P1 OSCE Calculations Exam at Chicago State University School of Pharmacy practice questions with answers

P1 OSCE Calculations Exam at Chicago State University School of Pharmacy

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Free P1 OSCE Calculations Exam at Chicago State University School of Pharmacy Questions

A metronidazole vaginal gel contains 0.75% of drug in 70-g tubes.
How many milligrams of drug will be contained in 10 g of application?
Round answer to the nearest whole number. Do not include units.

A 50
B 65
C 75
D 90

Explanation

A 0.75% w/w preparation contains 0.75 g per 100 g of product.
To determine how much drug is in 10 g:
0.75 g ÷ 100 g = 0.0075 g per gram.
0.0075 g × 10 g = 0.075 g, which equals 75 mg.
Therefore, 75 mg of metronidazole is contained in a 10-gram dose.

Correct Answer Is:

C. 75

A manufacturer wishes to prepare 3 L of sodium acetrizoate solution, 25.0% w/v. The specific gravity of this solution is 1.1.
How many milliliters of water will be required?
Round to the nearest whole number. Do not include units.

A 2200
B 2400
C 2550
D 2700

Explanation

To prepare 3 liters (3000 mL) of a 25% w/v solution, calculate the amount of solute first:
25 g per 100 mL × 3000 mL = 750 g of solute.
Next determine the total mass of the final solution using specific gravity:
3000 mL × 1.1 g/mL = 3300 g total.
Subtract the solute mass from the total mass to determine required water:
3300 g − 750 g = 2550 g, which equals 2550 mL of water.

Correct Answer Is:

C. 2550

Syrup is an 85% w/v solution of sucrose in water. It has a density of 1.313 g/mL.
How many milliliters of water should be used to make 300 mL of syrup?
Round to the nearest tenth. Do not include units.

A 120.5
B 138.9
C 150.0
D 165.2

Explanation

To find the amount of water needed, first determine the total mass of 300 mL of syrup:
300 mL × 1.313 g/mL = 393.9 g.
An 85% w/v syrup contains 85 g sucrose per 100 mL, so for 300 mL:
85 g × 3 = 255 g sucrose.
Subtract sucrose mass from total mass to find water mass:
393.9 g − 255 g = 138.9 g water.
Because water has a density of 1 g/mL, this equals 138.9 mL.

Correct Answer Is:

B. 138.9

A patient has a 200 g tube of topical lidocaine with a concentration of 2%. If the patient is applying 2 g three times daily, what is the total daily dose of lidocaine in milligrams? Round to the nearest whole number. Do not include units.

A 80
B 100
C 120
D 150

Explanation

A 2% w/w lidocaine preparation contains 2 g of lidocaine per 100 g of product, which equals 0.02 g per gram.
The patient applies 2 g per dose, and 2 g × 0.02 = 0.04 g lidocaine per dose.
Convert to milligrams: 0.04 g = 40 mg per dose.
Since the patient applies this three times daily:
40 mg × 3 = 120 mg per day.

Correct Answer Is:

C. 120

A prescription calls for 5 mg of morphine sulfate. Using a sensitivity requirement of 5 mg, how would you obtain the required amount of morphine sulfate with an error not greater than 5%?

A 125 mg of drug, 1675 mg of diluent, 80 mg of aliquot
B 80 mg of drug, 1920 mg of diluent, 80 mg of aliquot
C 100 mg of drug, 1900 mg of diluent, 100 mg of aliquot
D 120 mg of drug, 2000 mg of diluent, 100 mg of aliquot
E 150 mg of drug, 2250 mg of diluent, 100 mg of aliquot

Explanation

With a sensitivity requirement (SR) of 5 mg and a maximum error of 5%, the minimum quantity that can be weighed accurately is: (SR × 100) / % error = (5 × 100) / 5 = 100 mg. Therefore, the aliquot taken must be at least 100 mg. In option C, 100 mg drug + 1900 mg diluent = 2000 mg mixture. A 100 mg aliquot of this mixture contains (100/2000) × 100 mg = 5 mg morphine sulfate, meeting both the dose requirement and the weighing accuracy requirement.

Correct Answer Is:

C. 100 mg of drug, 1900 mg of diluent, 100 mg of aliquot

A prescription calls for 50 milligrams of chlorpheniramine maleate. Using a prescription balance with a sensitivity requirement of 6 milligrams, explain how you would obtain the required amount of chlorpheniramine maleate with an error not greater than 5%.

A Weigh 600 mg chlorpheniramine maleate, dilute with 850 mg of diluent to make 1600 mg, then weigh 1200 mg
B Weigh 150 mg chlorpheniramine maleate, dilute with 450 mg of diluent to make 600 mg, then weigh 200 mg
C Weigh 60 mg chlorpheniramine maleate, dilute with 50 mg of diluent to make 100 mg, then weigh 20 mg
D Weigh 760 mg chlorpheniramine maleate, dilute with 950 mg of diluent to make 6000 mg, then weigh 2000 mg
E Weigh 120 mg chlorpheniramine maleate, dilute with 780 mg of diluent to make 1000 mg, then weigh 200 mg

Explanation

With a sensitivity requirement of 6 mg and a maximum error of 5%, the minimum quantity that can be weighed accurately is (SR × 100) / %error = (6 × 100) / 5 = 120 mg. So both the initial drug weight and the final aliquot must be ≥ 120 mg. In option B, 150 mg drug (≥120 mg) is mixed with 450 mg diluent to make 600 mg total. The concentration is 150/600 = 0.25 (25%). Taking a 200 mg aliquot (also ≥120 mg) gives 0.25 × 200 = 50 mg of chlorpheniramine maleate, meeting the dose requirement while satisfying the balance’s accuracy limits.

Correct Answer Is:

B. Weigh 150 mg chlorpheniramine maleate, dilute with 450 mg of diluent to make 600 mg, then weigh 200 mg

The pharmacist is tasked with making an elixir that is 1:2000 (w/v) of phenobarbital. The pharmacy has 300 mL of a 0.5% elixir to be used for this. How many milliliters of the 1:2000 (w/v) elixir can be made? (Round to the nearest whole number. Do not include units.)

A 1500
B 2000
C 2500
D 3000

Explanation

A 0.5% w/v phenobarbital elixir contains 0.5 g per 100 mL, which is 500 mg per 100 mL, or 5 mg/mL.
With 300 mL available:
5 mg/mL × 300 mL = 1500 mg total drug.
A 1:2000 solution contains 1 g in 2000 mL, or 0.5 mg/mL.
To find the final volume that will contain 1500 mg at a concentration of 0.5 mg/mL:
1500 mg ÷ 0.5 mg/mL = 3000 mL. Therefore, 3000 mL of the 1:2000 elixir can be prepared.

Correct Answer Is:

D. 3000

A solution contains 0.4% w/v of active ingredient.
How many kilograms of active ingredient are needed to prepare 800 L of solution?
Round answer to the nearest tenth. Do not include units.

A 1.6
B 2.4
C 3.2
D 4.0

Explanation

A 0.4% w/v solution contains 0.4 g per 100 mL.
Convert 800 L to milliliters: 800 L = 800,000 mL.
Determine grams per milliliter: 0.4 g / 100 mL = 0.004 g/mL.
Multiply by total volume:
0.004 g/mL × 800,000 mL = 3200 g.
Convert to kilograms: 3200 g ÷ 1000 = 3.2 kg.

Correct Answer Is:

C. 3.2

The following is a formula for a testosterone nasal spray:
Testosterone 1 g
Alcohol 10 mL
Propylene glycol 20 mL
Benzalkonium chloride 15 mg
Purified water qs ad 100 mL

Benzalkonium chloride is available as a 1:500 w/v stock solution.
How many milliliters of the stock solution would provide the 15 mg needed in the prescription?
Round to the nearest tenth. Do not include units.

A 5.0
B 7.5
C 10.0
D 12.5

Explanation

A 1:500 w/v solution contains 1 gram (1000 mg) in 500 mL, which equals 2 mg/mL. To supply the required 15 mg, divide the needed amount by the concentration:
15 mg ÷ 2 mg/mL = 7.5 mL.
Therefore, 7.5 mL of the stock benzalkonium chloride solution will provide the exact required amount for the nasal spray formulation.

Correct Answer Is:

B. 7.5

10.

A liquid medicine is to be taken three times daily. If 180 mL are to be taken in 4 days, how many teaspoonfuls should be prescribed for each dose? Round to the nearest whole number.

A 2 teaspoonfuls
B 3 teaspoonfuls
C 4 teaspoonfuls
D 5 teaspoonfuls

Explanation

To calculate the correct dose, first determine how many total doses the patient will take over 4 days: 3 doses per day × 4 days = 12 doses. Then divide the total volume by the number of doses: 180 mL ÷ 12 = 15 mL per dose. Since one teaspoon equals 5 mL, convert by dividing 15 mL ÷ 5 mL = 3 teaspoonfuls per dose.

Correct Answer Is:

B. 3 teaspoonfuls

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P1 OSCE Calculations Exam at Chicago State University School of Pharmacy

P1 OSCE Calculations Exam at Chicago State University School of Pharmacy

P1 OSCE Calculations Exam at Chicago State University School of Pharmacy practice questions with answers | nursingprepplug.com

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