsalt

= 4.76 + Log 2.02.0

= 4.76 + Log 1

= 4.76 The addition of 0.04M of NaOH converts 0.04M of acetic acid to 0.2M concentration of sodium acetate, 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 baseSaltbaseSalt

−+

= 4.76 + Log 04.02.004.02.0

−+

Page 6 of 24

20 ml → x mg

mlx

mlmg

20120200

= then 3.33120

20020==

xx

15. 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. 160 ml. b. 200 ml. c. 400 ml. d. 600 ml. e. 800 ml.

One tablespoonful (tbsp) delivers 15 ml of liquid. In this prescription, the patient is receiving four doses per day for 10 days. Therefore, 15 ml x 4 x 10 days = 600 ml total.

16. How many capsules of 250mg of a drug should be used to prepare 100ml solution which has a concentration of 250mg/5ml?

a. 20 capsules. b. 15 capsules. c. 10 capsules. d. 5 capsules.

250mg (One capsule) -------------------- 5ml of solution ?? -------------------- 100ml of solution

Number of capsules = ml

capmlx5

1100 = 20 Capsules.

Page 5 of 24

9. A 10ml vial of concentrated stock solution for electrolyte replacement therapy contains 55g of calcium chloride. The amount of calcium in each ml is:

a. 25 mmole. b. 50 mmole. c. 100 mmole. d. 500 mmole.

10. Five subjects given a single IV dose of a drug have the following elimination half lives: 3, 9, 6, 5, and 4 hours.

The mean half-life is: a. 4.0 h. b. 5.0 h. c. 5.4 h. d. 5.8 h. e. 6.0 h.

Mean half-life= (3+9+6+5+4)/5 = 5.4

11. Solution of a substance is 1:10000, what is the percentage w/v in mg? a. 5mg% b. 10mg% c. 15mg% d. 100mg%

12. Ampicillin is mixed in a 5% dextrose bag. The degradation rate of ampicillin is 12mh/hr. What is T90 of the

ampoule? 13. Prepare 30ml solution of 2.5% Clindamycin using 150mg Clindamycin capsules. How many capsules are needed?

a. 3 Capsules. b. 5 Capsules. c. 7 Capsules. d. 10 Capsules.

2.5g in 100ml is equal to 2500mg in 100ml

150 mg → 1 Capsule 2500 mg → X

X = mgcapsulemgx

15012500

= 16.67 Capsules

16.67 Capsules → 100ml

Y → 30ml

Y = ml

mlcapx100

3067.16 = 5 Capsules.

14. 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.s. 120 ml Sig: 3 I t.i.d. p.c. & h.s.

a. 6.25 mg. b. 8.25 mg. c. 19 mg. d. 25 mg. e. 33 mg.

In today’s health practice, the symbol “i” is used to represent a 1 teaspoon dose. The symbol’s original meaning as a drachm (weight) or fluidrachm (volume) quantities is archaic and should not be used. Because a standard teaspoon is considered to be 5 ml, the patient in this prescription is receiving four daily doses for a total of 20 ml. 120 ml → 200 mg codeine phosphate

Page 4 of 24

7. How many ml of 2% iodine solution must be mixed with a 7% iodine solution to obtain 1 L of 5% strength? a. 200 ml. b. 300 ml. c. 400 ml. d. 500 ml. e. 600 ml.

This problem can be solved by the allegation alternate or simple parts method.

Thus, the final solution will contain 2 parts of 2% iodine for every 3 parts of 7% iodine.

mlparts

xparts

100052

=

x = mlx 4005

10002= of 2% iodine solution.

8. How many grams of pure hydrocortisone powder must be mixed with 60 g of 0.5% hydrocortisone cream if one

wishes to prepare a 2.0% w/w preparation? a. 0.90 g. b. 0.92 g. c. 0.30 g. d. 1.20 g. e. 1.53 g.

Because the amount of 0.5% hydrocortisone cream is exactly 60 g, the final weight of the cream will be greater when hydrocortisone powder is added. Therefore, the problem may be solved by allegation alternate method or by simple algebra.

partsx

partsg

5.19860

= then gx 92.098

605.1=

or, by algebra, let x = weight of 100% HC powder (x g) (100%) + (60 g) (0.5%) = (60 g + x g) (2%) x + 0.3 = 1.2 + 0.02 x x = 0.92 g

100%

2%

1.5 Parts of 100%

0.5% 98 Parts of 0.5%

7%

5%

3 Parts of 7%

2% 2 Parts of 2%

Page 3 of 24

NB. This system can be used to determine the relative concentration of three, four, or more different strengths required to prepare another requested strength.

5. A pharmacist wishes to prepare a 10% ointment of drug. He has some 50%, 20%, and 5% in stock. How should he mix to make the 10% product?

The total = 5 parts of 50% + 5 parts of 20% + 50 part of 5%

6. In what proportion must the following strengths be mixed to obtain a 10% mixture, 20%, 15%, 5%, and 3%?

Proportions are 7 parts of 20% + 5 parts of 15% + 5 parts of 5% + 10 parts of 3%

20%

10%

7 Parts of 20%

3% 10 Parts of 3%

15%

5%

5 Parts of 15%

5 Parts of 5%

20%

10%

5 Parts of 20%

5% 10 Parts of 5%

50%

10%

5 Parts of 50%

5% 40 Parts of 5%

Page 2 of 24

Low alcohol = 68

46473x = 320ml.

High alcohol = 68

22473x = 153ml.

4. In what proportion should 15% Boric acid solution be mixed with white petrolatum to produce 2% boric acid

ointment?

The final proportion is 2:13

15%

2%

2 Parts of Boric acid

0% 13 Part of petrolatum

78%

32%

22 Parts

10% 46 Parts

Page 1 of 24

"PHARMACEUTICAL CALCULATIONS"

"ALLIGATION ALTERNATE" To find out the relative amounts of solute or other substances of different strengths, you must use the X to make a mixture of a required strength.

1. A formula for a cosmetic cream requires 5g of an emulsifying blend consisting of Span 80 and Tween 80. If the

required HLB (Hydrophilic Lipophilic Balance) is 10.5, how many grams of each emulsifying agent should be used in preparing the emulsion? (HLB of span 80 is = 4.3 and Tween 80 is 15)

Span 80 4.3 4.5 parts of span 80 10.5

Tween 80 15 6.2 parts of tween 80

Span 80 = 5.105.45x

= 2.14 g.

Tween 80 = 5.102.65x

= 2.95 g

2. Which proportion of 95% alcohol and 50% alcohol should be used to make a solution of 70% alcohol?

The final proportions are 20:25 or (4:5 95%:50%)

3. A physician requested that an elixir containing 32% alcohol be prepared, how much low alcohol elixir (10%) and high alcohol elixir (78%) must be mixed to prepare 1 pint (473ml) of the requested elixir?

95%

70%

20 Part

50% 25 Part

High Concentration

ingredient

Desired concentration

Part of high concentration

ingredient needed.

Low concentration

ingredient.

Part of low concentration

ingredient needed.

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PEBC-Calculation Questions