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npl exam pharmacy year 1
Pharmaceutical Chemistry and Physicochemical Principles for Pharmacy (University of
Wolverhampton)
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FACULTY OF SCIENCE AND ENGINEERING
SCHOOL OF PHARMACY
SUMMATIVE ASSESSMENT
______________________________________________________________
Module Code :
4PY019
Module Title:
Pharmacy Stage 1
Module (year) leader:
David Gay
Pharmaceutical Calculations Assessment
Assessment title:
Sitting:
July 2020
Examiner:
Dr J Daly
Time Allowed:
72 hours
_____________________________________________________________
Instructions to candidates
This assessment is open book.
ALL questions should be attempted. There is no negative marking.
The assessment is not cut scored. The pass mark is 40%.
A calculator may be used for this assessment.
Working out space is provided for each question. You must complete your working out in the space
provided for each question and you must enter your final answer in the ‘Answer’ section at the bottom of
each question. You will not be required to fill in a GPhC-style answer grid. Write your final answer in
normal numerical notation, not standard form/scientific notation. For example, if your answer is one
thousand two hundred and thirty four, write it as 1234 not 1.234 x 103. Answers written in a form other
than normal numerical notation will be marked as incorrect. You may use a comma to separate digits into
groups of three for whole numbers greater than 1000, such as 1,234 rather than 1234.
It is an assessment requirement that you must show a logical working out process to arrive at a computed
answer. Accordingly, even where a submitted answer is correct, it will not be awarded a mark in the
absence of a logical working out method with supporting numerical data.
You will need to save the Word document, complete the questions within the allocated time period and
upload the completed Word document before the due date and time. No other file formats will be
accepted.
You must complete this assessment on your own. All submissions will be assessed for plagiarism and
collusion. Evidence of plagiarism or collusion will be reported for suspected academic misconduct.
University guidance on academic misconduct can be found at: https://www.wlv.ac.uk/currentstudents/conduct-and-appeals/academic-misconduct/
Student number: 1927921
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Question 1
Express 0.065 grams in nanograms
0.065grams x 1000 = 65milligrams
65milligrams x 1000 = 65,000micrograms
65,000micrograms x 1000 = 65,000,000nanograms
Answer: 65,000,000
nanograms(s)
Question 2
Express 0.0052 L (litres) in millilitres. Give your answer to 1 decimal place.
0.0052litres x 1000 = 5.2millilitres
Answer:
millilitre(s)
5.2
Question 3
A cream contains 4.5% w/w of active pharmaceutical ingredient (API). How many grams of API are
there in 220 g of the cream? Give your answer to 1 decimal place.
4.5g API in 100g
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220g x 4.5g = 990g
990g /100g = 9.9g
WORKING OUT
Answer:
9.9
gram(s)
Question 4
Drug A is available as a liquid containing 120 mg/5 mL. A prescription requests a dose of 144 mg four
times a day for ten days. How many millilitres of the liquid should be supplied?
(144mg x 5ml) / 120mg = 6ml
6ml x 4 times per day x 10 days =240ml
WORKING OUT
Answer:
240
millilitre(s)
Question 5
A patient requires 0.75 mg of drug B. It is available as a 1 in 10,000 solution. How many mLs of the
solution will contain the correct dose? Give your answer to 1 decimal place.
1/10,000 = 0.0001g
0.0001g x ? = 0.00075g
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0.00075/0.0001 = 7.5ml
WORKING OUT
Answer:
7.5
millilitre(s)
Question 6
A bottle of eye drops contains drug C at a concentration of 0.45% w/v. It is available in bottles
containing 3 mL. How many milligrams of drug C does one bottle contain? Give your answer to 1
decimal place.
(0.45g x 3ml) / 100ml = 0.0135g
0.0135g x 1000 = 13.5mg
WORKING OUT
Answer:
13.5
milligram(s)
Question 7
A baby weighing 5 kg requires an injection at a dose of 800 micrograms/kg. If the injection is
available as a solution containing 10 mg/mL, how many millilitres should be administered? Give your
answer to 1 decimal place
800 micrograms / 1000 = 0.8milligrams
0.8 milligrams x 5kg = 4 milligrams
(4mg x 1ml) / 10mg = 0.4ml
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Answer:
0.4
millilitre(s)
Question 8
Drug D is prescribed at a total daily dose of 315 mg, given in divided doses every 8 hours. If it is
available as 35 mg tablets, how many tablets are required for one dose?
24 hours / 8 = 3 doses
315mg / 3 = 105mg per dose
105mg / 35mg = 3 tables at a time
WORKING OUT
Answer:
3
tablet(s)
Question 9
How many millilitres of active ingredient are there in 8.5 L (litres) of a 6.4% v/v solution?
8.5L x 1000 = 8500ml
(8500ml x 6.4ml) / 100ml = 544ml
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WORKING OUT
Answer:
544
millilitre(s)
Question 10
A disinfectant solution is available at a concentration of 0.225% w/v. Express the amount of active
ingredient in mg/mL. Give your answer to 2 decimal places.
0.225% w/v = 0.225g / 100ml (grams x 1000 = mg)
= 225mg / 100ml (divide both sides by 100)
= 2.25mg / ml
WORKING OUT
Answer:
2.25
mg/mL
Question 11
How many milligrams of active pharmaceutical ingredient (API) are required to prepare 2.5 kg of
cream containing 0.35% w/w API?
(0.35g x 2500g) / 100g = 8.75g
=8750mg
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WORKING OUT
Answer:
8750
milligram(s)
Question 12
A doctor has prescribed 75 g of an ointment containing 0.7% w/w of the active pharmaceutical
ingredient (API). You have a 50 g tube of an ointment containing 3.5% w/w of the API in stock which can
be mixed with a suitable diluent to make the 0.7% w/w strength ointment required. How many grams
of 3.5% w/w ointment would you use to prepare the 0.7% w/w product the doctor prescribed?
3.5% x ? = 0.7% x 75g
(0.7 x 75) / 3.5 = 15g
WORKING OUT
Answer:
15
gram(s)
Question 13
A newborn baby is prescribed drug E at a dose of 30 micrograms/kg/day, given in two divided doses.
The baby weighs 3.5 kg. An oral solution contains drug E 25 micrograms in 5mL. How any millilitres of
the oral solution provide one dose? Give your answer to 1 decimal place.
(30 micrograms x 3.5 kg) / 2 doses = 52.5 micrograms per dose
(52.5micrograms / 25micrograms) x 5ml = 10.5ml
WORKING OUT
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Answer:
10.5
millilitre(s)
Question 14
What will be the resulting concentration of a glucose solution (expressed as a % w/v) if you mix
together 100 mL of 30 %w/v glucose, 200 mL of 20 % w/v and 150 mL of 5 %w/v glucose? Give your
answer to 3 significant figures.
30g | 40g | 7.5g | = 77.5g = 17.2g
100 | 200 | 150 | 450ml 100ml
WORKING OUT
Answer:
17.2
%w/v
Question 15
How many milligrams of active ingredient would be needed to prepare 3.5 L (litres) of a 1 in 8000
solution? Give your answer to 1 decimal place.
3.5L = 3500ml
(1g x 3500ml) / 8000ml =0.4375g = 437.5milligrams
WORKING OUT
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Answer:
437.5
milligram(s)
Question 16
A patient weighing 69 kg is prescribed drug F at a total daily dose of 4 mg/kg. This is given in divided
doses every 8 hours. How many milligrams of drug F are required for each dose?
4mg x 69kg = 276mg
24hours / 8 = 3 doses per day
276mg / 3 = 92 mg per dose
WORKING OUT
Answer:
92
milligram(s)
Question 17
A patient is prescribed drug G, at a dose of 225 mg three times a day for one week. Drug G is
available as 75 mg capsules. How many capsules are required in total?
225mg x 3 times per day x 7 days = 4725mg
4725mg / 75mg = 63 capsules
WORKING OUT
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Answer:
63
capsule(s)
Question 18
A prescription requires you to supply 300 mL of an aqueous oral solution containing 400 mg of active
ingredient per 5 mL dose. A stock solution containing 20% w/v is available. How many millilitres of
water should be added to the appropriate volume of the stock solution to prepare the required
product?
400mg / 1000 = 0.4g
20g x 100ml = 0.4g x ?
(20g x 100ml) / 0.4g = 5000ml
5000ml / 5ml = 1000ml
WORKING OUT
Answer:
1000
millilitre(s)
Question 19
How many millimoles of sodium bicarbonate (NaHCO 3) are there in a 200 mL -pack containing sodium
bicarbonate 4.2% w/v? (Ar for sodium = 23; Ar for hydrogen = 1; Ar for carbon = 12; Ar for oxygen = 16).
23 + 1 + 12 + (16 x 3) = 84
(4.2g x 200ml) / 100ml = 8.4g
8.4g / 84 = 0.1moles
0.1 x 1000 = 100millimoles
WORKING OUT
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Answer:
100
millimole(s)
Question 20
How many grams of potassium chloride (KCl) are required to prepare 150 mL of a solution containing
800 millimoles/L of potassium chloride? Give your answer to 2 decimal places. (A r for potassium = 39;
Ar for chloride is 35.5).
800 millimoles/L /1000 = 0.8moles/L
150mL /1000 = 0.15L
0.8moles/L x 0.15L = 0.12moles
0.12moles xx 74.5mr = 8.94g
WORKING OUT
Answer:
8.94
gram(s)
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