Sample/practice exam, questions and answers PDF

Title Sample/practice exam, questions and answers
Course Chemical Analysis
Institution Aston University
Pages 4
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CALCULATIONS PRACTICE EXERCISES (1) 1. Lassar’s paste contains zinc oxide 24%. How much zinc oxide would be required to make 375 g of lassar’s paste? 24% = 24 g in 100g = 12 g in 50 g = 6 g in 25 g Therefore for 375 g we need ((24 x 3) + 12 + 6) g = (72 + 12 + 6) g = 90 g OR 24% of 375 g = ((375 ÷ 100) x 24) g = (3.75 x 24) g = 90 g Answer = 90 g 2. What is the final strength, expressed as a percentage, when 300 ml of 35% v/v is added to 200 ml of 15% v/v? 35% v/v is 35 ml in 100 ml = 105 ml in 300 ml 15% v/v is 15 ml in 100 ml = 30 ml in 200 ml When added together there will be: (105 + 30) ml in (300 + 200) ml = 135 ml in 500 ml = (135 ÷ 5) ml in 100 ml = 27 ml in 100 ml = 27% v/v Answer = 27% v/v 3. What volume of a 40% stock solution should be used to prepare 150 ml of a 12% solution? C1 = 40% C2 = 12% C1 x V1 = C2 x V2 40 x V1 = 12 x 150 V1 = (12 x 150) ÷ 40 = 1800 ÷ 40 = 45 ml Answer = 45 ml

V1 =? V2 = 150 ml

4. You need to prepare 250 ml of an antiseptic solution which when diluted 1 in 25 gives a 0.02% solution. How much of a 25% stock solution is required? C1 = 25% C2 = 0.02% when diluted 1 in 25 Therefore before dilution it would be: (0.02 x 25)% = 0.5%

V1 =?

V2 = 250 ml

C1 x V1 = C2 x V2 25 x V1 = 0.5 x 250 V1 = (0.5 x 250) ÷ 25 V1 = 125 ÷ 25 =5 V1 = 5 ml Answer = 5 ml 5. A prescription for a 10 year old child reads: ‘Carbamazepine 100 mg/5 ml liquid – Give 250 mg bd’. What volume of carbamazepine liquid would be needed to provide 4 weeks supply? Carbamazepine liquid contains 100 mg/5 ml It therefore contains 20 mg in 1 ml And 250 mg in (250 ÷ 20) ml = 12.5 ml Directions are ‘250 mg bd’ for 4 weeks’ = 12.5 ml bd for 4 weeks = 25ml daily for 28 days = (25 x 28) ml = 700 ml OR 250 mg bd = 500 mg daily 500 mg daily for 4 weeks = (500 x28) mg = 14000 mg The liquid contains 20 mg in 1 ml And 14000 mg in (14000 ÷ 20) ml = 700 ml Answer = 700 ml

6. A patient who weighs 80 kg is prescribed an intravenous infusion of dopamine at a dose of 2.5 micrograms/kg/minute. If dopamine is available as a 3.2 mg/ml solution, at what rate in ml/hour should the infusion be set? The dose is 2.5 micrograms/kg/minute = (2.5 x 80) micrograms/minute = 200 micrograms/minute = (200 x 60) micrograms/hour = 12,000 micrograms/hour = 12 mg/hour Dopamine is available as a 3.2 mg/ml solution It therefore contains 1 mg in (1÷ 3.2) ml And 12 mg in ((1 ÷ 3.2) x 12) ml = 3.75 ml Therefore rate = 3.75 ml/hour Answer = 3.75 ml/hour

7. If a tablet contains 2.5 mg of active ingredient and has a total mass of 225 mg and a manufacturing batch run produces 500,000 tablets, what mass of excipient is required? Total mass of each tablet = 225 mg The active ingredient = 2.5 mg Therefore mass of excipient in each tablet = (225 – 2.5) mg = 222 .5 mg The batch run is 500,000 tablets Therefore mass of excipient in batch = (222.5 x 500,000) mg = (222.5 x 500) g = (22250 x 5) g = 111250 g or 111.25 kg Answer = 111250 g or 111.25 kg

8. A child weighing 6 kg is prescribed folic acid at a dose of 500 micrograms/kg/day. If the folic acid syrup available contains 75 mg of folic acid per 150 ml, what volume of this syrup should be given daily? The dose is 500 micrograms/kg daily = (500 x 6) micrograms daily = 3000 micrograms daily = 3 mg daily The syrup contains 75 mg in 150 ml It therefore contains 0.5 mg in 1ml And 3 mg in (3 ÷ 0.5) ml = 6 ml Answer = 6 ml

9. What weight of clobetasone butyrate, in milligrams, is present in 30 g of Eumovate cream? Eumovate cream contains clobetasone butyrate 0.05% (BNF 66 p.764) 0.05% = 0.05 g in 100 g = 0.05 g = 50 mg in 100 g = 5 mg in 10 g And 15 mg in 30 g Answer = 15 mg

10. A patient is prescribed Cacit-D3 sachets at a dose of one sachet twice daily. Calculate the weekly calcium intake in millimoles. One sachet twice daily for 1 week = 14 sachets Each sachet contains 12.5 mmol of calcium (BNF 66 p. 664) Therefore the weekly calcium intake = (12.5 x 14) mmol = 175 mmol Answer = 175 mmol...


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