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
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
What is the concentration, in ratio strength, of a trituration made by combining 100 mg of ibuprofen and 2.5 grams of talcum powder? Use the format 1:XX
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
Three 100 mL vials of calcium chloride 10% are added to normal saline to prepare an IV solution. How many milliequivalents of calcium are contained in this solution? Round answer to the nearest tenth. (MW of CaCl2 = 111)
Explanation
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 ✓
A pharmacist adds 5.3 grams of hydrocortisone to 150 grams of 2.5% hydrocortisone ointment. What is the percentage strength (w/w) of hydrocortisone in the finished product? Round to the nearest whole number. Do not include the percent sign.
Explanation
Correct Answer: 5
Step 1: Find grams of hydrocortisone already in the 150g ointment: 150 g × 2.5% = 3.75 g
Step 2: Add the additional hydrocortisone: 3.75 g + 5.3 g = 9.05 g
Step 3: Calculate new total weight: 150 g + 5.3 g = 155.3 g
Step 4: Calculate new percentage strength: 9.05 g ÷ 155.3 g × 100 = 5.8% ≈ 6%
Correction upon rounding: 5.82% rounds to 6 ✓
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.
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A 120.5
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B 138.9
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C 150.0
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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. 138.9How many grams of hydrocortisone should be added to 2500 g of a 1% hydrocortisone cream to prepare a product containing 2.5% of hydrocortisone. Round answer to the nearest TENTH. Do not include units.
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 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%.
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A Weigh 600 mg chlorpheniramine maleate, dilute with 850 mg of diluent to make 1600 mg, then weigh 1200 mg
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B Weigh 150 mg chlorpheniramine maleate, dilute with 450 mg of diluent to make 600 mg, then weigh 200 mg
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C Weigh 60 mg chlorpheniramine maleate, dilute with 50 mg of diluent to make 100 mg, then weigh 20 mg
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D Weigh 760 mg chlorpheniramine maleate, dilute with 950 mg of diluent to make 6000 mg, then weigh 2000 mg
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E Weigh 120 mg chlorpheniramine maleate, dilute with 780 mg of diluent to make 1000 mg, then weigh 200 mg
Explanation
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 mgHow many milliosmoles of Na₂SO₄ (M.W. = 142) are represented in 200 mL of a 5% (w/v) sodium sulfate solution? Round to the nearest TENTH. Do not include units.
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 ✓
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.
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A 2 teaspoonfuls
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B 3 teaspoonfuls
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C 4 teaspoonfuls
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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 teaspoonfulsThe 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.
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A 5.0
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B 7.5
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C 10.0
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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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