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| pharmaceutical calculations questions and answers |
Note:: First try your best to solve these pharmaceutical calculations. If you find them difficult you can consult the step-by-step solutions at the end of the MCQ questions.
51) How many mEq of KCI are present in a 500mL infusion bottle, if the bottle contains 11.729 grams of KCl? The molecular weight of KCl is 74.6.
a- 220 mEq
b- 314 mEq
c- 128 mEq
d- 250 mEq
e- 500 mEq
Solution:
The formula for milliequivalents calculation is;
mEq = moles x valence x 1000
As is clear from the above formula; to calculate the mEq of KCl present in the 500 mL infusion bottle, we first need to determine the number of moles of KCl.
Number of moles of KCl = Mass of KCl / Molecular weight of KCl
Number of moles of KCl = 11.729 g / 74.6 g/mol
Number of moles of KCl = 0.157 moles
To convert 0.157 moles of KCl into mEq, we need to know the valence of each ion in KCl. In this case, both K+ and Cl- have a valence of 1.
For KCl, the valence is 1 + (-1) = 0, since K+ has a valence of +1 and Cl- has a valence of -1. So, we cannot convert moles of KCl directly into mEq of KCl.
However, we can convert the moles of K+ and Cl- separately into mEq, since they are present in equal amounts in KCl.
mEq of K+ = 0.157 moles x 1 x 1000 = 157 mEq
mEq of Cl- = 0.157 moles x 1 x 1000 = 157 mEq
Therefore, the total mEq of KCl is equal to the sum of the mEq of K+ and Cl-:
mEq of KCl = (mEq of K+) + (mEq of Cl-)
mEq of KCl = 157 mEq + 157 mEq
mEq of KCl = 314 mEq
So, 0.157 moles of KCl is equivalent to 314 mEq of KCl.
Therefore, the correct answer is b.
52) The amount of 190 proof alcohol required to make 500cc of 70% alcohol is?
a- 150 cc
b- 220 cc
c- 284 cc
d- 368 cc
e- 184.2 cc
Solution:
190-proof alcohol is made up of 95 parts alcohol and 5 parts water. So the percentage of alcohol in 190 proof alcohol is 95%.
To determine the amount of 190 proof alcohol needed to make 500cc of 70% alcohol, we can use the following formula:
% strength1 x volume1 = % strength2 x volume2
where;
% strength1 is the percentage of alcohol in the 190 proof alcohol,
volume1 is the volume of 190 proof alcohol needed,
% strength2 is the desired percentage of alcohol in the final solution (70%), and
volume2 is the final volume of the solution (500cc).
We can rearrange this formula to solve for volume1:
volume1 = (% strength2 x volume2) / % strength1
Substituting the given values, we get:
volume1 = (70% x 500cc) / 95%
We need to convert the percentages to decimal form:
volume1 = (0.7 x 500cc) / 0.95
volume1 = 368.42cc
Therefore, we need 368.42cc of 190 proof alcohol to make 500cc of 70% alcohol.
Therefore, the correct answer is d.
53) If a child's weight is 60 pounds, what is the correct amount of a drug that should be given to the child? The recommended dose of the drug for the child is 1.2 milligrams per kilogram of body weight.
a- 8 mg
b- 12 mg
c- 32 mg
d- 50 mg
e- 76 mg
Solution:
To determine the appropriate dose of the drug for a child weighing 60 lb, we first need to convert their weight from pounds to kilograms, because the dose is given in mg/kg (milligrams per kilogram) of body weight.
We can use the conversion factor of;
2.2 lb = 1 Kg
1 lb = 1 / 2.2 Kg
60 lb = 60 x (1 / 2.2 Kg) = 27.27 kg
So, the child weighs 27.27 kg.
Next, we can calculate the appropriate dose of the drug for the child using the formula:
dose = weight (in kg) x dose per kg
Substituting the given values, we get:
dose = 27.27 kg x 1.2 mg/kg
dose = 32.7 mg
Therefore, the appropriate dose of the drug for a child weighing 60 lb is approximately 32.7 mg.
Therefore, the correct answer is c.
54) If a neonate with a body weight of 16 pounds is receiving Theophylline via an infusion at a rate of 0.08 milligrams per kilogram of body weight per hour, what is the total amount of the drug required for one daily bottle in milligrams?
a- 7 mg
b- 14 mg
c- 30 mg
d- 80 mg
e- 150 mg
Solution:
To determine how many milligrams (mg) of Theophylline are needed for one daily bottle for a neonate weighing 16 pounds, we need to use the following formula:
Total dose = infusion rate x time
where infusion rate is in mg/kg/hr,
time is in hours,
and the total dose is in milligrams.
First, we need to convert the weight of the neonate from pounds to kilograms:
2.2 lb = 1 Kg
1 lb = 1 / 2.2 Kg
16 lb = 16 x (1 / 2.2 Kg) = 7.27 kg
Next, we can calculate the total dose of Theophylline needed for one daily bottle using the given infusion rate and assuming a 24-hour period:
Total dose = 0.08 mg/kg/hr x 7.27 kg x 24 hr
Total dose = 13.96 mg
Therefore, approximately 13.96 mg of Theophylline is needed for one daily bottle for a neonate weighing 16 pounds, if the infusion rate is 0.08 mg/kg/hr.
Therefore, the correct answer is b.
55) What is the amount of aspirin in milligrams present in 12 tablets if three tablets contain 975 mg of aspirin?
a- 2200 mg
b- 3000 mg
c- 3900 mg
d- 4125 mg
e- 4400 mg
Solution:
We can use a proportion to solve this problem.
If three tablets contain 975 mg of aspirin, we can write:
3 tablets / 975 mg = 12 tablets / x
where x is the number of milligrams of aspirin in 12 tablets.
To solve for x, we can cross-multiply and simplify:
3 tablets * x = 975 mg * 12 tablets
3x = 11700 mg
x = 3900 mg
Therefore, 12 tablets would contain 3900 mg of aspirin.
Therefore, the correct answer is c.
56) What is the total amount of brompheniramine maleate in milligrams present in a 120-mL container of cough syrup if each 5-mL dose contains 2 mg of the agent?
a- 24 mg
b- 42 mg
c- 48 mg
d- 84 mg
e- 240 mg
Solution:
To calculate the total amount of brompheniramine maleate in a 120-mL container of the cough syrup, we can use a proportion.
If 2 mg of brompheniramine maleate is present in each 5-mL dose, we can write:
2 mg / 5 mL = x mg / 120 mL
where x is the amount of brompheniramine maleate in milligrams present in a 120-mL container.
To solve for x, we can cross-multiply and simplify:
2 mg * 120 mL = 5 mL * x
240 mg = 5x
x = 48 mg
Therefore, a 120-mL container of the cough syrup contains 48 mg of brompheniramine maleate.
Therefore, the correct answer is c.
57) What is the amount of dextrose in grams needed to make a 5% w/v solution with a volume of 4,000 mL?
a- 5 grams
b- 40 grams
c- 140 grams
d- 200 grams
e- 400 grams
Solution:
We can calculate the number of grams of dextrose required to prepare a 5% w/v solution in 4,000 mL of solution as;
The given concentration of dextrose is 5% w/v solution which means 5 grams of dextrose are there in 100ml solution.
Knowing that we can calculate the number of grams of dextrose required to prepare a 5% w/v solution in 4,000 mL of solution;
grams of dextrose = (5g / 100mL) x 4000mL
grams of dextrose = 200 grams
Therefore, the correct answer is d.
58) What is the equivalent amount of fluid ounces in 2.5 L?
a- 6 fluid ounces
b- 18 fluid ounces
c- 34 fluid ounces
d- 65 fluid ounces
e- 85 fluid ounces
Solution:
To convert liters to fluid ounces, we can use the conversion factor;
1 liter = 33.814 fluid ounces
Therefore, to convert 2.5 L to fluid ounces, we can multiply 2.5 by 33.814:
2.5 L × 33.814 fluid ounces/L = 84.535 fluid ounces
Therefore, 2.5 L is equal to approximately 84.535 fluid ounces.
Therefore, the correct answer is e.
59) What are the dimensions of a nicotine transdermal patch system in inches if its measurements are 4.7 cm by 4.8 cm?
a- 2.35 inches by 2.39 inches
b- 1.85 inches by 1.89 inches
c- 3.25 inches by 3.29 inches
d- 7.45 inches by 7.49 inches
e- 8.55 inches by 8.59 inches
Solution:
To convert centimeters to inches, we can use the conversion factor that 1 inch is equal to 2.54 centimeters.
Therefore, to convert the dimensions of a nicotine transdermal patch system from centimeters to inches, we can divide each dimension by 2.54:
Length in inches = 4.7 cm ÷ 2.54 cm/in = 1.85 in
Width in inches = 4.8 cm ÷ 2.54 cm/in = 1.89 in
Therefore, the dimensions of the nicotine transdermal patch system are approximately 1.85 inches by 1.89 inches.
Therefore, the correct answer is b.
60) If a pharmacist incorporates 1 pint of Alcohol USP into 5 L of a mouthwash formula that was initially labeled as 10% V/V, what would be the percentage of alcohol present in the resulting mixture?
a- 14.34 %
b- 17.34 %
c- 19.68 %
d- 22.43 %
e- 24.40 %
Solution:
First of all, lets know something about Alcohol USP;
Alcohol USP (United States Pharmacopeia) refers to a specific grade of alcohol that is used for medicinal purposes, such as for use as an antiseptic or as a solvent in the manufacturing of pharmaceuticals. The concentration of alcohol in Alcohol USP is typically 95% ethanol by volume, which means that it contains 95% ethanol and 5% water.
So, Alcohol USP is 95% Alcohol.
In the question it is 1 pint of Alcohol USP, so we need to find out the amount of alcohol in 1 pint of Alcohol USP;
1 pint = 473.18 ml
So, (95ml / 100ml) x 473.18 ml = 449.521 ml
Thus, 1 pint of Alcohol USP contains 449.521 ml of alcohol.
Now we need to find out the amount of alcohol in 5 L of 10% V/V;
10% V/V means 10 ml alcohol is there in 100 ml of solution;
so, in 5 L (5000ml) of solution, how much alcohol?
(10 ml / 100 ml) x 5000 ml = 500 ml
so 500 ml of alcohol is there in 5 L of 10% V/V solution.
After adding 1 pint of Alcohol USP into 5 L of 10% V/V formula;
The total volume of alcohol becomes;
449.521 ml + 500 ml = 949.521 ml
The total volume of the mixture becomes;
473.18 ml + 5000 ml = 5473.18 ml
Now lets calculate the percentage of alcohol present in the resulting mixture;
( 949.521 ml / 5473.18 ml ) x 100 = 17.34%
Therefore, the correct answer is b.
61) How many 10-mL vials of an injection that contains 5 mg/mL of diazepam would be required to produce the diazepam rectal gel formula given below?
Diazepam .................... 100 mg
Methylcellulose ............ 2.5 g
Methylparaben ............. 100 mg
Glycerin ........................ 5 g
Purified water to make ......100 mL
a- 2 vials
b- 5 vials
c- 6 vials
d- 8 vials
e- 10 vials
Solution:
To calculate the number of 10-mL vials of an injection containing 5 mg/mL of diazepam required to prepare the given diazepam rectal gel formula, we can use the following steps:
First, we need to calculate the total amount of diazepam required in the formula. From the given formula, we know that we need 100 mg of diazepam.
Next, we need to calculate the volume of the injection that contains 5 mg/mL of diazepam needed to obtain 100 mg of diazepam.
To do this, we can use the formula:
volume of injection = required amount of drug / concentration of drug in the injection
Plugging in the values, we get:
volume of injection = 100 mg / 5 mg/mL = 20 mL
Therefore, we need 20 mL of the injection containing 5 mg/mL of diazepam to prepare the given formula.
Since each vial of the injection contains 10 mL of the solution, we need:
Number of vials = total volume of injection / volume of each vial
Number of vials = 20 mL / 10 mL/vial = 2 vials
Therefore, we need 2 vials of an injection containing 5 mg/mL of diazepam to produce the given diazepam rectal gel formula.
Therefore, the correct answer is a.
62) If a patient is receiving IV fluid at a rate of 90 mL per hour for a duration of 4 and a half hours, what will be the total amount of fluid, in milliliters, that the patient will receive?
a- 380 mL
b- 400 mL
c- 405 mL
d- 420 mL
e- 460 mL
Solution:
To calculate the total amount of IV fluid that a patient will receive if infused at a rate of 90 mL per hour over 4 and a half hours, we can use the formula:
Total amount of IV fluid = infusion rate x infusion time
Converting 4 and a half hours to decimal form:
4 and a half hours = 4.5 hours
Now we can plug in the numbers:
Total amount of IV fluid = 90 mL/hr x 4.5 hours
Total amount of IV fluid = 405 mL
Therefore, the patient will receive 405 mL of IV fluid if infused at a rate of 90 mL per hour over 4 and a half hours.
Therefore, the correct answer is c.
63) If the drop factor is 16 gtt/mL, what would be the infusion rate in drops per minute for a patient who is to receive 1.5 liters of IV fluid over a 12-hour period?
a- 16 gtt/min
b- 21 gtt/min
c- 25 gtt/min
d- 31 gtt/min
e- 33 gtt/min
Solution:
To calculate the infusion rate in gtt/min for a patient receiving 1.5 L of IV fluid over a 12-hour period with a drop factor of 16 gtt/mL, we can use the following formula:
Infusion rate = (Volume to be infused x Drop factor) / Time in minutes
First, we need to convert the volume to be infused from liters to milliliters:
1.5 L = 1500 mL
Next, we need to convert the time in hours to minutes:
12 hours = 12 x 60 minutes = 720 minutes
Now we can plug in the numbers:
Infusion rate = (1500 mL x 16 gtt/mL) / 720 minutes
Infusion rate = 33.3 gtt/min (rounded to one decimal place)
Therefore, the infusion rate in gtt/min for this patient will be approximately 33.3 gtt/min if they receive 1.5 L of IV fluid over a 12-hour period with a drop factor of 16 gtt/mL.
Therefore, the correct answer is e.
64) What is the calibrated volume of the mold used by a pharmacist to prepare ten suppositories from a polyethylene glycol base with a density of 1.18 g/mL, given that the total weight of the suppositories is 22.5 g?
a- 1.91 mL
b- 1.95 mL
c- 6.98 mL
d- 11.72 mL
e- 19.02 mL
Solution:
A suppository mold is a device used by pharmacists to form suppositories. The pharmacist fills the mold with a suppository base, like polyethylene glycol, and allows it to harden. Then the finished suppositories are popped out of the mold and are ready for use.
To calculate the calibrated volume of the mold used to prepare ten suppositories with a total weight of 22.5 g from a polyethylene glycol base with a density of 1.18 g/mL, we can use the following formula:
Total volume of suppositories (mL) = Total weight of suppositories (g) / Density of base (g/mL)
Total volume of suppositories (mL) = 22.5 g / 1.18 g/mL
Total volume of suppositories (mL) = 19.07 mL
Since we are to calculate the calibrated volume of the mold used to prepare each suppository, we need to divide the total volume of suppositories by the number of suppositories.
Calibrated volume of the mold = 19.07 mL / 10
Calibrated volume of the mold = 1.91 mL
Therefore, the calibrated volume of the mold used to prepare each suppository is approximately 1.91 mL.
Thus, the correct answer is a.
65) Zinc sulfate dissociates into two ions (Zn2+ and SO42-) in weak solutions. 40% of the zinc sulfate dissociates in weak solutions. Calculate the dissociation factor for zinc sulfate?
a- 0.4
b- 0.8
c- 1.2
d- 1.4
e- 2
Solution:
The formula for dissociation factor is;
dissociation factor = total number of particles after dissociation divided by the original number of particles
Zinc sulfate dissociates into 2 ions in weak solutions: Zn2+ and SO42-. So it is a 2-ion electrolyte.
40% of the zinc sulfate dissociates in weak solutions. This means 40% of the particles break apart into ions.
If there are 100 original zinc sulfate (ZnSO4) particles, then 40 of them dissociate into ions. So there are now 40 Zn2+ ions and 40 SO42- ions, for a total of 80 ions.
In addition to the 80 ions, 60 undissociated ZnSO4 particles remain (100 original - 40 dissociated).
So the total number of particles after dissociation are;
80 + 60 = 140 particles
Plugging in the values into the formula;
dissociation factor = 140 particles / 100 original particles = 1.4
So the dissociation factor is 1.4.
Thus, the correct answer is d.
66) A physician prescribed 60 milliliters (mL) of phenoxymethyl penicillin solution for oral administration which contains 4,800,000 units of penicillin. How many penicillin units will be in each teaspoonful dose of this solution?
a- 40,000
b- 200,000
c- 400,000
d- 450,000
e- 500,000
Solution:
To calculate the number of penicillin units in each teaspoonful dose of the prepared suspension, we need to know the total number of teaspoonful doses that can be made from the 60 mL of the suspension, and then divide the total number of units by the number of doses.
First, to find the total number of teaspoonful doses that can be made from the 60 mL of the suspension;
1 teaspoonful = 5 mL
Total number of teaspoonful doses = 60 / 5 = 12
Now to find the number of penicillin units in each teaspoonful dose, we need to divide the total number of units by the number of doses;
4,800,000 units / 12 = 400,000 units/teaspoonful
Therefore, each teaspoonful dose of the prepared suspension would contain 400,000 units of penicillin.
Thus, the correct answer is c.
67) What volume of 90% (v/v) alcohol solution is required to prepare 2 pints of a 60% (v/v) solution?
a- 200 mL
b- 300 mL
c- 370 mL
d- 460 mL
e- 630 mL
Solution:
To prepare 2 pints of a 60% (v/v) solution from a 90% (v/v) alcohol solution, we need to calculate how much of the 90% (v/v) solution to mix with water to obtain the desired concentration.
First, we need to determine the volume of alcohol (90% v/v) required to prepare the solution. We can use the formula:
V1 x C1 = V2 x C2
where,
V1 is the volume of the 90% (v/v) alcohol solution,
C1 is the concentration of the 90% (v/v) alcohol solution,
V2 is the final volume of the 60% (v/v) solution, and
C2 is the desired concentration of the final solution.
Using this formula, we can solve for V1:
After converting percentages into decimal forms;
V1 x 0.90 = 2 pints x 0.60
V1 = (2 pints x 0.60) / 0.90
V1 = 1.333 pints
So we need 1.333 pints of the 90% (v/v) alcohol solution.
Now we need to convert this volume to milliliters (mL) since we are working with volume in mL.
One pint is equal to 473.176 mL, so:
1.333 pints x 473.176 mL/pint = 630.7 mL
Therefore, we need to use 630.7 mL of the 90% (v/v) alcohol solution to prepare 2 pints of the 60% (v/v) solution.
Thus, the correct answer is e.
68) What is the maximum number of colchicine tablets, each containing 600 mcg, that can be produced from 50 g of colchicine?
a- 25,000 tablets
b- 50,000 tablets
c- 75,000 tablets
d- 83,333 tablets
e- 100,000 tablets
Solution:
To determine the maximum number of colchicine tablets that can be produced from 50 g of colchicine, we need to know the amount of colchicine in each tablet.
If each tablet contains 600 mcg of colchicine, which is equivalent to 0.6 mg, then the number of tablets that can be produced from 50 g of colchicine can be calculated as follows:
First, we need to convert 50 g to milligrams (mg), since the amount of colchicine in the tablets is in mg:
50 g x 1000 mg/g = 50,000 mg
Next, we divide the total amount of colchicine by the amount of colchicine in each tablet:
50,000 mg ÷ 0.6 mg/tablet = 83,333 tablets
Therefore, the maximum number of colchicine tablets, each containing 600 mcg or 0.6 mg of colchicine, that can be produced from 50 g of colchicine is 83,333.
Thus, the correct answer is d.
69) A patient has been directed by the physician to take 5mg of Coumadin on Monday, Wednesday, and Friday, while on Tuesday, Thursday, Saturday, and Sunday, the patient is required to take 2 1/2 mg. What is the total amount of Coumadin, in milligrams, that the patient will consume in a week?
a- 20 mg
b- 25 mg
c- 35 mg
d- 50 mg
e- 57 mg
Solution:
The patient is to take Coumadin 5mg on Monday, Wednesday, and Friday, which adds up to 5mg x 3 = 15mg per week for these three days.
On Tuesday, Thursday, Saturday, and Sunday, the patient is to receive 2 1/2 mg, which is equivalent to 2.5mg x 4 = 10mg per week for these four days.
To find the total amount of Coumadin the patient will take in one week, we add the amounts taken on the three days (15mg) to the amounts taken on the four days (10mg), which gives us a total of 15mg + 10mg = 25mg per week.
Therefore, the patient will take 25mg of Coumadin in one week.
Thus, the correct answer is b.
70) If a patient is required to take 250 mg of medication thrice a day for a week, and the tablets available are of 62.5 mg strength, how many tablets will the pharmacist need to dispense to meet the prescribed dosage?
a- 44 tablets
b- 56 tablets
c- 75 tablets
d- 84 tablets
e- 96 tablets
Solution:
If the patient is prescribed to take 250mg of medication thrice a day, the total daily dose would be 250mg x 3 = 750mg.
Since the tablets available are of 62.5mg strength, the number of tablets needed to meet the prescribed dosage can be calculated by dividing the total daily dose by the strength of a single tablet:
750mg ÷ 62.5mg per tablet = 12 tablets per day
Therefore, the pharmacist would need to dispense 12 tablets per day to meet the prescribed dosage. For a week's supply (7 days), the total number of tablets required would be:
12 tablets per day x 7 days = 84 tablets for the patient to take 250mg thrice a day for a week.
Thus, the correct answer is d.
71) The following is a prescription to the compounding pharmacist;
Rx:
Hydrocortisone bitartrate ........... 0.34 g
Phenacetin .................................. 3.6 g
Aspirin ........................................ 6.0 g
Caffeine ...................................... 0.6 g
M ft. Caps no 45
Sig. i cap tid prn Pain
What would be the amount of Hydrocortisone bitartrate in milligrams present in a single capsule?
a- 34 mg
b- 340 mg
c- 7.55 mg
d- 45 mg
e- 79 mg
Solution:
This prescription is for a medication in capsule form that contains hydrocortisone bitartrate, phenacetin, aspirin, and caffeine.
Hydrocortisone is a steroid medication that can help reduce inflammation and relieve pain. It's often used topically, but in this case, it's in the form of hydrocortisone bitartrate, which is designed for oral use.
Phenacetin and aspirin are both pain relievers, and caffeine can enhance their effectiveness.
The prescription instructs the pharmacist to make 45 capsules. The dosage is "i cap tid prn Pain," which means to take one capsule three times a day as needed for pain.
M ft. - Misce fiat (Latin for mix and make), indicating a compound medicine is to be prepared.
Caps - Abbreviation for capsules
no - Number
45 - Refers to the quantity, in this case, 45 capsules.
So in full, this prescription notation means - Prepare 45 capsules of the compound medicine described in the full prescription details.
To find out the amount of Hydrocortisone bitartrate in milligrams present in a single capsule;
The formula given in the prescription seems to be a master formula meaning the quantities given are actually for the whole batch of medication. In this case, 45.
So we need to divide the quantity of Hydrocortisone bitartrate by the batch size, thus
0.34 g = 340 mg
340 mg / 45 = 7.555 mg
Therefore, the amount of Hydrocortisone bitartrate in milligrams present in a single capsule is 7.555 mg.
Thus the correct option is c.
72) Following is the data of heights in cm of 13 pediatric patients:
95, 103, 107, 110, 129, 124, 129, 136, 95, 147, 142, 140, 95
Measuring the central tendency, what does the value '124' represent?
A)Mode
B)Mean
C)Median
D)Confidence interval
E) Average
Solution:
The central tendency for a data set is measured through the mean, median, and mode.
The mean is the average of all the values in the data set:
(95+103+107+110+129+124+129+136+95+147+142+140+95)/13
= 119.38
Thus the value '124' does not represent the mean.
The median is the middle value when the data set is arranged in order from least to greatest or vice versa:
The numbers in the given data are;
95, 103, 107, 110, 129, 124, 129, 136, 95, 147, 142, 140, 95
Lets arrange them in ascending order;
95, 95, 95, 103, 107, 110, 124, 129, 129, 136, 140, 142, 147
The middle value seems to be 124 in this case. So the value '124' represents the median of this data set.
The mode is the value that appears most frequently:
The most frequent value in this data set is 95. Thus the value '124' does not represent the mode also.
Thus the correct option is c.
73) Given an estimated surface area of 0.6 m2 for a 3 1/2 years old child, what would be the appropriate dosage in mg of a drug that has an adult dose of 125 mg?
a- 18 mg
b- 24 mg
c- 29 mg
d- 36 mg
e- 43 mg
Solution:
This calculation problem uses a method called body surface area (BSA) dosing to estimate a child's dosage based on their surface area. The BSA dosing method assumes that a child's dose should be proportional to their surface area, which is a measure of their body size.
Assuming an adult dose of 125 mg, the child's dose can be calculated using the following formula:
Child's dose = (child's surface area / 1.73 m2) x adult dose
In this case, the child's surface area is 0.6 m2. Putting these values into the formula gives us:
Child's dose = (0.6 / 1.73) x 125 mg
Child's dose = 43.35 mg
However, it's important to note that this calculation is based solely on the child's surface area and does not take into account other factors such as age, weight, and medical history. Therefore, it is not appropriate to provide a dosage recommendation without a thorough medical evaluation by a qualified healthcare professional.
However, the correct option is e.
74) What amount of hydrocortisone powder (in grams) is required to be mixed with 1 lb of 2% hydrocortisone ointment to produce a 5% w/w ointment? (Hint: Solve by Alligation Method)
a- 4 grams
b- 8 grams
c- 14 grams
d- 19 grams
e- 24 grams
Solution:
This calculation problem can be solved with the Alligation method.
The following figure shows the flow of the Alligation method.
First, we draw a 9 boxes grid as shown above.
The highest concentration is placed in the top left box.
The lowest concentration is placed in the bottom left box.
The desired concentration is placed in the very central box.
We get parts of the highest concentration by subtracting desired concentration and lowest concentration from each other.
Similarly, we get parts of the lowest, concentration by subtracting desired concentration and highest concentration from each other.
Adding parts of the highest concentration and parts of the lowest concentration will give us the total parts.
Now, solving our problem;
Since hydrocortisone powder is there in pure form, we can take its concentration to be 100%, and thus the highest concentration.
The lowest concentration is 2% and the desired concentration is 5%.
So, the resulting Alligation diagram will be;
100% 3 parts
5%
2% 95 parts
Since we have 1 lb of 2% hydrocortisone ointment; converting pounds into grams;
1 lb = 1 / 2.2 kg = 0.454 kg = 454 grams
Now from the Alligation diagram, it can be concluded;
To produce the desired 5% w/w hydrocortisone ointment, we will have to take 3 parts of the 100% hydrocortisone powder and 95 parts of the 2% hydrocortisone ointment and mix them together.
But we need to convert the parts into grams as asked in the question.
So,
The amount of the 2% hydrocortisone ointment is 1 lb or 454 grams.
Thus,
3 parts of HC powder x grams
------------------------------ = ------------
95 parts of 2% HC Oint. 454 grams
x = 14.336 grams
Thus we need 14.336 grams of hydrocortisone powder to be mixed with 1 lb of 2% hydrocortisone ointment to produce a 5% w/w ointment.
Thus, the correct option is c.
75) How many grams of anhydrous magnesium sulfate (MgSO4) are needed to obtain 40 g of hydrous magnesium sulfate (MgSO4.7H2O)?
a- 12.8 grams
b- 15.5 grams
c- 19.53 grams
d- 23.3 grams
e- 27 grams
Solution:
To determine how many grams of anhydrous magnesium sulfate are needed to obtain 40 g of hydrous magnesium sulfate, we can use the following equation;
Mass of MgSO4 mass of MgSO4·7H2O
----------------------------- = -----------------------------------
molar mass of MgSO4 molar mass of MgSO4·7H2O
Lets first find the molar mass of MgSO4;
24.31 + 32.06 + (4 x 16) = 120.37 g/mol
And also find the molar mass of MgSO4·7H2O;
24.31 + 32.06 + (4 x 16.00) + (7 x 18.02) = 246.48 g/mol
Mass of MgSO4 is to be found.
Mass of MgSO4·7H2O is 40 grams.
Putting the values into the equation, we get
x grams 40 grams
----------------- = -----------------
120.37 g/mol 246.48 g/mol
x = 19.53 grams
Therefore, 19.53 grams of anhydrous magnesium sulfate (MgSO4) are needed to obtain 40 g of hydrous magnesium sulfate (MgSO4.7H2O)
Thus, the correct option is c.
