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. 138.9A 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. 120A 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
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. 2550Explanation
Correct Answer: 270.3
Step 1: Total volume of CaCl2 solution: 3 vials × 100 mL = 300 mL
Step 2: Grams of CaCl2 in 300 mL of 10% solution: 300 mL × 0.10 = 30 g = 30,000 mg
Step 3: CaCl2 dissociates providing Ca²⁺ (valence = 2): mEq = (mg × valence) ÷ MW mEq = (30,000 × 2) ÷ 111 = 540.5 mEq
Note: If asking specifically for calcium ion only (valence 2, atomic weight 40): 30,000 mg ÷ 111 × 2 = 540.5 mEq of calcium ✓
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 ✓
Explanation
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
Explanation
Correct Answer: 38.5
Step 1: Grams of hydrocortisone in original 2500 g of 1% cream: 2500 × 0.01 = 25 g
Step 2: Let x = grams of hydrocortisone to add. Final weight = 2500 + x Final concentration must be 2.5%: (25 + x) ÷ (2500 + x) = 0.025
Step 3: Solve for x: 25 + x = 0.025(2500 + x) 25 + x = 62.5 + 0.025x 0.975x = 37.5 x = 37.5 ÷ 0.975 = 38.5 g ✓
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. 3.2Explanation
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 ✓
Rx: 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
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