Dose calculations -_january_2021_-_upload_slide_share

Govt. Medical College, Datia, MP
DOSE CALCULATIONS
Set A
Department of Pharmacology
Dr. S. K. BOUDDH
MD Pharmacology
Professor & HOD
Email: sk85bouddh@gmail.com

Basics
WEIGHTS AND MEASURESMETRIC SYSTEM
• Weight:
• Measures:
1 Kilogram 1000 grams
1 gram 1000 milligram
1 milligram 1000 microgram
1 microgram 1/1000 Milligram
1 Litre 1000 Millilitres
1000 Litres 1 kilolitres

Basics
Relationship between Metric System and Imperial System
• Weights (Approximately equivalent for all practical purposes)
• Measures:
1 kilogram 2.2 Pounds
1 Pound 450 Grams
1 Gram 15 Grains
1 Gram 60 Milligrams
1 Ounce 30 Grams
1 Millilitre 15 Minims
1 pint 600 Millilitres
1 Ounce 30 Millilitres
1 Drachm 4 Millilitres

Basics
Domestic Measures
Domestic Measures Metric equivalent Imperial equivalent
1 Drop 0.05ml 1 minim
1Tea spoonful 4ml 1 Fluid drachm
1 Dessert spoonful 8ml 2 Fluid drachms
1 Table spoonful 15ml 4 Fluid drachms
1 Tumbler full 300ml 10 Ounces

Problem no. 1
A patient is admitted to the casualty with chest pain of sudden onset. On
examination he had cold, clammy skin and sweating. The body weight is 60 kg;
blood pressure is 74 mm Hg (systolic), bilateral basal crepitations were present.
ECG was suggestive of Inferior myocardial infarction. An I.V. drip of dopamine
has to be set for this patient to treat cardiogenic shock.
Note:
• The dose of dopamine required by this patient is 5mcg/kg/min
• Dopamine is supplied in ampoules of 200mg/5ml
• A drip rate of 20 drops/minute delivers 1 ml
Q. 1. Calculate the dose of the drug required.
Q. 2. Calculate the volume of the drug required.
Q. 3. Selection of fluid.
Q. 4. Calculate rate of infusion.

Solution of the Problem 1
Q. 1. Calculate the dose of the drug required.
Solution:
Dose of Dopamine require 5 mcg/kg/min
Weight of the patient 60 kg
So, the require dose will be 5 mcg/kg/min X 60 kg
= 300 mcg / min

Q. 2. Calculate the Calculate the volume of the drug required.
Solution:
Dopamine is given diluted with 500 ml of Normal Saline as an slow I. V.
Infusion.
Since, Dopamine is supplied in ampoules of 200 mg/5ml,
So, 200 mg dopamine dissolved in 500 ml of NS,
So, 200 X 1000 = 200000 microgram dissolved in 500 ml of NS
500 ml X 300 microgram / min
Thus, for 300 microgram of dopamine will require = ------------------------------
200000 microgram
= 15 / 2 ml / min
= 0.75 ml / min

Q. 3. Selection of fluid.
Solution:
Normal saline.

Q. 4. Calculate rate of infusion.
Solution:
To deliver 1 ml needs 20 drops / minute
So, for 0.75 ml/ min will require
= 20 drops / min X 0.75 ml / min
= 15 drops / min will be drip rate.

Additional Problem
In a 50 kg adult male patient with hemorrhagic shock, dopamine is
infused until arrangements are made for blood transfusion.
Note:
• The available dopamine hydrochloride is 40 mg/ml in 5 ml ampoule.
• Normal saline 500 ml.
• The dose of dopamine required by this patient is 4 mcg/kg/min
• A drip rate of 1 ml = 16 drops.
Q. 1. Calculate the number of drops per minute to be infused.
Q.2. What will be the total duration of the time required to infuse
1 ampoule of the dissolved dopamine at the same rate
4 mcg/kg/min.

Solution of the additional Problem
Q. 1. Calculate the number of drops per
minute to be infused.
Solution:
 Total volume of solution = 500 ml NS, i.e. 495 ml NS + 5 ml
dopamine vial.
 Dopamine = 500 ml solution.
 The total amount of dopamine hydrochloride = 40 X 5 = 200
mg.
 Since 500 ml solution contains 200 mg of dopamine,
 So, 1 ml solution will contains = 200 /500 = 0.4 mg of
dopamine = 400 microgram of dopamine.
 So, Concentration of the solution will be 400 microgram / ml.

Solution:
Desired concentration in μg / kg / minute X Weight of the patient.
 Infusion Rate = ---------------------------------------------
Concentration of the solution in μg / ml
4 μg / kg / minute X 50 kg
 Infusion Rate = ---------------------------------------------
400 μg / kg / minute
= 0.5 ml/min

 Since, 1 ml = 16 drops,
 So, for 0.5 ml = 8 drops.
 So, 8 drops per minute to be infused for the patient.

Q.2. What will be the total duration of the
time required to infuse 1 ampoule of the
dissolved dopamine at the same rate
4 mcg/kg/min.
 Since, time required for 0.5 ml infusion = 1 minute.
 So, for 500 ml infusion = 500 ml / 0.5 ml
= 1000 minute.
= 16 hours 66 minutes.

Problem no. 2
Mr. Hari Ram, 62 years old male admitted in ICU with
acute myocardial infarction. After treatment with nitrates
and other measures patient suddenly developed
ventricular arrhythmias for which Inj. Lignocaine IV
infusion was started at a rate of 2mg/min.
• The drip set is adjusted with a micro-drip set at a rate of
60 drops/ml. Lignocaine vial contains 1gm in 5 ml to
be dissolved in 250 ml of normal saline infusion.
• Q. Calculate the drip rate per minute.

Solution of the Problem no. 2
Q. 1. Calculate the Calculate the drip rate per
minute.
Solution:
 Total volume of solution = 250 ml NS, i.e. 245 ml NS + 5 ml
lignocaine vial.
 Lignocaine solution = 250 ml solution.
 The total amount of Lignocaine = 5 gm in 5 ml
= 5000 mg in 5 ml
= 1000 mg / ml Lignocaine.
 Since 250 ml solution contains 1000 mg of lignocaine.
 So, 1 ml solution will contains = 1000 /250
= 4 mg / ml of lignocaine.
 So, Concentration of the solution will be 4000 microgram / ml.

minute.
Desired concentration in μg / kg / minute X Weight of the patient.
 Infusion Rate = ---------------------------------------------
Concentration of the solution in μg / ml
2000 μg / minute X ABC kg
 Infusion Rate = ---------------------------------------------
4000 μg /ml
= 0.5 ml/min

minute.
 Since, 1 ml = 60 drops,
 So, for 0.5 ml = 30 drops.
 So, 30 drops per minute to be infused for the
patient.

Q.2. What will be the total duration of the
time required to infuse 1 ampoule of the
dissolved lignocaine at the same rate
60 drops / ml.
 Since, time required for 0.5 ml infusion = 1 minute.
 So, for 250 ml infusion = 250 ml / 0.5 ml
= 500 minute.
= 8 hours 33 minutes.

Problem no. 3
Mr. Varma a 40 years old male with body weight
50 kg came with complaints of tiredness, not
able to do daily activities with poor
concentration on work. Diagnosis of Iron
deficiency anemia was made. His hemoglobin
level was 8 gms/dl (normal – 13mg/dl).
Note: One ampoule (2ml) of Iron dextran
contains 100 mg of iron.
Q. Calculate the parental iron therapy for this
patient.

Q. 1. Calculate the parental iron therapy for
this patient.
 Iron requirement (mg) = 4.4 X Body weight (Kg) X Hb deficit (g/dl)
 Given, Body weight = 50 kg, Hb deficit = (13 – 8 )
= 5 gm/dl.
 So, Iron requirement (mg) = 4.4 X 50 kg X 5 gm/dl
= 1,100 gm / dl parental iron will require.

Q. 1. Calculate the parental iron therapy for
this patient.
 Since, 1100 gm parental iron in 100 ml require, so 1 ml will contain
11 gm parental iron.
 Since, 100 mg Iron dextran in 2 ml of ampoule.
 So, for 11 gm = 1100 mg will require = 2 ml X 1100 mg / 100 mg =
22 ml, means 11 ampoule.
 The test dose prescribed earlier is no longer recommended, because
the life threatening anaphylactic reaction can occur even when a
previous dose or test dose has been well tolerated.
 So, total 11 ampoule Iron dextran is diluted in 500 ml of DNS / NS
solution and infused I.V. over 8 hours under constant observation.
 Injection should be terminated if the patient complains of giddiness,
paresthesias or tightness in the chest. And keep ready the emergency
tray for the treatment of anaphylactic reactions.

Additional Problem
A seven year old male child of 20 kg body weight was brought to the casualty in a
semiconscious state. The parents told that the child had consumed around 40 iron tablets (the
container showed that the preparation was ferrous sulfate). After confirming the diagnosis,
you planned to give the specific antidote desferrioxamine.
Note:
The strength of ferrous sulfate tablet is 150 mg. The elemental iron content of
ferrous sulfate is 20 %. Each 85 mg of elemental iron needs 1000 mg of
desferrioxamine for neutralization.
Q1. Calculate the total amount of the specific antidote.
Q2. What is the route of administration of desferrioxamine?
Q3. Mention the other antidote and its route of administration for iron poisoning?

Solution of the Additional Problem
Q1. Calculate the total amount of the specific antidote.
Solution:
 1 Iron tablet = 150 mg,
 So, for total amount of iron in 40 tablet will be = 6000 mg.
 Since, the elemental iron content of ferrous sulfate is 20 %.
 So, the amount of elemental iron will be = 6000 X 20/100 = 1200 mg.
 Since, for neutralization of each 85 mg of elemental iron needs 1000 mg of
desferrioxamine.
1000 mg X 1200 mg
 So, 1200 mg elemental iron required = ----------------------- mg of desferrioxamine.
85 mg
= 14117 mg ~ 14 gm of desferrioxamine will be
required.

Solution of the Additional Problem
Q2. What is the route of administration of desferrioxamine?
 Oral route ( Cap. Deferiprone mainly) to bind the iron present in the gut.
 Desferrioxamine (Inj. Desferal 0.5 gm/vial) 0.5 – 1 gm / day i.m. helps to
excrete the chronic overload.
 It may also be infused i.v. concurrently with blood transfusion 2 gm per
unit of blood.
Q3. Mention the other antidote and its route of administration for iron
poisoning?
 Other antidote is Deferiprone. It is an orally active iron chelator. Used for
treatemt of transfusion siderosis in thalassemia patient, removal of iron
overload in hepatic cirrhosis. It is less effective than Desferrioxamine.
Dose: 50 – 100 mg mg/kg daily in 2 – 4 divided dosed. P/p: (Cap. Kelfer
250 mg, 500 mg).

Problem no. 4
A 60 year old male Mr. Vishwanathan, weight
70 kg came to the hospital with Diabetic
ketoacidosis. Regular Insulin infusion was
started in a dose of 0.1U/kg/hour. The given vial
contains plain insulin 40 U/ml. The drug to be
dissolved in 250ml of normal saline.
Q. Calculate drip rate in drops per minutes.

Q2. Calculate drip rate in drops per minutes.
Solution:
 Regular Insulin infusion dose is 0.1U/kg/hour,
 The patient’s body weight is 70 kg,
 So, 7 U / hour Regular Insulin infusion required for this patient.
 The given Insulin vial contains plain (regular) insulin 40 U/ml.
 And, it is dissolved in 250 ml of normal saline.
 So, 1 ml solution will contain 40 U / 250 ml = 0.16 U / ml strength.
 Thus, 0.16 U Regular Insulin strength solution in 1 ml , so for 7 U, = 1 ml X 7 U /
0.16 U = 43.75 ml.
 As, we know 1 ml = 16 drops, so, for 43.75 ml = 43.75 X 16 drops
= 700 drops / hour, and = 700 / 60 minutes = 11.66 drops ~ 12 drops / minutes.

DOSE CALCULATIONS
Percentage and Ratio Strength
Dr. S. K. BOUDDH
MD Pharmacology
Professor & HOD

Problem no. 1
Q.1 How many milligrams of noradrenaline
are in 2 ml of 1: 1000 solution?
Solution:
 Since 1 : 1000 solution strength,
 Means, it contains 1gm noradrenaline in 1000 ml,
 So, for 1 ml = 1 mg / ml, thus, 2 ml ampoule will
have 2 mg of noradrenaline of 1:1000 strength.

Problem no. 2
Q.2 How many mg of lignocaine are in a 10 ml
ampoule of 2 % lignocaine ?
Solution:
 Since 2 % lignocaine ampoule,
 Means, 2 gm lignocaine in 100 ml,
 So, for 1 ml = 2000 mg / 100 ml, = 20 mg / ml lignocaine,
 Since, total 10 ml ampoule volume, so, 20 mg X 10
= 200 mg lignocaine in a 10 ml ampoule of 2% strength.

Problem no. 3
Q.3 Convert 1 : 500 solution to mg / ml.
Solution:
 Since 1 : 500 solution strength,
 Means, it contains 1gm drug in 500 ml,
 So, for 1 ml = 1000 mg / 500 ml, = 2 mg / ml

Problem no. 4
Q.4 How many ml of a 1:10,000 solution
would you need to obtain 1 mg of adrenaline?
Solution:
 Since 1 : 10,000 solution strength,
 Means, it contains 1gm adrenaline in 10,000 ml,
 So, for 1 ml = 1000 mg / 10,000 ml, = 0.1 mg of
adrenaline / ml.

Problem no. 5
Q.5 Convert 8 % pilocarpine hydrochloride to mg/ml ?
Solution:
 Since 8 % pilocarpine hydrochloride ampoule,
 Means, 8 gm pilocarpine hydrochloride in 100 ml,
 So, in 1 ml = 8000 mg / 100 ml, = 80 mg / ml of
pilocarpine hydrochloride.

Problem no. 6
Q.6 Calculate the amount of sodium chloride required
to prepare 500 ml of normal saline?
Solution:
 As, we know Normal Saline = 0.9 % NaCl,
 Means, 0. 9 gm NaCl in 100 ml,
 So, in 500 ml = 0.9 gm X 500 ml / 100 ml, = 4.5 gm
NaCl will be required in 500 ml.

Problem no. 7
Q.7 The Unit head orders intravenous administration of 10 % (W/V)
solution of mannitol in a dose of 0.5 gm/kg body weight. How many
milliliter of mannitol should be administered to a patient weighing 60
Kg ?
Solution:
 10 % Weight upon Volume, means 10 gm Mannitol in 100 ml of
solution, = 10000 mg in 100 ml = 100 mg / ml of solution strength.
 The require dose is 0.5 gm / kg Body weight, and the Patient’s weight
is 60 kg, so 0.5 gm X 60 kg = 30 gm = 30000 mg of mannitol require,
 Since, 100 mg solution strength per ml, so for 30000 mg of mannitol =
1 ml X 30000 mg / 100 mg = 300 ml mannitol should be
administered.

Problem no. 8
Q.8 A physician order to add 5 ml of 25 % (W/V)
solution of potassium iodide to 1000 ml of D5W for
parenteral infusion. What is the percentage strength of
potassium iodide in the infusion solution?
Solution:
 25 % Weight upon Volume = 25 gm Potassium Iodide
in 100 ml of solution, = 0.25 gm / ml,
 So, for 1005 ml of D5W ml = 0.25 gm of Potassium Iodide, So,
for 1 ml will be = 0.25 gm / 1005 ml, and for percentage strength
of KI will be = 0.25 gm X 100 / 1005 ml = 0.025 % of KI will
be required.

Problem no. 9
Q.9 Calculate the amount of KMnO4 required to prepare 60 ml of
stock solution. So, that 25 ml of which diluted to 50 ml to give rise
to strength of 1 : 1000 solution ?
Solution:
1:1000
1 gm = 1000 ml
So, per ml = 1000 mg / 1000 ml = 1 mg / ml
So, 25 ml will contain = 25 mg,
And 25 ml diluted in 50 ml , so total So, quantity of the solution is 75 ml,
So, to prepare 60 ml of stock solution will require = 25 mg X 60
ml / 75 ml = 20 mg KMnO4 will be required.

Problem no. 10
Q.10 If a physician orders a 25 mg test dose of sodium thiopental prior
to anaesthesia, how many milliliters of 2.5% (W/V) solution of sodium
thiopental should be used?
Solution:
2.5 % w/v = 2.5 gm in 100 ml
= 25 mg / ml
To be Required dose = 25 mg
Since, 1 ml require 25 mg
So, for 25 mg = 25 mg / 25 mg = 1 ml of 2.5% w/v solution of sodium
thiopental will be required.

Problem no. 11
Q.11 How much water should be mixed with 3000 ml of
90 % (V/V) alcohol to make 25% (V/V) alcohol ?
Solution:
N1 . V1 = N2 . V2
N1 = 90 % , N2 = 25 % , V1 = 3000 ml , V2 = ?
90 % X 3000 ml = 25 % X V2
V2 = 90 % X 3000 ml / 25 % = 90 X 120 ml = 10800 ml
 So, 10,800 ml (~ 11 Liters) water should be mixed.

Problem no. 12
Q.12 How many milliliters of purified water should be added
to 25 ml of 1:500 solution of Benzalkonium chloride to
prepare a 1:5000 Benzalkonium solution for wet dressing ?
Solution:
N1 . V1 = N2 . V2
N1 = 1:500 , N2 = 1:5000 , V1 = 25 ml , V2 = ?
1 gm / 500 ml X 25 ml = 1 gm / 5000 ml X V2
V2 = 1 gm / 500 ml X 25 ml X 5000 ml / 1 gm
 So, V2 = 250 ml of purified water.

Problem no. 13
Q.13 A physician order calls for the I.V. administration of 0.005% concentration
of nitroglycerine in 300 ml dextrose saline, how many milliliters of 8 mg/ml
nitroglycerine should be added to obtain the final concentration ?
Solution:
0.005 % Nitroglycerine
= 0.005 gm Nitroglycerine in 100 ml of solution
So, 1 ml will contain = 0.005 X 1000 mg / 100 ml = 0.05 mg / ml
So, in 300 ml Dextrose saline = 0.05 X 300 ml = 15 mg NTG in 300 ml
Since, 8 mg in 1 ml
So, for 15 mg = 1 ml X 15 mg / 8 mg = 1.875 ml should be added to
obtain the final concentration.

Set B
DRUGS DOSAGE CALCULATION
Dr. S. K. BOUDDH
MD Pharmacology
Professor & HOD

Brief about the Dose calculations
 Cancer patients receive chemotherapy.
 Pediatric patients with the exception of premature new born,
whose immature renal and liver functions needs an additional
assessment of dosing.
 Young’s rule Age/Age+12 * Adult dose
• Cushing’s rule Age at next birthday (in years)*adult dose/24 = Dose for child
• Fried’s rule Age in months * adult dose/150 = Dose for Infant (1 month to 1 year)
• Clark’s rule Wt. in lb * Adult dose /150lbs = Dose for child
• Based on BSA BSA of child (in m2)*Adult Dose

• For precise calculation of dose based on BSA, one
should refer to a standard normogram given that includes
weight and height as a factor affecting BSA.
• In this normogram, the BSA in square meter (m2) is
indicated where a straight line drawn to connect the
height and weight of the child intersects the surface area
column.

Problem no. 1
Q.1 The daily dose of cefadroxil 50 mg/kg/day to be given at 12 hours interval.
How many teaspoons of reconstituted cefadroxil suspension containing
250 mg/5ml should be administrated to a pediatric patient weighing 10 kg ?
Solution:
 For pediatric patient weighing 10 kg,
 Daily dose = 50 mg/kg/day,
 So, dose require 50 mg X 10 kg = 500 mg / day,
 Since, the cefadroxil suspension strength is 250 mg / 5 ml = 50 mg / ml.
 Since, 50 mg cefadroxil in 1 ml, so for 500 mg cefadroxil will
require 1 ml X 500 mg / 50 mg = 10 ml
 As we know, 1 TSF = 5 ml, so for 2 TSF = 10 ml
Cefadroxil will be required.

Problem no. 2
Q.2 A medication order calls of taxol 135mg/m2 for the
management of metastatic cancer. How much should be given to
a patient weighing 48 kg and 150 cm tall ?
Solution:
 Height of the patient is 150 cm,
 Body weight is 48 kg,
 So, Body Surface Area Height (cm) X Weight (kg)
(m2 ) will be = √ ---------------------------------
3600
 So, BSA for the patient will be = √ 2 = 1.4 m2,
 Since, the adult dose of Taxol require 135mg/m2
 So, Dose of the drug = BSA of child (m2) X Adult Dose.
 So, total = 1.4 X 135 = 189 mg of Cefadroxil will be required.

Problem no. 3
Q.3 A The daily dose of Diphenhydramine for a child is 5 mg/kg
of 150 mg/m2. Calculate the dose on each basis, how much should
be given for a child weighing 20 kg and measuring 92 cm?
Solution:
 Since, Diphenhydramine strength is 5 mg/kg, for 20 kg, require =
5 X 20 = 100 mg of Diphenhydramine.
 Height of the patient is 92 cm,
 Body weight is 20 kg,
 So, Body Surface Area Height (cm) X Weight (kg)
( m2 ) will be = √ ---------------------------------
3600
 So, BSA for the patient will be = √ 0.5 = 0.7 m2,
 Since, the adult dose of Diphenhydramine require 150 mg/m2.
 So, Dose of the drug = BSA of child (m2) X Adult Dose.
 So, = 0.7 X 150 mg/m2 = 105 mg of Diphenhydramine will be required.

Problem no. 4
Q.4 The dose of Gentamicin is 2.5 mg/kg administrated
12 hourly. What would be the daily dose for the new born
weighing 2.6 kg?
Solution:
 Since, for 1 kg require 2.5 mg of Gentamicin,
 But, for 2.6 kg will require 2.5 X 2.6 = 6.5 mg of Gentamicin for
12 hours.
 And, for 24 hours will require 6.5 X 2 = 13 mg of Gentamicin.

Problem no. 5
Q.5 On the basis of renal impairment a medication order calls
to administer 40% of the usual dose of Gentamicin in a patient
weighing 60 kg, If the usual dose of Gentamicin is 3mg/kg/day in
three equally divided doses, how much should be administrated
per dose?
Solution:
 Here, the given dose of Gentamicin is 3mg/kg/day in three equally
divided doses,
 So, single dose will be 1 mg / kg/day, and for 60 kg weigh of the
patient will be 60 mg / day single dose of Gentamicin,
 But, we should administer only 40 % of the required dose,
 So, 100 mg of the 40 mg, so for 60 mg = 40 mg X 60 mg / 100 mg
= 24 mg single dose of Gentamicin will be required.

Problem no. 6
Q.6 A child weighing 20 kg is suffering from fever. Calculate the
dose of paracetamol for that child?
Solution:
Weight in lb (pound) X Adult Dose
 Clark’s Rule, Dose for child = ------------------------------------------
150 lbs
 Since, weight of the child is 20 kg, so, in pound = 44.1 lbs.
 Adult dose of Paracetamol is 325 to 650 mg (for children 10 – 15 mg/kg) 3 – 4 times a day.
44.1 lbs X 325 mg
 Clark’s Rule, Dose for child = ----------------------- = 95.55 ~ 100 mg
150 lbs
44.1 lbs X 625 mg
 Clark’s Rule, Dose for child = ----------------------- = 191. 1 ~ 200 mg
150 lbs
 So, Paracetamol range can be given = 95.55 - 191. 1 or ~ 100 to
200 mg for this child.

Problem no. 7
Q.7 Calculate the dose of Diazepam for a child aged
2 years weighing 10 kg, suffering from febrile
convulsions.
Solution:
Weight in lb (pound) X Adult Dose
 Clark’s Rule, Dose for child = ------------------------------------------
150 lbs
 Since, weight of the child is 20 kg, so, in pound = 44.1 lbs.
 Adult dose of Paracetamol is 325 to 650 mg (for children 10 – 15 mg/kg) 3 – 4 times a day.
44.1 lbs X 325 mg
 Clark’s Rule, Dose for child = ----------------------- = 95.55 ~ 100 mg
150 lbs
44.1 lbs X 625 mg
 Clark’s Rule, Dose for child = ----------------------- = 191. 1 ~ 200 mg
150 lbs

Problem no. 8
Q.8 Calculate the iron required for a patient weighing
60 kg diagnosed as severe anaemia with the Hb level
6gm/dl ?
Solution:
 Iron requirement (mg)= 4.4 X Body weight (Kg) X Hb deficit (g/dl)
 Given, Body weight = 60 kg, Hb deficit = (13 – 6 )
= 7 gm/dl.
 So, Iron requirement (mg) = 4.4 X 60 kg X 7 gm/dl
= 1,848 gm / dl parental iron will require.

Problem no. 9
Q.9 A medication order calls of I.V. administration of
dopamine at the rate of 5 mcg/kg/minute for a 75 kg
patient. The clinical pharmacist adds one ampoule of
dopamine containing 200 mg in 5 ml to a 250 ml bottle
of D5W. What drip rate should be run in drops per
minute using mini-drip set that delivers 60 drops per
ml?

Solution:
 Require dopamine for 75 kg weight patient will be
= 75 X 5 mcg/kg/minute = 375 mcg / minute
= 0.375 mg / minute.
 Since, 200 mg dissolved in 255 ml of D5W,
 So, for 0.375 mg = 0.375 mg X 255 ml / 200 mg
= 95.625 / 200 = 0.48 ml of dopamine.
 Since, delivery of dopamine require 60 drops / minutes,
and we know 1 ml = 60 drops micro-drips,
 Thus, for 0.48 ml will require = 60 drops X 0.48 = 28.8
drops, i.e., 29 micro-drips / minutes.

Problem no. 10
Q.10 A medication order calls for 1000 ml of normal
saline to be administered over 5 hours. If the venoclysis
set calibrates 25 drops per ml, at what rate of flow in
drops per minute should be the set adjusted to
administer the solution in stipulated time interval?
Solution:
 1 ml = 25 drops,
 So, for 1000 ml = 25000 drops,
 Since, in 5 hours (300 minutes) should infuse 25000
drops, so in 1 minute = 25000 / 300 = 83.33 ~ 83 micro-
drips / minute.

Problem no. 11
Q.11 A physician has ordered an i.v infusion of 1000 ml
of saline to run over 24 hrs using macro drip. Calculate
the drip rate ?
Solution:
 As we know, 1 ml = 16 macro drop,
 So, for 1000 ml = 16000 macro-drops,
 Since, in 16 macro drops deliver in 1 minute,
 So, in 24 hours (1440 minutes) should infuse 16000
macro drops, so in 1 minute = 16000 / 1440 = 11.11
macro drops ~ 11 macro drops / minute.

Problem no. 12
Q.12 A 30 kg child received 8 mg of phenytoin per
kilogram of body weight as an anticonvulsant.
How many milliliters of phenytoin suspension
containing 60 mg per 5 ml child should receive ?
Solution:
 Dose of Phenytoin is 8 mg / kg body weight,
 So, for 30 kg child = 30 X 8 = 240 mg phenytoin require,
 Since, Phenytoin suspension is of 60 mg per 5 ml, so for
240 mg = 5 ml X 240 mg / 60 mg = 20 ml phenytoin
suspension will be required.

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