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
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
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. 138.9Explanation
Correct Answer: 5
Step 1: 1 tablespoon = 15 mL Step 2: Grams of MgSO4 in 15 mL of 2% solution: 15 mL × 0.02 = 0.3 g = 300 mg
Step 3: Milliequivalents of Mg²⁺ (valence = 2): mEq = (mg × valence) ÷ MW mEq = (300 × 2) ÷ 120 = 5 mEq ✓
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
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. 120Rx: Oxymetazoline Hydrochloride (E=0.20) — 0.5% Boric Acid Solution (E=0.52) — q.s. Purified Water ad — 15 mL Make isoton. Sol. Sig: For the nose, as decongestant. Round to the nearest hundredth. Do not include units.
Explanation
Correct Answer: 4.33
Step 1: Calculate NaCl equivalents already contributed by oxymetazoline: 0.5% in 15 mL = 0.075 g × E(0.20) = 0.015 g NaCl equivalent
Step 2: NaCl needed for isotonicity in 15 mL: 15 mL × 0.9% = 0.135 g NaCl
Step 3: Remaining NaCl needed from boric acid: 0.135 − 0.015 = 0.12 g NaCl equivalent
Step 4: Grams of boric acid needed: 0.12 ÷ 0.52 = 0.2308 g boric acid
Step 5: Volume of 5% boric acid solution needed: 0.2308 g ÷ 0.05 g/mL = 4.62 mL
Explanation
Correct Answer: 211.3
Step 1: Grams in 200 mL of 5% solution: 5 g/100 mL × 200 mL = 10 g
Step 2: Convert to millimoles: 10,000 mg ÷ 142 mg/mmol = 70.4 mmol
Step 3: Na₂SO₄ dissociates into 3 particles (2 Na⁺ + 1 SO₄²⁻): 70.4 mmol × 3 = 211.3 mOsmol ✓
Explanation
Correct Answer: 9
Step 1: Convert weight: 165 ÷ 2.2 = 75 kg
Step 2: Calculate dose: 75 kg × 6 mg/kg = 450 mg
Step 3: Calculate volume: 450 mg ÷ 50 mg/mL = 9 mL ✓
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
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 teaspoonfulsExplanation
Correct Answer: 1:26
Total mixture = 100 mg + 2,500 mg = 2,600 mg 100 mg / 2,600 mg = 1/26 Ratio strength = 1:26
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
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. 3.2The 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
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. 7.5How to Order
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