Handout 07 Calculations PDF

Australian Statutory Education License: OPP 2024 ELITE EXPERTISE PTY. LTD HANDOUT 07 CALCULATIONS Presented By ELITE EXPERTISE Phone +61 407 177 175 +91 63616 87479 Email [email protected] Address 103/11 Goodson street Doncaster, VIC 3108 eliteexpertise.com.au © 2024 Elite...

Australian Statutory Education License: OPP 2024 ELITE EXPERTISE PTY. LTD HANDOUT 07 CALCULATIONS Presented By ELITE EXPERTISE Phone +61 407 177 175 +91 63616 87479 Email [email protected] Address 103/11 Goodson street Doncaster, VIC 3108 eliteexpertise.com.au © 2024 Elite Expertise  Dear students, these handouts will assist you in swiftly recalling the topics covered in our lectures prior to the examination. They shall not be used in place of lectures because lectures include extensive explanations of all the important points for KAPS preparation. We strongly advise you to re-listen to lectures, take notes and practice multiple choice questions in order to pass the KAPS examination. © 2024 Elite Expertise ELITEEXPERTISE.COM.AU | [email protected] Calculations Units to remember 1 kilogram=1000grams 1 gram=1000milligrams 1milligram(mg)=1000micrograms 1 microgram=10-6 grams (0.000001) 1 nanogram=10-9grams (0.000000001) 1 litre=1000ml= 10 dL (decilitre) 1 Litre=1000cc Ratios: If ratio is given as follows: W/W or 1% W/W then it is 1gm of solute in 100 gm of solvent W/V or 1% W/V then it is 1gm of solute in 100ml solvent V/V or 1% V/V then it is 1ml of solute in 100ml of solvent Units of pharmacokinetic parameters: Area under the curve: microgram.hr/ml Total body clearance CLT=L/hr Vd (volume of distribution) is L(litres) Peak plasma concentration Cmax=mg/L Plasma concentration Cp=mg/L Steady state concentration Css=mg/L Elimination half life t1/2 is in hr F (bioavailability) is normally expressed in % to convert it divide it by hundred to get a decimal value. List of formulas you need to remember for KAPS: Half-life (t1/2) for first order kinetics =0.693/k where k is elimination rate constant Shelf life (t90) =0.105/k Vd (volume of distribution) = dose/Cp (plasma concentration) F (bioavailability)= AUC (area under the curve) oral/ AUC IV © 2024 Elite Expertise 001 ELITEEXPERTISE.COM.AU | [email protected] Absolute bioavailability of batch2/relative bioavailability of batch 2 = Absolute bioavailability of batch 1/relative bioavailability of batch 1 CLT (total body clearance) =Vd*K Or CLT=vd*0.693/t1/2 Hepatic clearance (CLH)=CLT (total clearance)-CLR (renal clearance) CLH=Q*ER where Q is liver perfusion rate and ER is hepatic excretion ratio ER=Ca-Cv/Ca where Ca is arterial drug concentration and Cv is venous drug concentration F=f *(1-ER) where f is the fraction of drug absorbed across the gut Css (steady state concentration) = R/vd*k or R/CLT where R is infusion rate Loading dose= vd*desired plasma drug concentration = vd*cp bioavailability F units of a loading dose is mg (if Iv dose is given then F=1 as IV route has 100% bioavailability) Maintenance dose rate = CL*desired plasma concentration = CL*cp bioavailability F Units of maintenance dose is mg/min Dose=CL*Cp/F CP= F*D D is the dose required; Vd*K*ĩ Ĩ is frequency or time interval © 2024 Elite Expertise 002 ELITEEXPERTISE.COM.AU | [email protected] BMI definition :Weight(kg)/height(m) square No of moles=wt(gm)/MWT(molecular weight) Molarity(M)=no of moles/volume (litres) milliequivalent(Meq)=wt (mg)/equivalent weight Equivalent weight= molecular weight(mwt)/ valence Milli osmole means the number of particles in a solution: If you place Nacl in water it dissociates into Na+ and Cl- Then you say 1 millimole of Nacl=2 milliosmole Cacl2 gives Ca+ 2 Cl then you say 1 millimole of calc2 is equal to 3 milli osmole If you place dextrose in a solution it doesn’t dissociate so that means 1millimole of dextrose is equal to 1 milli osmole for such calculations you calculate millimoles first and then convert it into milli osmoles depending on the dissociation © 2024 Elite Expertise 003 ELITEEXPERTISE.COM.AU | [email protected] for such calculations you calculate millimoles first and then convert it into milli osmoles depending on the dissociation Salts with valence 1 have same molarity and normality The valence in salts means the number of metal ions present Valence 1 salts example: Nacl, Hcl, kcl, Li2Co3, Na2co3, NaHco3 Valence 2 salts example: caco3, cacl2, mgcl2, mg(oH)2, Zncl2 Aluminium citrate has valence 3-- C6H5AlO7 PH means the potential of hydrogen PH 7 is neutral PH7 is basic PH= -log H+ Or PH=14+log OH- If PKW value is not given always PKW is 14 (water at 25 degrees) Remember most asked log10=1 Antilog1=10 © 2024 Elite Expertise 004 ELITEEXPERTISE.COM.AU | [email protected] Density =mass/volume © 2024 Elite Expertise 005 ELITEEXPERTISE.COM.AU | [email protected] Practice calculations: 1. Data obtained from two different batches of tablets of a new hypoglycaemic agent undergoing clinical trial revealed that Batch 1 contains 200 mg of the drug and has an absolute bioavailability of 55%. Batch 2 contains 200 mg of the drug and relative to the first batch has a bioavailability of 165%. What is the absolute bioavailability of the second batch? 1. A 33.33% 2. B 61.6% 3. C 90.75% 4. D 181.5% 5. E 300% 2. The volume of distribution of drug c was determined to be 20L. What would be the expected plasma drug concentration immediately after an intravenous dose of 2mg is given to a 70kg male patient? A 0.01 microgram/dL B 0.1micrograms/dL C 1 micrograms/dL D 10micrograms/dL 3. A 54 year old male patient weighing 65kg requires a loading dose of a drug to treat infection. What is the suitable intravenous loading dose of this drug for the patient to achieve a plasma drug concentration between 7 to 9mg/L (vd=0.25L/kg) A75mg B 100mg C 125mg D 85mg E150mg © 2024 Elite Expertise 006 ELITEEXPERTISE.COM.AU | [email protected] Determine the loading dose of aminophylline needed- in a 55-year-old male patient weighing 70 kg if the targeted theophylline plasma level is 10 mg/L. The patient’s estimated volume of distribution is 0.5 L/kg. 1. (A) 150 mg 2. (B) 350 mg 3. (C) 400 mg 4. D 500mg 5. It was determined that 95% of an oral 80-mg dose of verapamil was absorbed in a 70 kg test subject. However, because of extensive biotransformation during its first pass through the portal circulation, the bioavailability of verapamil was only 25%. Assuming a liver blood flow of 1500 mL/min, the hepatic clearance of verapamil in this situation was: 1. A 60 mL/min 2. B 375 mL/min 3. C 740 mL/min 4. D 1110 mL/min 5. E 1425 mL/min 6. If the total body clearance for carbenicillin in a patient X is 6237ml/hr and the renal clearance is 86ml/min calculate the hepatic clearance 1. A:108ml/hr 2. B. 1077ml/hr 3. c. 3840ml/hr 4. d.5160ml/hr 7. If the extraction ratio of a drug X is 0.6 and the absorption from GIT is 40% what is the bioavailability of the drug X 1. A.16% 2. B.52% 3. C.33% 4. D. 22% 8. For morphine if we assume f of 1 and ER of 0.6 then what is the oral bioavailability of morphine 1. A.40% 2. B.20% 3. C.30% 4. D. 60% © 2024 Elite Expertise 007 ELITEEXPERTISE.COM.AU | [email protected] 9. The elimination half-life of penicillin G is approximately 0.5 hours. Thus when administered IV every six hours 1. A Half the administered dose will be eliminated in the urine at one hour 2. B Little accumulation of penicillin in the body will occur after repeated administration 3. C Each dose will only act for half an hour 4. D It will take an hour for each dose to be eliminated 5. E All of the administered dose will be metabolized one hour after taking the dose 10. Given that: (i) ascorbic acid (Vitamin C), in an aqueous mixture decomposes by first order kinetics and (ii) the half-life of ascorbic acid in this mixture at shelf temperature is 8 months, What is the acceptable shelf-life of ascorbic acid in aqueous solution assuming that it is unacceptable when its initial concentration is reduced by 1/8? 1. A 0.2 month 2. B 0.5 month 3. C 1.5 months 4. D 3.0 months 5. E 4.0 months 11. Cocaine has two ester bonds that hydrolyse to produce either benzoylecgonine or ecgonine methyl ester. At 40 °C and pH 7.25, cocaine has a half-life of 5 h. If the total extent of degradation is not allowed to be more than 10% what would the shelf-life be for cocaine under those conditions? 1. A 0.76 h 2. B 1.52 h 3. C 3.0 h 4. D 7.5 h 12. The elimination half-life for ranitidine is approximately 2 hours. What percentage of this drug would be eliminated from the body 4 hours after an intravenous (IV) bolus dose? 1. A 12.5% 2. B 25% 3. C 50% 4. D 75% 5. E 87.5% 13. Dopamine 200 mg in 500 mL of normal saline at 5 μg/kg/min is ordered for a 70 kg patient. At what rate (mL/min) should the solution be infused to deliver the desired dose of 5 μg/kg/min? 1. A 0.35 2. B 0.40 3. C 0.88 4. D 2.0 5. E 5.0 © 2024 Elite Expertise 008 ELITEEXPERTISE.COM.AU | [email protected] 9. The elimination half-life of penicillin G is approximately 0.5 hours. Thus when administered IV every six hours 1. A Half the administered dose will be eliminated in the urine at one hour 2. B Little accumulation of penicillin in the body will occur after repeated administration 3. C Each dose will only act for half an hour 4. D It will take an hour for each dose to be eliminated 5. E All of the administered dose will be metabolized one hour after taking the dose 10. Given that: (i) ascorbic acid (Vitamin C), in an aqueous mixture decomposes by first order kinetics and (ii) the half-life of ascorbic acid in this mixture at shelf temperature is 8 months, What is the acceptable shelf-life of ascorbic acid in aqueous solution assuming that it is unacceptable when its initial concentration is reduced by 1/8? 1. A 0.2 month 2. B 0.5 month 3. C 1.5 months 4. D 3.0 months 5. E 4.0 months 11. Cocaine has two ester bonds that hydrolyse to produce either benzoylecgonine or ecgonine methyl ester. At 40 °C and pH 7.25, cocaine has a half-life of 5 h. If the total extent of degradation is not allowed to be more than 10% what would the shelf-life be for cocaine under those conditions? 1. A 0.76 h 2. B 1.52 h 3. C 3.0 h 4. D 7.5 h 12. The elimination half-life for ranitidine is approximately 2 hours. What percentage of this drug would be eliminated from the body 4 hours after an intravenous (IV) bolus dose? 1. A 12.5% 2. B 25% 3. C 50% 4. D 75% 5. E 87.5% 13. Dopamine 200 mg in 500 mL of normal saline at 5 μg/kg/min is ordered for a 70 kg patient. At what rate (mL/min) should the solution be infused to deliver the desired dose of 5 μg/kg/min? 1. A 0.35 2. B 0.40 3. C 0.88 4. D 2.0 5. E 5.0 © 2024 Elite Expertise 009 ELITEEXPERTISE.COM.AU | [email protected] 15. SULFACETAMIDE EYE-DROPS Sulphacetamide sodium 10 g Sodium metabisulphite 0.1 g Disodium edetate 0.05 g Phenylmercuric nitrate 0.002 g Water for injections to 100 mL The phenylmercuric nitrate is available as a sterile aqueous solution containing 3 mg in 10 mL. The volume of this solution required to prepare 15 mL of the above formula is: 1. A 0.3 mL 2. B 1.0 mL 3. C 1.5 mL 4. D 5.0 mL 5. E 10.0 mL 16. A solution with a concentration of 0.05 per cent may also be described as containing 1. A 1 part in 100 parts 2. B 1 part in 200 parts 3. C 1 part in 500 parts 4. D 1 part in 2000 parts 5. E 1 part in 5000 parts 17. How much of a 10% injection of a drug is required to make 100 mL of a mixture containing 7.5 mg in 2.5 mL? 1. A 0.03 mL 2. B 0.3 mL 3. C 3 mL 4. D 5 mL 5. E 30 mL © 2024 Elite Expertise 010 ELITEEXPERTISE.COM.AU | [email protected] 18. A new cardiac glycoside has been developed for oral and intravenous (IV) administration. The drug has an elimination half-life of 24 hours and an apparent volume of distribution (Vd) of 3 L/kg. The effective drug concentration in plasma is 1.5 ng/mL. Toxic effects of the drug are observed at drug concentrations above 4 ng/mL. The drug is 75% bioavailable after an oral dose. Calculate an oral maintenance dose to be given once a day for a 65 kg male patient (age 68 years) with congestive heart failure and normal renal function. The aim is to achieve an average plasma concentration of 1.5 ng/mL. The dose should be 1. A 0.125 mg 2. B 0.180 mg 3. C 0.203 mg 4. D 0.270 mg 5. E 0.333 mg 19. Lignocaine hydrochloride is available for intravenous injection as a 2% solution packed in 5 mL disposable syringes. The dose is 1 mg per kg of body weight by slow intravenous injection. The volume of solution that should be used for a patient weighing 60 kg is 1. A 1.2 mL 2. B 2.5 mL 3. C 3.0 mL 4. D 4.0 mL 5. E 6.0 mL 20. A solution with a concentration of 0.0125 % may also be described as containing 1. A 1 part in 80 parts 2. B 1 part in 800 parts 3. C 1 part in 8000 parts 4. D 1 part in 12500 parts 5. E 1 part in 25000 parts 21. Drug t1/2 is 16 hours how long does it take for at least 90 percent to be excreted— 1. 64 hours 2. 12 hours 3. 15 hours 4. 72 hours 5. 92 hours 22. Morphine in a 70 kg patient taking 20 mg sustained-release capsules twice daily was found to have a steady-state concentration of 4 ng/mL. Assuming a bioavailability of 25%, the clearance of morphine from this patient was 1. A 30 L/h 2. B 52 L/h 3. C 104 L/h 4. D 208 L/h © 2024 Elite Expertise 011 ELITEEXPERTISE.COM.AU | [email protected] 23. Dose of drug is 500mg assuming no elimination occurring and plasma conc is 2 microgram/ml and plasma volume is 3 l what is the total proportion of drug found in plasma 1. 1.2 % 2. 1.5% 3. 2% 4. 4% 5. 6% 24. Each Slow-K dosage unit contains 8 mmoL of potassium as potassium chloride. How many mg of potassium chloride are in each Slow-K tablet? [Atomic Weight K = 39, Cl = 35] 1. A 296 mg 2. B 312 mg 3. C 592 mg 4. D 740 mg 5. E 872 mg 25. Calculate the number of moles of sodium chloride present in 10g ( Atomic weight of Na= 23 and CL=35.5) 1. A.0.17 moles 2. B. 2 moles 3. C. 3 moles 4. D. 4moles 5. E 1.8 moles 26. To prepare 100ml of 1M Nacl how many grams of Nacl are needed (Atomic weight of Na=23 and CL=35.5) 1. A 5.85g 2. B 6.2g 3. C 7.9g 4. D 8.5g 5. E 9.1g 27. How many milliequivalents of Na ion is in 92mg/ml of Nacl salt(mwt of Na is 23) 1. A1mEq 2. B 2mEq 3. C 3mEq 4. D 4mEq 28. What is the concentration in mg/ml of a solution containing 2mEq of kcl per ml( kcl mwt=74.5) 1. A 74.5mg/ml 2. B 149mg/ml 3. C 39 mg/ml 4. D 75mg/ml © 2024 Elite Expertise 012 ELITEEXPERTISE.COM.AU | [email protected] 29. Describe the concentration of the following solution in mEq/L..when a solution contains 10mg% of Ca+2 (mwt=40) 1. 5 mEq/L 2. 10 mEq/L 3. 20 mEq/L 4. 30 mEq/L 5. 42 mEq/L 30. How many mEq of kcl are present in a 15ml dose of 10% w/v kcl elixir( mwt=74.5g) A 20 mEq B 30 mEq C 40 mEq D 50 mEq C 22 mEq 31. For a weak base, according to the Henderson-Hasselbalch equation: When the pKa of a base is equal to 9 and the ratio of the non-ionised species to the ionized species is 10:1, the pH equals A8 B9 C 10 D 11 E 12 32. What is the pH of a solution containing 0.01 moles of acetic acid and 0.1 mole of sodium acetate per litre of solution? [Pka = 4.15] A 5.15 B 3.15 C 6.15 D 7.15 E 8.1 33. What is the molar ratio of salt/acid required to prepare a buffer having pH = 5 and pKa = 4.00? A 10:1 B 9:1 C 8:1 D 7:1 E 11:1 © 2024 Elite Expertise 013 ELITEEXPERTISE.COM.AU | [email protected] 34. What is the pH of the solution containing 0.5 mol's of ephedrine and 0.05 moles of ephedrine HCI per litre of solution? [ PKb =4.56) A 10.44 B 11.44 C 11.44 D 9.44 E 11.22 35. The ratio of ionized to unionized acetic acid in a solution is 100:1. What is the pH of a solution having pKa 5.5? A 7.5 B 2.5 C 3.5 D 4.5 E 11.5 36. Calculate the dose of paracetamol for a 25kg child if the adult dose is 3000mg daily A 1200mg B 1100mg C1000mg D 3500mg 37. Calculate the dose of paracetamol for a 2 month baby if the adults dose is 3000mg A 40mg B 50mg C 60mg D 70mg E 20 mg 38. If the adult dose is 600mg every 6 hours the dose of a 3-year-old child is? A 100mg B 120mg C 150mg D 180 mg E 200mg © 2024 Elite Expertise 014 ELITEEXPERTISE.COM.AU | [email protected] 39. The adult dose of a drug is 750 mg, what is the dose for child weighing 20 lbs? A 100mg B 200mg C 150mg D 75mg E 96mg 40. An adult dose of drug is 325 mg, what is the dose Of a drug for a 15-month-old infant? A32.5mg B 42.5mg C 22.2mg D 100mg E 22.8mg 41. If an adult dose of a drug is 100 mg, what would be the dose for a child having a body surface area of 2m2? A 115mg B 12.8mg C 9.67mg D 5.2mg 42. If a 1000ml of 15%v/v solution is diluted to 2000ml what is the resultant percentage strength of the solution A 7.5% v/v 8 8 % v/v C 2.5% v/v D 9.3 %v/v E 3.2%v/v 43. A pharmacist adds 240ml of aluminium acetate solution containing 5%w/v of aluminium acetate to 2 litres of water for irrigation. What is the final percentage w/v concentration of aluminium acetate present in the solution? A 0.53% B 0.28 % C 2.5% D 2.2% E 3.3% © 2024 Elite Expertise 015 ELITEEXPERTISE.COM.AU | [email protected] 44. Calculate the quantity (Q2), in g, of diluent that must be added to 60 g of a 10% (w/w) ointment to make a 5% (w/w) ointment A 60g B 120g C 20g D 100g E 70g 45. How much diluent should be added to 10 g of a trituration (1 in 100) to make a mixture that contains 1 mg of drug in each 10 g of final mixture? A 990g B 220g C 330g D 1000g E 500g 46. How many cc of 75% alcohol should be mixed with 10% of 1000cc alcohol to prepare 30% of 500 cc alcohol solution? A 346.16cc B 234.43cc C 153cc D 121 cc 47. In what proportion should 5% of salicylic acid be mixed with 1000 grams of 1% salicylic acid to prepare 500 grams of 3% salicylic acid? a. 250 gm b. 490 gm c. 125 gm d. 375 gm 48. In what proportion should 3% acetic acid and 1% acetic acid be mixed to prepare 200 cc of 2.5% acetic acid? a.150 and 50 cc b.100 and 100 cc c.75 cc and 125 cc d.175 cc and 25 cc © 2024 Elite Expertise 016 ELITEEXPERTISE.COM.AU | [email protected] 49. How much ointment having a 12% drug concentration and how much ointment having a 16% drug concentration must be used to make 1 kg of a preparation containing a 12.5% drug concentration? A 875g and 125g B 125g and 875g C 225g and 176g D 200g and 230g 50. In what proportion should 20% benzocaine ointment be mixed with an ointment base to produce a 2.5% benzocaine ointment? A 1:7 B 7:1 C3:1 D 4:1 E 5:1 51. How much 50% w/v dextrose solution (in milliliters) and how much 5% w/v dextrose solution (in milliliters) are required to prepare 4500 mL of a 10% w/v solution? A 500 and 4000ml B 20 and 30ml C4000 and 55ml D 12 and 16ml E 33 and 90ml 52. How many grams of 2.5% hydrocortisone be mixed with 360g of 0.25% cream to make a 1% hydrocortisone cream A 180g B 120g C 100g D 75g E 95g 53. How much ichthammol would be required to be added to a 5% w/w ichthammol in yellow soft paraffin ointment to prepare 200 g of a 17.5% w/w ichthammol in yellow soft paraffin ointment? A 26.3g B 52.8g C 13.2g D 11.5g E 12.2g © 2024 Elite Expertise 017 ELITEEXPERTISE.COM.AU | [email protected] 54. How much coal tar (in grams) should be added to 3200 g of 5% coal tar ointment to prepare an ointment containing 20% coal tar? A 600g B 1200g C 800g D 700g E 120g 55. How much urea is required to be added to a 64g of 4% liquid paraffin ointment to make the resultant strength 10% A 2.26gm B 4.26gm C 3.82g D 4.7g E 33.3g 56. You want to prepare olive oil in cetrimide cream base. If 40g of olive oil is to be added to this preparation what is the volume of olive oil to be added (density of olive oil=0.910g/ml) A 44ml B 72ml C 20ml D 18ml E 10ml 57. The density of an oil is 0.9232g/ml if you need 3.5kg of oil for your cream preparation what volume of oil should you measure A 3791ml B 5000ml C 2000ml D 3500ml E 4234ml 58. You have an ointment jar of 100ml capacity to fill the jar what weight of liquid paraffin do you need if the density of liquid paraffin is 0.93g/ml A 94g B 930g C 230g D 1000g E 500g © 2024 Elite Expertise 018 ELITEEXPERTISE.COM.AU | [email protected] 59. In your lab your dispensing balance has a sensitivity of 20mg. Find out what is the minimum weight which will incur a 5% error? A 25mg B 55mg C 400mg D 18mg E 33mg 60. If the sensitivity of the balance is 4mg what is the minimum permissible quantity that can be weighed on a balance (an error of +5 or -5 is permissible)? A 80mg B 100mg C 120mg D 233mg E 350mg © 2024 Elite Expertise 019 ELITEEXPERTISE.COM.AU | [email protected]  1. Formuala : AB2(x) = AB1(55) RB2(165) RB1(100) 165*55/100 ans: C 90.75% 2. Direct vd of patient p is 20l Dose given to him is 2mg What is plasma drug concentration Vd=dose/Cp 20=2/x X=2/20 =1/10 mg/L 1mg is 1000 micrograms 1L is 10 DL So 1*1000 =10micrograms/DL Ans:D 10*10 3. Ans :C (approximately) Patient is IS 65kg AND HIS VD IS 0.25L/KG…. SO TOTAL VD IS(0.25*65)= 16.25L WHAT DOSE OF DRUG SHOULD BE GIVEN TO him (X) TO ACHIEVE CP=8MG/L Vd=dose/Cp 16.25=x/8 so x =16.25*8=130mg …. 4. A 70KG patient HAS VD OF 0.5L/KG SO HOW MUCH IS THE TOTAL VD… 70*0.5=35L WHAT DOSE OF AMINOPHYLLNE TO BE GIVEN TO her SO THAT HER PLASMA DRUG CONCNETRATION WILL BE 10MG/L VD=DOSE TO BE GIVEN /PLASMA DRUG CONCENTRATION 35= X/10 X=35*10 =350mg ANS :B © 2024 Elite Expertise 020 ELITEEXPERTISE.COM.AU | [email protected]  1. Formuala : AB2(x) = AB1(55) RB2(165) RB1(100) 165*55/100 ans: C 90.75% 2. Direct vd of patient p is 20l Dose given to him is 2mg What is plasma drug concentration Vd=dose/Cp 20=2/x X=2/20 =1/10 mg/L 1mg is 1000 micrograms 1L is 10 DL So 1*1000 =10micrograms/DL Ans:D 10*10 3. Ans :C (approximately) Patient is IS 65kg AND HIS VD IS 0.25L/KG…. SO TOTAL VD IS(0.25*65)= 16.25L WHAT DOSE OF DRUG SHOULD BE GIVEN TO him (X) TO ACHIEVE CP=8MG/L Vd=dose/Cp 16.25=x/8 so x =16.25*8=130mg …. 4. A 70KG patient HAS VD OF 0.5L/KG SO HOW MUCH IS THE TOTAL VD… 70*0.5=35L WHAT DOSE OF AMINOPHYLLNE TO BE GIVEN TO her SO THAT HER PLASMA DRUG CONCNETRATION WILL BE 10MG/L VD=DOSE TO BE GIVEN /PLASMA DRUG CONCENTRATION 35= X/10 X=35*10 =350mg ANS :B © 2024 Elite Expertise 021 ELITEEXPERTISE.COM.AU | [email protected] 5. CLH=Q*ER ER=CA-CV CA CA is the amount of drug entering the liver; as 95%0f 80mg entered the liver CA=95*80 100 So CA is 76mg CV is the amount of drug leaving liver as the bioavailability of drug is calculated from dose administered, Cv =25*80/100 (given the biovailability of the drug is 25%) So Cv=20mg ER=76-20/76 =0.73 CLH=Q*ER =1500*0.73 =1095 ml/min (answer is approximately) Ans: D 6. Hepatic clearance=CLT-CLR CLT=6237ml/hr CLR=86ml/min= 86*60=5160ml/hr So hepatic clearance= 6237-5160= 1077ml/hr Ans: B 7. F = f · (1 - ER ) F=40 (1-0.6) F=16 (ans A) 8. F = f · (1 - ER ) F=1 (1-0.6) F=40% (Ans A) © 2024 Elite Expertise 022 ELITEEXPERTISE.COM.AU | [email protected] 9. ans B Lets imagine 100mg of penicillin is injected So 10 50 25 12.5 6.25 3.125 1.25 So for each half life it becomes half of what it is initially was if you consider each arrow is a t1/2 by 6t1/2 it becomes almost zero T1/2 is given as 0.5 hours 6t1/2=6*0.5=3 hrs that means within 3 hours it becomes almost zero, by the time it is administered every 6 hours no penicillin will be remaining in the patient’s body. 10. t90 = 0.105/k t1/2=0.693/k T1/2=0.693/k 8(months)= 0.693/k K =0.693/8 K=0.08 T90=0.105/0.08 Ans C(approximately 1.5months) 11. t90 = 0.105/k t1/2=0.693/k 5=0.693/k K=0.693/5=0.1386 T90=0.105/0.1386 T90=0.75 hours (approximately ans A) 12. Given elimination half life t1/2=2h Lets say 100mg of drug was given 100 50 25 Each arrow is a t1/2 that means in 4 hours the drug becomes 25% of what it was; that means the remaining 75% is eliminated so answer is D © 2024 Elite Expertise 023 ELITEEXPERTISE.COM.AU | [email protected] 13. dose given is 5 micrograms/kg/min---patient is 70kg so dose is 70*5=350 micrograms 350 micrograms is 350*0.001mg =0.35mg The solution of dopamine is available in strength of 200mg in 500ml 200mg-----500ml 0.35mg---? 0.35*500/200=0.875ml/min (ans C) 14. 1g* 8 tablets there is 8g of sucralfate in 120ml of water 8g----120ml ?..... 10ml 10*8/120 =0.66g (ans C) 15. the phenyl mercuric acetate in the formulation is 0.002g=2mg/100ml 2mg---100ml ?...........15ml 2*15/100=0.3mg is in 15ml of formulation The available solution in the market is 3mg in 10ml For 0.3mg prepeartion how much of this solution do we need? 3mg….10ml 0.3mg….? 0.3*10/3 =1ml of this solution is required to prepare the above formulation (Ans B) 16. 0.05 percent is 0.05/100=0.0005 1/2000 is 0.0005 so option D is the answer © 2024 Elite Expertise 024 ELITEEXPERTISE.COM.AU | [email protected] 17. in 7.5mg-----2.5ml ?.....................100ml 100*7.5/2.5 =300mg So given there is a 10% solution that means 10g in 100ml 10 g is 10000mg 10000mg-----100ml 300mg………….? 300*100/10000= 3ml (ans C) 18. CP= F*D D is the dose required; K=0.693/t1/2 Vd*K*ĩ Ĩ is frequency or time interval Cp=1.5ng/ml =1.5*10-6 mg/ml =1.5*0.000001/0.001 mg/litre =0.0015mg/litre F=75%=75/100=0.75 D=? Vd=3L/kg=3* 65=195L K=0.693/t1/2 0.693/24= 0.0288 Frequency Ĩ is 24 hours CP= F*D =0.75*D Vd*K*ĩ 195*0.0288*24 D=0.0015*195*0.0288*24/0.75 Ans-D 0.270mg (approximate) © 2024 Elite Expertise 025 ELITEEXPERTISE.COM.AU | [email protected] 19. the dose is 1mg/kg and the patient if 60kg so the dose is 60*1=60mg The available solution in the market is 2% =2g in 100ml =2000mg in 100ml 2000mg-----100ml 60mg………….? 60*100/2000 =3ml (ans C) 20. 0.0125% =0.0125/100=0.000125 1/8000 will give 0.000125 so option c is the answer 21. lets say initially 100mg of drug was given 100 50 25 12.5 6.25 3.125 Each arrow is a t1/2 90 % Is to be eliminated (so 50+25+12.5+6.25) By 4t1/2 90% will be eliminated T1/2 is 16 hours 4t1/2 is 16*4=64 (ans A) 22. Dose=CL*CP/F Cp=4ng/ml = 4*0.000001/0.001 (mg/L) =0.004mg/L F=25%=25/100=0.25 Dose =20mg*2=40mg 40=CL*0.004/0.25 CL=40*0.25/0.004 =2500L/day Each day has 24hours so 2500/24= 104 L/h (ans C) 23. CP given is 2microgram/ml =2*0.001/0.001 =2mg/L The Vd is 3L so CP in total is 2*3= 6mg/L 6*100/500=1.2%(ans A) © 2024 Elite Expertise 026 ELITEEXPERTISE.COM.AU | [email protected] 24. NO of moles=wt.(gm)/MWT Mwt=Atomic Weight K = 39+ Cl = 35 Mwt=74 8mmoles=8*0.001 moles 0.008=X/74 X=0.008*74=0.592g=592mg( Ans C) 25.. NO of moles=wt.(gm)/MWT MWt=23+35.5=58.5 No of moles=10/58.5=0.17 moles (ans A) 26. Molarity(M)=no of moles/volume (litres) NO of moles=wt.(gm)/MWT Molarity(M)=no of moles/volume (litres) (volume given is 100ml so 0.1L) 1=x/0.1= 0.1moles NO of moles=wt.(gm)/MWT 0.1=Wt(gm)/58.5 Wt (gm)=58.5*0.1=5.85g(ans A) 27. milliequivalent(Meq)=wt (mg)/equivalent weight Equivalent weight= molecular weight(mwt)/ valence Valence of Na is 1 Equivalent wt=23/1=23 mEq=92/23=4mEq(ans D) © 2024 Elite Expertise 027 ELITEEXPERTISE.COM.AU | [email protected] 28. milliequivalent (Meq)=wt (mg)/equivalent weight Equivalent weight= molecular weight(mwt)/ valence Valence of kcl=1 Equivalent weight=74.5/1=74.5 milliequivalent (Meq)=wt (mg)/equivalent weight 2=x/74.5 X=74.5*2 =149mg/ml (ans B) 29. valence of calcium is 2 Equivalent weight= molecular weight(mwt)/ valence Equivalent weight= 40/2=20 The concentration of solution is 10mg% that is 10mg/100ml 10mg----------100ml ? -------------------1000ml(1 litre) 1000*10/100=100mg/litre milliequivalent (Meq)=wt (mg)/equivalent weight MEq= 100/20=5mEq/L (ans A) 30. 10%w/v means 10g in 100ml 10g-------100ml ? -------------15ml 15*10/100=1.5g=1500mg Equivalent weight= molecular weight(mwt)/ valence Valence of kcl=1 Equivalent weight=74.5/1=74.5 milliequivalent (Meq)=wt (mg)/equivalent weight MEq= 1500/74.5=20mEq(Ans A) © 2024 Elite Expertise 028 ELITEEXPERTISE.COM.AU | [email protected] 31. For weak base: PH=pka+ log {base} or unionized {salt} ionized PH=9+log 10/1 Log 10=1 PH=9+1= 10 (Ans C) 32. For weak acid: PH=pka+log {salt} or ionized {acid} unionized PH=4.15+log0.1/0.01 PH=4.15+1=5.15 (Ans A) 33. For weak acid: PH=pka+log {salt} or ionized {acid} unionized 5=4+logx/y 5-4=logx/y 1=log x/y x/y=10:1 (Ans A) 34. For weak base: PH= PKw-pkb + log {Base} {salt} PKw is always 14 PH=14-4.56+log 0.5/0.05 PH=9.44+log10 PH=9.44+1=10.44 (Ans A) © 2024 Elite Expertise 029 ELITEEXPERTISE.COM.AU | [email protected] 35. For weak acid: PH=pka+log {salt} or ionized {acid} unionized PH=5.5+log100 Log 100=2 PH=5.5+2=7.5 (Ans A) 36. Clark’s rule is child dose= weight of child in lb* adult dose average adult weight (150 lb) 25kg is 25*2.2=55lb Dose= 55*3000/150 =1100mg (Ans B) 37. Frieds rule child dose =Adult dose* Age in months/150 Ans: A (40mg) 38. Child dose= age of child in years *adult dose child’s age in years+ 12 Ans B (120mg) 39. Clark’s rule is child dose= weight of child in lb* adult dose average adult weight (150 lb) Ans A (100mg) 40. Frieds rule child dose =Adult dose* Age in months/150 Ans A (32.5mg) 41. Child dose= Ault dose * surface area of child in m2 1.7 m2 Ans A (approximately 117mg) © 2024 Elite Expertise 030 ELITEEXPERTISE.COM.AU | [email protected] 42. C1V1=C2V2 15*1000=C2*2000 C2=15*1000/2000=7.5% (Ans A) 43. V1=240ml V2=2000+240=2240ml C1V1=C2V2 5*240=C2*2240 C2=5*240/2240=0.53%(Ans A) 44. Q1C1=Q2C2 60*10=Q2*5 Q2=600/5=120g Q2 is the final volume which is 120g in which 60g is active ingredient and 60g is diluent Ans: A 45. Q1C1=Q2C2 C2 is 1mg in 10g =0.001g/10g 10*1/100 = Q2*0.001/10 Q2=1000g which is the final quantity. But the solution already has 10g of active ingredient in it so 1000-10=990g of diluent must be added. (Ans A) 20+45= 65 parts For 65 parts---------------20 (75% alcohol) 500……………………………? 500*20/65=153cc (Ans C) © 2024 Elite Expertise 031 ELITEEXPERTISE.COM.AU | [email protected] 47. 2+2=4 parts 4 parts has 2 parts of (5% salicylic acid) 500g……………? 500*2/4=250g (Ans A) 48. 1.5+0.5=2 For 2 parts 1.5 parts is 3% alcohol For 200cc……? 200*1.5/2 =150cc So in 200cc 150cc is 3% and 50 cc is 1% (Ans A) 49. 3.5+0.5=4 parts In 4 parts 0.5 parts was 16% In 1kg(1000g) …...? 1000*0.5/4 =125 So in 1000grams 125 is 16% and 875 is 12% (Ans A) © 2024 Elite Expertise 032 ELITEEXPERTISE.COM.AU | [email protected] 50. Proportion is 2.5:17.5 =1:7 (Ans A) 51. 40+5=45 In 45 parts 5 parts is 50% In 4500…...? 4500*5/45 =500 parts is 5% Therefore, in 4500ml 500 is 50% and 4000 is 5% (Ans A) 52. X=360*0.75/1.5= 180g (Ans A) © 2024 Elite Expertise 033 ELITEEXPERTISE.COM.AU | [email protected] 53. 82.5+12.5=95 In 95 parts 12.5 is 100% In 200…….? 200*12.5/95 =26.3g (Ans A) 54. X=3200*15/80 =600g (Ans A) 55. X=64*6/90 =4.2g (Ans B) 56. Density=mass/volume 0.910=40/x X=40/0.910 =43ml ( Ans approximately A) © 2024 Elite Expertise 034 ELITEEXPERTISE.COM.AU | [email protected] 57. 3.5kg is 3500g Density=mass/volume 0.9232=3500/x X=3791ml (Ans A) 58. Density=mass/volume 0.93=X/100= 93g (Ans approximately A) 59. % error=sensitivity * 100 weight desired 5=20*100/x X=20*100/5 =400mg (Ans C) 60. % error=sensitivity * 100 weight desired 5=4*100/x X=400/5= 80 mg (Ans A) © 2024 Elite Expertise 035 ELITEEXPERTISE.COM.AU | [email protected] ThankYou! © 2024 Elite Expertise