Pharmacy Exam Review

Calculation

Last updated on: Oct 5th, 2020
Abbreviation

Here I am not going to list the very common ones like BID, TID, BIW (twice a week), as you should've already learned from pharm school, so I am not going to waste your precious time reading through those. =)

  • AC: before meals
  • PC: after meals
  • AU: both ears (AD= right ear, AS = left ear),A for auditory
  • OU: both eyes (OD= righteye,  OS = left eye), O for ocular
  • NR: no refill
  • C or w/ : with
  • S or w/o: without
  • Top: topically
  • Wa: while awake
  • Stat: immediately
  • Sup/supp: suppository
 
Common Units
  • 1 tsp = 5 ml, 1 tbsp = 15 ml
  • 1 fl oz = 30ml, 1 oz = 28.4 g ***some exam uses 30g, no need to be precise
  • 1 lb = 16 oz = 454 g; 1 kg = 2 lbs
  • 1 inch = 2.54 cm
  • 1 cup = 8 oz (240ml), 1 pint = 473ml, 1 quart = 2 pints, 1 gallon = 4 quarts
  • Start with 1 cup → 1 pint → 1 quart → 1 gallon. Double cup to get pint, double up to get quart, lastly, multiply by 4 to get gallon
  • 1 grain (gr) = 65 mg
  • Specific gravity (SG) = weight (g) /volume (mL), think about density! Anything floats on water would have <1 SG, anything that sinks into the bottom would have SG > 1.
  • Body mass index (BMI) is a measure of body fat based on height and weight = kg/m2

 

Less frequent units
  • F to C: C = (F- 32)/1.8, C to F: F= (CX1.8) +32
  • BSA () = Square Root of [(Height in cmx Weight in kg)/ 3600]
  • Example: Calculate the BSA of AB who is 5'8" and weighs 190 lbs.
    • Height: 68 inches x 2.54 = 172.72 cm; Weight: 190 lbs / 2.2 = 86.36 kg
    • BSA (m²) = Square Root of [(172.72 cmx 86.36 kg)/ 3600] = 2.03 m²
 

Concentrations

  • The body regulates concentrations of solutes in order to maintain homeostasis, all calculations are essentially concentration manifested in different forms: whether it is as obvious as mass percentage (0.9 %), ratio (e.g: 1:20), parts per million (ppm), or in ways we were not used to see but frequently shows up in TPN and lab values (mEq, mOsmol per L), electrolytes are presented in tiny amounts and is impossible to measure their weight directly as mg or g, they dissociate in solvent (aqueous body environment) and become cations (Na+, K+, Ca2+) or anions (PO43- , Cl-) that are more reasonably expressed as mEq/L, or mOsmol/L; Or let it be to calculate infusion rate, insulin dose conversion, allegations, antibiotic reconstitution errors.
  • Percentage: % w/v = g/100ml; % v/v = ml/100ml; % w/w = g/100g
    • The % stays on the top is either gram (for weight) or ml (for liquid). The bottom is always 100 (either ml for volume, or g for weight)
    • Know the concentration strength of most common fluids. E.g: NS = 0.9% = 0.9 g/ 100 mL; D5W = Dextrose 5% = 5 g/ 100 mL. 
  • PPM (parts per million) = 1mg/1L, think as divide 1 liter into a million parts, and take one part.

 

The confusing ones: mEq, mOsmol/L, normality
  • Milliequivalents (mEq): the amount of solute’s gram equivalent
    • Think about mg, but since valence matters, we take into account of the number of ionic charges. Valence charge examples: Na bicarb = 1, K gluconate = 1, Mg sulfate = 2.
    • mEq = (mg /MW) x valence
    • Or = mmol x valence; mmol (millimole) = mg/MW
    • Monovalent species, mEq = mmol because the multiply factor would be 1 (Na, K: 1 mEq = 1 mmol); but for divalent species such as Ca: 1 mEq = 2 mmols.
  • Osmolarity (Osmol or mOsmol per L): measures # of particles per liter.
    • # of particles are proportional to its osmotic pressure, the more # of particles, the higher the osmolarity. The particles are usually measured in milliosmoles.
    • Solutes can be either ionic (such as NaCl, which dissociates into 2 solutes), or non-ionic (do not dissociate, such as glucose, urea).
    • Osmol = (g / MW) x number of particles
    • mOsmol = (mg / MW) x number of particles
    • g: CaCl2 dissociates into 3 particles, 10g of CaCl2 has 10/147 x 3 = 0.204 Osmole.
    • mOsmol/L = (substance weight g/L / MW g/mol) x number of particles x 1000
  • Practice: Calculate mmol, mEq, and mOsmol of 10g CaCl2 (MW = 147) in a 1 L container.
    • mol or mmol = 10g/147 = 0.068 mol or 68 mmol.
    • mEq = 10000mg/147 x 2 = 136 mEq, or = 68mmol x 2 = 136 mEq
    • mOsmol = 10000mg/147 x 3 = 204 mOsmol or = 68mmol x 3 = 204 mOsmol
  • Very often, the calculations are to convert between mg (the easily manifested weight) with mEq, or mOsmol using molecular weight (the ones that many dread).
  • Less frequently, a notion like normality (N) = Equivalent weight (molecular weight) in g / L.
    • Gram equivalent weights of solute in 1 L solution. e.g: 1N of sodium bicarb means 84 g in 1000 ml solution.

 

Cockcroft-Gault equation
  • Commonly used to calculate creatinine clearance (mL/min).Creatinine clearance (CrCl) is often used as an estimate of GFR.
  • Creatinine Clearance= [[140 - age] * weight] / [72 * serum Cr(mg/dL)] (multiply by 0.85 for women) **** Must memorize!
  • Use actual body weight (TBW) for underweight patients; ideal body weight and an adjusted body weight for normal-weighted patients.
  • Ideal body weight (IBW): Some drugs with a narrow therapeutic index (aminophylline, theophylline, acyclovir) are dosed based on IBW to avoid toxicity.
    • Male: IBW (kg) = 50 kg + (2.3 kg x height in inches over 60 inches)
    • Female: IBW (kg) = 45.5 kg + (2.3 kg x height in inches over 60 inches)
  • Adjusted Body Weight (AdjBW): for overweight or obese patients; Use for aminoglycoside
    • AdjBW = IBW + 0.4(TBW — IBW)

 

Buffer and ionization

  • This is a low yield topic, but just in case they appear on the exam, have a general idea of what buffer system is and its purpose, what ionized and unionized mean (it helps to determine whether a small or large percentage of drugs ionized).
  • Buffer systems are usually composed of a weak acid and salt of the acid (acetic acid and Na acetate), or weak base and salt of the base. Buffer system minimizes pH fluctuation and limits the harm.
  • An acid releases (donates) protons, a base accepts (binds) the protons (e.g: NH3 picks up a proton becomes NH4+).
  • Henderson-Hasselbalch equation:
    • pH = pKa + log [conjugated base/acid]. (pKa = pKw – pKb)
  • Ionized Vs. unionized: ionized drugs are soluble but cannot readily cross lipid membranes, unionized drugs are not soluble but can easily cross membranes, most drugs are weak acids (they are soluble and can pick up a proton to cross the lipid membrane).
  • pKa refers to the acid loses the proton to give to the base (pKa > pKb: acid; pKb > pKa: base). Strong acid or base are 100% dissociation, weak acid or base means very little dissociation.
    • pH = Pka: 50% is ionized (not protonated) and 50% unionized (protonated).
    • pH > pKa: more acid is ionized; more conjugate base is unionized.
    • pH < pKa: more acid is unionized; more conjugate base is ionized.
  • percentage of ionization:
    • % of weak acid = 100/ [1+ 10 ^ (pKa-pH)]
    • % of weak base = 100/ [1+ 10 ^ (pH - pKa)]
  • Practices:

Calculate the % ionization of sodium pentothal (pKa = 7.4) at a pH of 6.4.

  • Weak acid formula = 100/ [1+ 10 ^ (pKa-pH)] = 100/ [ 1+10^1] = 9.09% ionized. (in this case pH < pKa: more acid is unionized).

Calculate the % unionization of ASA (pKa = 3.4) at a pH of 4.4.

  • Weak acid formula = 100/ [1+ 10^ (-1)] = 90.9% of ionization
  • % unionization = 1 – 90.9% = 9.1% unionized. (in this case pH > pKa: more acid is ionized).

Calculate the % ionization of amitriptyline (pKa = 9.4) at a pH of 7.4.

  • Weak base formula = 100/ [1+ 10^ (-2)] = 100/1.01 = 99% ionized. (in this case, pKa >> pH, more conjugate base is ionized).

 

Alligation method

Why Alligation? To dilute an existing known strength to achieve the finished product (instead of prepare a new one which may involve weighing, heating, and extensive mixing).

When can it be used?

1) To calculate the amount of diluent to be added to a known higher strength preparation to form a lower strength.

2) For mixing two products of different strengths to form a product with a desired intermediate strength.

 

Practice 1

You have on hand a70% alcoholic elixir and a 20%alcoholic elixir. The prescription calls for a 30% alcoholic elixir. In what proportion must the 70% elixir and the 20% elixir be combined to make a 30% elixir?

  1. A. 2 parts 70% / 1 part 20%
     B. 3 parts 70% / 4 parts 20%
     C. 1 part 70% / 4 parts 20%
     D. 4 parts 70% / 2 parts 20%
1

Answer C. with one part of 70% elixir is mixed with four parts of 20% elixir, it will yield five parts of 30% elixir. 

 

Practice 2

How many mL of water must be added to 300 mL of 70% alcohol solution to make a 40% alcohol solution?

  • A. 225mL
    B. 300mL
     C. 180mL
     D. 275mL
  • 2

    Answer: A. 40% is the desired strength and must be placed in the center of the matrix. The next step is to see if a higher or lower strength is given. 70% is a higher strength and must be placed in the upper left-hand corner. If no lower strength is given, it can be assumed to be 0%.

     
    Practice 3

    A prescription calls for an elixir that contains 45% alcohol. On hand, you have 10% alcoholic elixir and a 75% alcoholic elixir. How many mL of the 75% elixir must be combined to make 1000 mL of 45% alcoholic elixir?

    1. 538mL
       B. 355mL
       C. 500mL
       D. 462mL
    3

    Answer A. 1) To solve for the higher 75%, place the X on the same level as 75%. The only other known factor is that 1000 mL of the 45% must be prepared. The 1000 mL is  placed on the same level as the 45%. The arrow indicates that the bottom line is the 45% line, solve for X first. In this case we just found the volume of the 75% elixir to be 538.46 mL. What's the volume of 10% elixir to be added? Subtract this figure from the final volume of 1000 mL to get 461.54 mL.

     
    Practice 4

    How many cc of 75 % alcohol should mix with 10 % of 1000cc alcohol to prepare 30% of 500cc alcohol solution? 
    a. 346.16 cc 
    b. 234.43 cc 
    c. 153.84 cc
    d. 121.12 cc

    Answer: (c) To solve this type of problem, use alligation method.
    75 20 (75%)  
       30
    10 45 (10%)  1000cc
    Total parts 65 (30%)
    To prepare 65 (30%), 20 parts (75%) needed
    To prepare 500 (30%) ? = (500 /65) x 20 = 153.84cc (75%) alcohol

    If we mixed 153.84 cc of 75% alcohol with 346.16cc [500cc - 153.84] of 10% alcohol, then we can get 500cc of 30% alcohol solution.

     

    Calculation strategies

    Keep practicing, even if one question still takes you15 mins to solve (maybe on the exam you only have 3 mins), as long as you have comprehended the problems, figured out what info is needed and what is missing, they will become more habitual and intuitive! And in the future, you will develop clear visualization of the steps needed 1--2--3.

    1. Remember almost all questions evolve around concentration (%), and most of them are weight/volume such as ?ml of solution contains ?g of sucrose (much less frequently v/v or w/w, and would be obvious to and specified out for you).
    1. Giveaway: given patient weight and administering rate (x mcg/kg/min), should prompt you to calculate dose/time, results in weight/min (e.g: 60kg x 2 mcg/kg/min). The unit is simplified: mcg/kg/min ---- mcg/min. We always want to simplify and unite our units.
    1. Infusion problems: infusion rates are either weight/time (g/min, mg/min, drops/min) or volume/time (ml/min, ml/hr). Figure out the time should be no brainer (min - hr or hr to min), so it means you only want to find out the weight or volume. If you tackle down the problem to mg, converting to ml is just one step away using concentration! Conversely, if you boil down to ml, you can convert to mg with ease. 

     

    Quiz

    1. What is the ratio strength (w/v) of 60mL containing a 1:20 (w/v) ammonia solution diluted to 1 L?

     

    2. How much Lidocaine is required to prepare 1 : 1000, 30 cc solution of Lidocaine? 
    a. 10 mg 
    b. 0.03 mg 
    c. 30 mg
    d. 300 mg

     

    3. If 60 g of 1% hydrocortisone is mixed with 80 gm of 2.5% of hydrocortisone, what is the % of hydrocortisone in final mixture? 
    a. 2.2 % w/w 
    b. 1.85 % w/w 
    c. 0.25 % w/w 
    d. 1.75 % w/w

     

    4. If 1000 tablets of Risperdal 1 mg cost $ 2250 and % mark up on prescription is 20, what would be the retail price of 30 tablets? 
    a. $ 150 
    b. $ 17 
    c. $ 500 
    d. $ 81

     

    5. If the ratio of ionized to unionized species of drug is 10^3 and PKa = 2.2, what is the PH of the solution? 
    a. 2.2 
    b. 0.8 
    c. 5.2 
    d. 3.0

     

    6. If dropper is calibrated to deliver 325 mg of iron sulfate in 0.6 cc and adult dose of drug is 325 mg, what is the dose of a drug in cc for a 15 months-old infant? 
    a. 1.2 cc 
    b. 0.3 cc 
    c. 0.06 cc
    d. 0.01 cc

     

    7. If the dose of a drug is 10 mg/kg/day, how many 250 mg/100cc ready infusion-bags require to fill above order? Patient's weight is 156 lbs. 
    a. 1 bag 
    b. 2 bags 
    c. 3 bags 
    d. 5 bags

     

    8.How much of atropine is required to prepare 240cc in such way that when 1 teaspoonful of the solution is diluted to 1 pint gives 1 in 500 solution? 
    a. 2.25 gm 
    b. 46.08 gm 
    c. 35.15 gm 
    d. 25.35 gm

     

    9. How much sodium bicarbonate powder is required to prepare 240cc of 0.10 N solution of sodium bicarbonate?

    a. 1.35 g 
    b. 3.25 g 
    c. 4.81 g 
    d. 2.016 g

    10. How many mEq of Na+ are present in 0.9% 250cc normal saline solution? (Na = 23, Cl = 35.5)
    a. 23.12 mEq
    b. 15.17 mEq
    c. 53.15 mEq
    d. 38.46 mEq

     

    11. If the probability of success in Null hypothesis is 0.6, what is the probability of failure? 
    a. 0.3 
    b. 0.9 
    c. 0.6 
    d. 0.4

     

    12. How many grams of cocabutter are required to dispense 12 suppositories of tannic acid each weighing 2 g and contain 400 mg of tannic acid? 
    a. 21.23 gm 
    b. 18.67 gm 
    c. 14.12 gm 
    d. 13.25 gm

     

    13. What is the "mode" or "median" of the following values?
    190, 120, 135, 140, 118, 175, 105, 115
    a. 135 
    b. 118 
    c. 127.5 
    d. 175

     

    14. If the concentration of reactant M is half in a reaction that is third order in M, by what factor will rate of reaction change? 
    a. 1/8 times 
    b. 1/4 times 
    c. 8 times 
    d. 4 times

     

    15. What is the rate of constant after 90 minutes if the initial concentration of drug is 500mg/cc and 50mg/cc after 90 minutes? (First Order Kinetic) 
    a. 0.051 min-1 
    b. 0.025 min-1 
    c. 0.35 min-1 
    d. 0.86 min-1

     

    16. If the total body clearance of the patient is 2100cc/hr and hepatic clearance is 300cc /hr, what is the status of renal function in a patient? 
    a. Excellent 
    b. Normal 
    c. Moderately impaired 
    d. Severely impaired

     

    17. How many grams of sodium chloride are required to prepare 250cc of 1% boric acid solution to isotonic with eye tears? 
    a. 250 mg 
    b. 325 mg 
    c. 991 mg 
    d. 1221 mg

     

    18. Which of the following correctly describes the use of Tween (polysorbate)? (Select ALL that apply.)

    A. It is used commonly as a thickening agent.

    B. Tween is a surfactant.

    C. Surfactants are compounds with polar sides and nonpolar sides.

    D. Surfactants are used as wetting agents.

    E. Surfactants are used often in ointment compounds.

     

    19. Interpret the following codes taken from prescriptions:

    ·       Propranolol HCl 10 mg po tid ac & hs

    ·       1 tsp q6h × 10 days

    ·       Flurazepam 30 mg at hs prn sleep

    ·       Caps i tid pc

     

    20. An injection for dental anesthesia contains 4% (w/v) of prilocaine hydrochloride and 1:200,000 (w/v) of epinephrine. Express the concentration of prilocaine as a ratio strength and the concentration of epinephrine as a percentage.

     

    21. Norgestrel and ethinyl estradiol tablets are available containing 0.25 mg of norgestrel and 75 μg of ethinyl estradiol. If a manufacturer wants to make 15,000 tablets, how many grams of each ingredient would be needed?

     

    22. A prescription balance has a sensitivity requirement of 6 mg. Explain how you would weigh 20 mg of acetaminophen with an error not more than 5%.

     

    23. A pharmacy technician attempts to weigh 350 mg of morphine sulfate on a balance of unknown accuracy. When the pharmacist checked the weight on a highly accurate balance, the actual weight is discovered to be 375 mg. Calculate the percentage error of the inaccurate balance.

     

    24. An injection contains 60 μg of drug A, 0.05 mg of drug B, and 8.18 mg of drug C in each milliliter. Calculate the milligrams percent (w/v) of drug A and the percentage (w/v) of drug B in the injection.

     

    25. What is the volume, in ml, of 3 lb of mineral oil with a specific gravity of 0.85?

     

    26. A vial contains 10 g of a powdered drug for reconstitution prior to use in an IV infusion. The label states that when 18.5 mL of diluent is added, the concentration of the resulting solution is 500 mg/mL. A medication order calls for a drug concentration of 300 mg/mL. How many milliliters of diluent should be added to the vial?

     

    27. How many grams of a 4% w/w hydrocortisone ointment must be mixed with a 0.5% w/w hydrocortisone ointment to achieve 30 g of an ointment with 2.5% strength?

     

    28. A pharmacist added 6 g of salicylic acid to 30 g of an ointment containing 4% salicylic acid, what would be the final concentration of the resulting mixture?

     

    29. If 500 mL of a 35% w/v solution is diluted to 1.75 L, what would be the resulting percentage strength?

     

    30. The phenytoin loading dose in children is 20 mg/kg infused at a rate of 0.5 mg/kg/min. Over how many minutes should the dose be administered to a 50-lb child?

     

    31. A doctor ordered 3 g of Cefotetan to be added to 100ml NS for a 58 y/o female with an anaerobic infection. Using a reconstituted injection containing 154mg/ml, how many ml should be added to prepare that order?

     

    32. JH is a 48 y/o male hospitalized for a knee replacement surgery. He was given unfractionated heparin and unfortunately developed heparin-induced thrombocytopenia. Argatroban was ordered at a dose of 2mcg/kg/min. The pharmacist mixes 100mg argatroban in 250ml of D5W. JH weighs 188 lbs. At what rate (ml/hr) should the nurse infuse argatroban to provide the desired dose? Round to the nearest whole number.

    33. A patient is scheduled to receive Tygacil dosed at 5mg/kg. Pt weights 140 lbs. The nurse labels Tygacil 500mg/50ml NS. The drug will be infused over 30 mins. The IV tubing unit is set to deliver 15 drops per ml. What is the correct rate of flow in drops per min? Round to the nearest drop.

     

    34. A 163 lb patient is to receive 250ml dopamine drop at a rate of 17 mcg/kg/min. The pharmacist has dopamine premixed in 3.2mg/ml of D5W. What is the infusion rate in ml/min? Round to the nearest tenth.

     

    35. A patient on clozapine visits the clinic for a routine blood work. Lab shows WBC = 4,500 cells/mm3 with 45% segs and 5% bands. Calculate the patient’s ANC.

     

    36. A patient receives D51/2 NS with KCl at 20 drops/min. After 10 hr, the pt has received a total of 40mEq of KCl using tubing delivers 15 drops/ml. What’s the % concentration of KCl in pt’s IV? Round to 2 decimal places (MW of K = 39, WM of Cl = 36)

     

    37. An order writes “Begin dopamine drop at 3mcg/kg/min, titrate by 5mcg/kg/min every 5 minds to achieve SBP > 100mmHg. Page the doctor for additional orders if max dose of 20 mcg/kg/min is reached”. Pt weighs 165 lbs and ICU stocks premixed dopamine drips (400mg/250ml) in the automated dispensing cabinet. Calculate the rate in ml/hr for the dopamine drip be started. Round to the nearest whole number.

     

    38. The patient from the above question is started on dopamine drip. Later on, the pharmacist checks on the pt and notes the drip is running at 70ml/hr but the patient’s SBP is still < 100 mmHg. According to the original order, should the doctor be paged?

     

    Osmolite Nutrition facts

    Serving size 8 fl oz (237ml)

    Calories 275 cal

    Total fat: 17.8g

    Protein 13.2g

    Total carbohydrate 37.4g

     39. From the above label, how many calories will someone receive from a 7.4% fat component of 1 (8 fl oz) shake? Round to the nearest whole number.

     

    40. According to the nutrition facts label above, how many calories will John receive from the carbohydrate component in 4 fl oz? Round to the nearest whole number.

     

    41. A nurse administered 1 can of Osmolite to John when she accidentally spilled 2 fl oz onto the floor. The remaining amount was administered to John. How many calories did he actually receive? Do not round the answer.

     

    42. The pharmacist is asked to convert Mrs. Gullivan’s lab values to different units so her case can be compared to a published report. Her serum K level is 4.5 mEq/L, convert that to mg/dL (MW of K = 39)

     

    43. A pharmacist reconstitutes an Augmentin suspension to provide 2000mg daily dose for a child for ear infection. The dose will be divided BID. To prepare 250mg/5ml Augmentin suspension, pharmacist needs to add 86ml water for a final volume of 100ml. the pharmacist adds ½ of water first, shake vigorously and add the remaining half. But he mistakenly adds too much water, the final volume is 110ml. The pharmacy has no other bottles of Augmentin, so the pharmacist has to dispense the bottle with extra water added. How many ml should the patient take BID to receive the correct dose? Round to the nearest whole number.

     

    1. 60mL x 1/20 = 3g
      3g/1000mL = 3: 1000

     

    1. (c) 1: 1000 means 1 g in 1000cc solution. The amount of lidocaine in 30cc of 1:1000 solution = 30 x 1/1000 = 0.03 g = 30 mg.

     

    1. (b) Amount of Hydrocortisone in 60 gm, 1% = 60/100 = 0.6 g.
      Amount of hydrocortisone in 80 gm, 2.5% = 80 x 2.5/100 = 2 g
      % amount of hydrocortisone in final mixture = 100 x 2.6 (2g + 0.6g)/140 (80g + 60g) = 1.85% w/w.

     

    1. (d) 1000 tablets of Risperdal 1mg cost $ 2250. The % mark up on prescription is 20%.
      Retail price of 1000 tablets = 120 x 2250/100 = $ 2700
    2. For each $ 100 cost = $120 retail cost
      Price for 30 tabs = 30 x 2700/1000 = $ 81

      5: (c) pH = pKa + log ionize/unionize = 2.2 + log 10^3 = 2.2 + 3 = 5.2

     

    1. (c) 0.06cc. According to Fried’s rule: = age in months/150 x adult dose = 15 x 325/150 = 32.5 mg
      The dropper is calibrated to deliver 325 mg of Iron sulfate in 0.6 cc = 0.6 x 32.5/325 = 0.06cc

     

    1. (c) Patient weight is 156 lbs, therefore weight in Kg would be 156/2.2 = 70.9 kg
      A normal therapeutically recommended dose of drug is 10mg/kg/day, therefore dose in above patient = 10 x 70.9 = 709 mg
      Each ready-infusion-bag contains 250 mg drug, so number of bags require to fill order = 709/250 = 2.83 = 3 bags.

     

    1. (b) To solve this kind of problem, we must first find out the amount of drug present in final solution. Amount of atropine in 1 pint, 1 in 500 sol’n = 480 x 1 /500= 0.96 g of atropine.
      Now, 0.96 g of drug must be present in 1 tsp of drug solution, therefore we can say: 5mL (1 teaspoonful) contains 0.96 g of drug.
      240mL solution requires ?= 240 x 0.96/5 = 46.08 g atropine.

     

    1. (d) Gram equivalent weights of solute in 1 L solution is defined as normality, therefore 1N solution of sodium bicarb contains 84 g in 1000 ml. Find quantity of sodium bicarbonate in 240cc, 0.1 N solution: 
      1 N solution contains 84 g of solute
      0.1 N solution contains ? = 0.1 x 84 = 8.4 g/1000 ml.
      240ml solution will contain: = 240 x 8.4/1000 = 2.016 g of NaHCO3

     

    1. (d) An amount of NaCl in 250 ml of 0.9% NaCl = 250 x 0.9/100 = 2.25 g
      Total equivalents Na+ = weight in g
      equivalent wt = 2.25/58.5 = 0.03846 equivalents

     

      (d) The sum of probability of success and failure would be equal to 1 in Null hypothesis:
      p + q = 1, where p = probability of success
      q = probability of failure
      q = 1 - p= 1- 0.6= 0.4

     

    1. (b) We want to dispense 10 suppositories each weighing 2g and containing 400mg of tannic acid:  Amount of coca butter = 2 g x 10= 20 g
      Amount of tannic acid = 0.4 g x 10= 4 g
      Displacement value of tannic acid is 0.9, therefore:= 4/0.9 = 4.44 g of base will displace 4g
      of tannic acid = 4.44 g coca butter
      Amount of coca butter = 20 g – 4.44 g= 15.56 g

     

    1. (c) Median or Mode is generally expressed as the middle value, if number of values are even, then average of middle values should be considered. To find median or mode of experiment data, first arrange the data in order.
      105, 115, 118, 120, 135, 140, 175, 190
      = (120 + 135)/2 = 127.50

     

    1. (a) The concentration of reactant M is half in a reaction that is third order in kinetic: 
      now M = M/2
      = k (M/2)3
      = 1 /8 k (M)3

     

    1. (b) 0.025 min-1
      For the first order kinetic,
      K = 2.303/t x log Co/C = 2.303/90 x log 500/50 = 2.303/90 x log 10 = 0.025 min-1

     

    1. (d) The status of renal function impairment can be expressed by creatinine clearance.
      ClT = ClH + ClR
      ClT = Total body clearance
      ClH = Hepatic clearance
      ClR = Renal clearance
      2100 = 300 + ClR
      ClR = 1800 ml/hr
      = 30 ml/min
      The normal creatinine clearance generally lies between 80 to 120 ml/min. A creatinine clearance in patient is 30 ml/min which will be considered severely impaired.

     

    1. (c) Blood serum freezes at 0.520 C, and all solutions having this freezing point are isotonic with blood serum. 0.9% sodium chloride have the same freezing point that of blood serum.
      FP provides by 1% Boric acid = -0.29o C
      FP of blood = - 0.52o C
      FP (needed) by NaCl = (0.52-0.29)
      = 0.23o C
      Now as we know that FP provides by 1% NaCl would be -0.58o C therefore one can say,
      For FP 0.58o needs 1 % NaCl
      For FP 0.23o needs ?
      = (0.23 x 1)/0.58 = 0.396 % NaCl
      0.396 gm of NaCl/ 100cc.
      The amount of NaCl needed for 250cc,
      = (250 x 0.396)/100 = 0.991 gm NaCl = 991 mg

     

    1. B, C, D. Tween is a commonly used hydrophilic (water-loving), non-ionic (the head is not charged) surfactant. Tweens are usually used for oil in water emulsions. Surfactants decrease the surface tension of a liquid or between a liquid and a solid. They are used as detergents, wetting agents, emulsifiers and dispersants. Surfactants have polar and non-polar sides.

     

    1. 10 mg of propranolol HCl by mouth three times a day before meals and at bedtime.
      Take 1 teaspoonful (5 mL) every 6 hours for 10 days.
      30 mg of flurazepam at bedtime as needed for sleep.
      Take 1 capsule three times a day after meals.

     

    1. P: 4% = 4/100 = 1/25, so ratio is 1:25 w/v
      E: 1/200,000 = 0.000005  x 100% = 0.0005% w/v

     

    1. N: 0.25 x 10^-3 g x 15000 = 3.75g
      E: 75x10^-6 x 15000 = 1.125g

     

    1. Least weighable quantity: 6mg/5% = 120mg
      20mg / 120   = 120/ x,  x =720mg
      Weigh 120 mg of acetaminophen, dilute with 600 mg diluent to make 720 mg of mixture, and take 120 mg of mixture to contain 20 mg of acetaminophen needed.

     

    1. % error = (375– 350) / 350 = 7.14%

     

    1. A in mg%: 60 x 10^-3 mg/1ml = 0.06mg = 6% mg, 
      B in %: 0.05 x 10^-3 g/1ml = 0.00005mg = 0.005%

     

    1. 3 x 454 = 1362g
      1362 g/ x = 0.85, x = 1602.35 ml

     

    1. 10,000mg/x = 500mg/ml, x = 20ml, so powder volume = 20-18.5 = 1.5ml
      300mg/ml = 10,000/x, x = 33.3ml, so diluent volume = 33.3-1.5 = 31.8ml

     

    1. Alligation
    2.  

      1. 30g x 4% = 1.2g salicylic acid
        (1.2+ 6 ) / (30+6) = 7.2 / 36 = 2 = 20%

       

      29:    500mlx35% = 1750ml*x, x = 10% w/v

       

      30:

      • 50 lb / 2.2 = 22.7 kg
      • Total loading dose: 22.7 kg x 20mg/kg = 454 mg
      • Infusion rate: 0.5 mg/kg/min x 22.7 kg= 11.35 mg/min
      • Time: 454 mg/11.35 = 40 mins

       

      1. 3000 mg x (1ml/154mg) = 19.5 ml

       

      32.

      1. Total drug needed: 2mcg/kg/min x 85.45kg = 170.9 mcg/min
      2. Rate convert to hr: 170.9 x 60 = 10254 mcg/hr
      3. Infusion bag concentration in desired units: 100mg/250ml = 100,000 mcg/ 250ml
      4. 10254mcg/100,000mcg x 250ml = 25.6 around to 26 ml/hr

       

      33:

      1. Total dose: 140/2.2 = 63.63kg x 5mg/kg = 318.18mg
      2. Volume required: 318.18mg x 50ml/500mg = 31.818ml
      3. Flow rate: 31.818ml/30 min x 15 drops/ml = 15.9 around to 16 drops/min

       

      34:

      1. 163 / 2.2 = 74.1 kg
      2. Amount of drug in 250ml bag = 250 x 3.2 mg/ml = 800 mg
      3. Rate by weight: 17 mcg/kg/min = 17 x 74.1kg = 1259.7 mcg/min = 1.26mg/min
      4. (1.26mg/min)/800 x 250ml = 0.4ml/min

       

      35:            ANC = 4500 x 【(45%+ 5%)/100】 = 4500 x 0.5 = 2250 cells/mm3

       

      36:

      1. Convert mEq to g: 40 mEq = xmg/ (39+36) x = 3000mg  = 3g
      2. Drops infused: 20 drops/min x 60 min/hr x 10 hr = 12000 drops
      3. Volume infused over 10 hr: 15 drops / ml = 12000 drops/ x ml, x= 800ml
      4. Concentration of KCl: 3g/800ml = 0.00375= 0.38%

       

      37:

      1. Total starting dose per min: 3mcg/kg/min x 75 kg = 225 mcg/min = 0.225mg/min
      2. Total starting dose per hr = 0.225 x 60 = 13.5 mg/hr
      3. Rate in ml/hr: (13.5 mg/hr) / 400mg x 250ml = 8.44 ml/hr around to 8ml/hr

       

      38:

      1. Our goal is to convert max rate per order from mcg/kg/min to ml/hr
      2. 20 mcg/kg/min x 75kg = 1500 mcg/min
      3. 1500 mcg/min x 60 = 90,000 mcg/hr = 90 mg/hr
      4. (90mg / 400mg) x 250ml = 56.25 ml/hr is our max rate in desired unit.
      5. 70 ml/hr already exceeded 56, so yes, page doctor.

       

      39:

      1. 4% x 240ml = 17.8 g fat
      2. 8 g x 9 cal/g = 160.2 cal around to 160

       

      40:

      1. total cal per can: 37.4 g x 3.4 cal/g = 127.16 cal
      2. Per ½ can (4 fl oz): 127.16 / 2 = 63.58 around to 64 cal

       

      41:             275 x (6oz/8oz) = 206.25 cal

       

      42:          Review equivalent is the number of moles of an ion in a solution, multiplied by the valence of that ion. If 1 mol of NaCl and 1 mol of CaCl2 dissolve in a solution, there is 1 equiv Na, 2  equiv Ca, and 3 equiv Cl in that solution. For example, 1 mmol (0.001 mol) of Na+ is equal 1 meq, while 1 mmol of Ca++ is equal 2 meq.

      1. 5 mEq/L = x mg/ 39 x= 175.5 mg/L
      2. 10 dL = 1 L so patient’s K level is: 175.5 / 10 = 17.75 mg/dL

       

      43:

      1. The prescribed mL/dose if the suspension was reconstituted correctly:

      2000mg/x = 250mg/5ml, x = 40ml,  40ml/2 = 20 ml BID

      1. In the bottle with extra water, still presents same amount of active drug (Augmentin), which is:

      250mg/5ml x 100ml = 5000 mg  (1 bottle of suspension contains 5000mg active drug)

      1. How many ml will contain the prescribed dose of 1000mg?

      5000mg/110 ml = 1000mg / xml, x = 22mL

      So the patient would need to take slightly more than the prescribed volume since the mixture is more diluted than the pharmacist intends. (step 1 is optional)




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