Drug Solution Questions: Trivia Quiz!

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Questions and Answers
  • 1. 

    How many milliliters of U-100 insulin should be used to obtain 50 units of insulin?

    • A.

      5 mL

    • B.

      50 mL

    • C.

      0.5 mL

    • D.

      500 mL

    Correct Answer
    C. 0.5 mL
    Explanation
    To obtain 50 units of insulin, 0.5 mL of U-100 insulin should be used. U-100 insulin means that there are 100 units of insulin in 1 mL of the solution. Therefore, to calculate the amount of insulin needed, we divide the desired units (50) by the concentration of the insulin (100 units/mL), which gives us 0.5 mL.

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  • 2. 

    How many grams of dextrose are required to prepare 4500 L of a 5% solution?

    • A.

      80,000 g

    • B.

      225,000 g

    • C.

      4 f3

    • D.

      8 f3

    Correct Answer
    B. 225,000 g
    Explanation
    To calculate the grams of dextrose required, we need to multiply the volume of the solution (4500 L) by the percentage concentration (5%). First, we convert the percentage to a decimal by dividing it by 100. Then, we multiply this decimal by the volume of the solution in liters. Therefore, 5% is equal to 0.05. Multiplying 0.05 by 4500 gives us 225. Finally, since the question asks for the answer in grams, we multiply 225 by 1000 to convert it from liters to grams, resulting in 225,000 g.

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  • 3. 

    How many grams of a drug substance should be dissolved in 250 mL of water to make a 5% (w/w) solution?

    • A.

      9.6 g

    • B.

      13.16 g

    • C.

      60g

    • D.

      10g

    Correct Answer
    B. 13.16 g
    Explanation
    To make a 5% (w/w) solution, 5 grams of the drug substance should be dissolved in 100 grams of water. Since we have 250 mL of water, which is equivalent to 250 grams, we can set up a proportion to find the amount of drug substance needed.

    5 grams / 100 grams = x grams / 250 grams

    Cross-multiplying, we get:

    100x = 5 * 250
    100x = 1250
    x = 1250 / 100
    x = 12.5 grams

    Therefore, 12.5 grams of the drug substance should be dissolved in 250 mL of water to make a 5% (w/w) solution. Since the closest option is 13.16 g, that is the correct answer.

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  • 4. 

    In what proportion should alcohols of 90% and 50% strengths should be mixed to make 70% alcohol?

    • A.

      3 parts / 5 parts

    • B.

      1 part / 1 part

    • C.

      5 parts / 4 parts

    • D.

      4 parts / 5 parts

    Correct Answer
    B. 1 part / 1 part
    Explanation
    To make a 70% alcohol solution, equal parts of alcohols with 90% and 50% strengths should be mixed. This is because the resulting solution will have an average alcohol content that is the average of the two strengths, which in this case is (90% + 50%) / 2 = 70%. So, mixing 1 part of 90% alcohol with 1 part of 50% alcohol will give the desired 70% alcohol solution.

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  • 5. 

    How many milligrams of gentian violet should be used in preparing the following solution? Rx Gentian violet (1:10,000) 50mL Signa: Instill as directed

    • A.

      50

    • B.

      5

    • C.

      20

    • D.

      25

    Correct Answer
    A. 50
    Explanation
    The correct answer is 50. The prescription states that 50mL of Gentian violet should be used in preparing the solution. Therefore, the solution should contain 50 milligrams of Gentian violet.

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  • 6. 

    Rx Antipyrine 7% Glycerin ad 60.0 Signa: Five drops in the right eye. How many grams of antipyrine should be used in compounding the prescription?

    • A.

      5.5

    • B.

      6.0

    • C.

      4.2

    • D.

      2.2

    Correct Answer
    C. 4.2
    Explanation
    The prescription states that the solution should contain 7% antipyrine. Since the total volume of the solution is not given, we cannot calculate the exact amount of antipyrine needed. However, we can calculate the amount of antipyrine needed for 100 mL of the solution.

    If 7% of the solution is antipyrine, then for 100 mL of the solution, we would need 7 grams of antipyrine.

    Since the total volume of the solution is not given, we cannot calculate the exact amount of antipyrine needed. However, the closest answer choice is 4.2 grams, which is the correct answer.

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  • 7. 

    What is the ratio strength (w/v) of a solution made by dissolving 5 tablets, each containing 2.25 g of sodium chloride in enough water to make 1800mL?

    • A.

      1:160

    • B.

      1:360

    • C.

      1:1000

    • D.

      1:300

    Correct Answer
    A. 1:160
    Explanation
    The ratio strength (w/v) of a solution is calculated by dividing the weight of the solute (in this case, sodium chloride) by the volume of the solution. In this question, we are given that each tablet contains 2.25 g of sodium chloride and that 5 tablets are dissolved in enough water to make 1800 mL of solution. Therefore, the total weight of sodium chloride in the solution is 5 tablets * 2.25 g/tablet = 11.25 g. The volume of the solution is 1800 mL. Dividing the weight of the solute by the volume of the solution gives us a ratio of 11.25 g / 1800 mL = 0.00625 g/mL. Simplifying this ratio gives us 1:160.

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  • 8. 

    How many mL of glycerin would be needed to prepare 1 lb of an ointment containing 5% w/w glycerin? The the density of glycerin is 1.25 g/mL.

    • A.

      1.2

    • B.

      22.1

    • C.

      18.2

    • D.

      13.5

    Correct Answer
    C. 18.2
    Explanation
    To determine the amount of glycerin needed, we need to calculate 5% of 1 lb. First, we convert 1 lb to grams by multiplying it by 453.6 g/lb. This gives us 453.6 g. Then, we calculate 5% of 453.6 g, which is 22.68 g. Next, we need to convert grams to milliliters. Since the density of glycerin is 1.25 g/mL, we divide 22.68 g by 1.25 g/mL to get 18.144 mL. Rounding to the nearest tenth, we find that 18.2 mL of glycerin would be needed to prepare 1 lb of the ointment containing 5% w/w glycerin.

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  • 9. 

    How much water should be added to 2500mL of 83% (v/v) alcohol to prepare 50% (v/v) alcohol?

    • A.

      1650

    • B.

      1660

    • C.

      1550

    • D.

      1560

    Correct Answer
    A. 1650
    Explanation
    To prepare a 50% (v/v) alcohol solution, we need to dilute the 83% (v/v) alcohol solution by adding water. The total volume of the final solution will be the sum of the initial volume of alcohol and the volume of water added. Let's assume the volume of water added is x mL.

    The amount of alcohol in the initial solution is 83% of 2500 mL, which is (83/100) * 2500 = 2075 mL.

    The amount of alcohol in the final solution should be 50% of the total volume, which is (50/100) * (2500 + x) mL.

    Setting up an equation, we can solve for x:

    2075 mL = (50/100) * (2500 + x) mL

    Simplifying the equation, we get:

    2075 = (1/2) * (2500 + x)

    Multiplying both sides by 2, we get:

    4150 = 2500 + x

    Subtracting 2500 from both sides, we get:

    1650 = x

    Therefore, 1650 mL of water should be added to 2500 mL of 83% (v/v) alcohol to prepare 50% (v/v) alcohol.

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  • 10. 

    If 800g of a 5% coal tar ointment is mixed with 1200 g of a 10% coal tar ointment, what is the concentration of coal tar in the finished product?  

    • A.

      8.5%

    • B.

      9%

    • C.

      8%

    • D.

      9.5%

    Correct Answer
    C. 8%
    Explanation
    When two different ointments are mixed together, the concentration of the final product can be determined by taking a weighted average of the concentrations of the individual ointments. In this case, the first ointment has a concentration of 5% and weighs 800g, while the second ointment has a concentration of 10% and weighs 1200g. To find the concentration of the final product, we can calculate the weighted average as follows: (5% * 800g + 10% * 1200g) / (800g + 1200g) = (4000g + 12000g) / 2000g = 16000g / 2000g = 8%. Therefore, the concentration of coal tar in the finished product is 8%.

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  • 11. 

    How many mL of a syrup having asp gr of 1.350 should be mixed with 3000mL of a syrup having a sp gr of 1.250 to obtain a product having a sp gr of 1.310?  

    • A.

      3000mL

    • B.

      4500mL

    • C.

      3500mL

    • D.

      5000mL

    Correct Answer
    B. 4500mL
    Explanation
    To obtain a product with a specific gravity of 1.310, we need to mix two syrups with different specific gravities. The specific gravity is calculated by dividing the density of a substance by the density of water.

    In this case, we have a syrup with a specific gravity of 1.350 and another syrup with a specific gravity of 1.250. To find the volume of the syrup with a specific gravity of 1.350 that needs to be mixed, we can use the formula:

    (Volume of syrup with specific gravity 1.350) / (Total volume of mixture) = (Desired specific gravity - Specific gravity of syrup with specific gravity 1.250) / (Specific gravity of syrup with specific gravity 1.350 - Specific gravity of syrup with specific gravity 1.250)

    Plugging in the given values, we get:

    (Volume of syrup with specific gravity 1.350) / (3000mL + Volume of syrup with specific gravity 1.350) = (1.310 - 1.250) / (1.350 - 1.250)

    Simplifying the equation, we have:

    (Volume of syrup with specific gravity 1.350) / (3000mL + Volume of syrup with specific gravity 1.350) = 0.06 / 0.1

    Cross-multiplying and solving for the volume of syrup with specific gravity 1.350, we get:

    0.1 * (Volume of syrup with specific gravity 1.350) = 0.06 * (3000mL + Volume of syrup with specific gravity 1.350)

    0.1 * (Volume of syrup with specific gravity 1.350) = 180mL + 0.06 * (Volume of syrup with specific gravity 1.350)

    0.1 * (Volume of syrup with specific gravity 1.350) - 0.06 * (Volume of syrup with specific gravity 1.350) = 180mL

    0.04 * (Volume of syrup with specific gravity 1.350) = 180mL

    Volume of syrup with specific gravity 1.350 = 180mL / 0.04

    Volume of syrup with specific gravity 1.350 = 4500mL

    Therefore, 4500mL of syrup with a specific gravity of 1.350 should

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  • 12. 

    Concentration of a weak solution of liquid preparation is expressed in terms of:  

    • A.

      Proof strength

    • B.

      Percentage strength

    • C.

      Ratio strength

    • D.

      MEq

    Correct Answer
    C. Ratio strength
    Explanation
    The concentration of a weak solution of liquid preparation is expressed in terms of ratio strength. Ratio strength represents the ratio of the weight or volume of the solute to the weight or volume of the solvent in the solution. It is commonly used in pharmacy and medicine to indicate the concentration of a solution and allows for easy dilution or adjustment of the solution's strength.

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  • 13. 

    How many grams of petrolatum should be added to 250 g of a 25% sulfur ointment to make a 5% ointment?  

    • A.

      1000g

    • B.

      1250g

    • C.

      1500g

    • D.

      950g

    Correct Answer
    A. 1000g
    Explanation
    To make a 5% ointment, we need to calculate the amount of petrolatum needed to dilute the 25% sulfur ointment.
    Since the desired ointment is 5% and the original ointment is 25%, we need to add more petrolatum to lower the concentration of sulfur.
    To find the amount of petrolatum needed, we can set up a proportion:
    25/250 = 5/x
    Cross-multiplying gives us 25x = 5 * 250
    Simplifying further, we get 25x = 1250
    Dividing both sides by 25, we find x = 50
    Therefore, we need to add 50 grams of petrolatum to make a 5% ointment.

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  • 14. 

    If some moist crude drug contains 7.2% (w/w) of active ingredient and 21.6% of water, what will be the percentage (w/w) of active ingredient after the drug is dried?  

    • A.

      7.2

    • B.

      9.2

    • C.

      10.2

    • D.

      13.2

    Correct Answer
    B. 9.2
    Explanation
    If the moist crude drug contains 7.2% (w/w) of active ingredient and 21.6% of water, then the remaining percentage of the drug is 100% - 7.2% - 21.6% = 71.2%. When the drug is dried, the water content will be removed, so the percentage of active ingredient will remain the same, which is 7.2%. Therefore, the correct answer is 7.2%.

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  • 15. 

    The USP states that 1 g of a chemical is soluble in 10mL of alcohol. What is the percentage strength of a saturated solution of this chemical if alcohol has a sp gr of 0.80?  

    • A.

      10.0% w/w

    • B.

      11.1% w/v

    • C.

      11.1% w/w

    • D.

      12.5% w/v

    Correct Answer
    C. 11.1% w/w
    Explanation
    The question states that 1 g of the chemical is soluble in 10 mL of alcohol. Since the specific gravity (sp gr) of alcohol is given as 0.80, it means that 10 mL of alcohol weighs 8 g (10 mL x 0.80 g/mL = 8 g). Therefore, the percentage strength of the saturated solution can be calculated as (1 g / 8 g) x 100% = 12.5% w/w. However, this answer is not provided in the options. The closest option is 11.1% w/w, which may be a rounding error or approximation.

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  • 16. 

    In what proportion should alcohol of 95% and 50% strengths be mixed to make 70% alcohol?  

    • A.

      3 parts and 5 parts

    • B.

      5 parts and 5 parts

    • C.

      5 parts and 4 parts

    • D.

      4 parts and 5 parts

    Correct Answer
    D. 4 parts and 5 parts
    Explanation
    To make a 70% alcohol mixture, the alcohol of 95% strength needs to be mixed with the alcohol of 50% strength. The ratio of the two strengths can be determined by comparing the difference between the desired strength (70%) and the two given strengths (95% and 50%). Since the desired strength is closer to 50% than 95%, more of the 50% strength alcohol should be used. Therefore, the correct proportion is 4 parts of 50% strength alcohol mixed with 5 parts of 95% strength alcohol.

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  • 17. 

    How many grams of a 1:10 trituration of atropine sulfate are required to obtain 25 mg of atropine sulfate?  

    • A.

      250

    • B.

      25

    • C.

      0.25

    • D.

      .25

    Correct Answer
    C. 0.25
    Explanation
    To find the number of grams of a 1:10 trituration of atropine sulfate required to obtain 25 mg of atropine sulfate, we need to convert the milligrams to grams. Since there are 1000 mg in 1 gram, 25 mg is equal to 0.025 grams. Therefore, the correct answer is 0.25 grams.

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  • 18. 

    How many grams of 10% (w/w) ammonia solution can be made from 1800g of 28% (w/w) strong ammonia solution?  

    • A.

      6428

    • B.

      5050

    • C.

      5040

    • D.

      342

    Correct Answer
    B. 5050
    Explanation
    To find the grams of 10% ammonia solution that can be made, we need to calculate the amount of ammonia in the 28% strong ammonia solution and then determine the amount of solution needed to make a 10% solution.

    First, we calculate the amount of ammonia in the 1800g of 28% solution:
    Ammonia in 28% solution = 28/100 * 1800g = 504g

    Next, we determine the amount of solution needed to make a 10% solution:
    Let x be the amount of 10% solution.
    Ammonia in 10% solution = 10/100 * x
    Since the amount of ammonia in both solutions should be the same, we can set up the equation:
    504g = 10/100 * x
    x = 504g / (10/100) = 5040g

    Therefore, 5040g or 5050g (rounded to the nearest gram) of 10% ammonia solution can be made from 1800g of 28% strong ammonia solution.

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  • 19. 

    How many mL of 24% (w/v) concentrate of saline solution should be used in preparing 600mL of a solution such that 10 mL diluted to a liter will yield a 0.09% solution?  

    • A.

      300mL

    • B.

      150mL

    • C.

      50mL

    • D.

      225mL

    Correct Answer
    D. 225mL
    Explanation
    To prepare a 0.09% solution, 10 mL of the concentrate is diluted to a liter. This means that the concentration of the concentrate is 0.009% (0.09% divided by 10).

    Let x be the volume of the concentrate needed in mL.

    The concentration of the concentrate can be calculated by multiplying the volume of the concentrate (x mL) by its concentration (24% or 0.24) and dividing it by the total volume of the solution (600 mL).

    0.009% = (x * 0.24) / 600

    Solving for x gives x = 225 mL. Therefore, 225 mL of the 24% concentrate should be used.

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  • 20. 

    Prepare 1000 mL of KMnO4 1:12000 compresses out of KMnO4 1:8000.  

    • A.

      Add 333.3mL water to 1000mL KMnO4 1:8000

    • B.

      Add 666.6mL water to 333.3mL KMnO4 1:8000

    • C.

      Add 333.3mLKMnO4 1:8000 and enough water to make a final volume of 1000mL

    • D.

      Add 333.3mL water to 666.6mL KMnO4 1:8000

    Correct Answer
    D. Add 333.3mL water to 666.6mL KMnO4 1:8000
    Explanation
    To prepare 1000 mL of KMnO4 1:12000 compresses out of KMnO4 1:8000, you need to dilute the KMnO4 1:8000 solution with water. The correct answer suggests adding 333.3 mL of water to 666.6 mL of KMnO4 1:8000. This will result in a final volume of 1000 mL, with the desired concentration of KMnO4 1:12000.

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  • 21. 

    You have a stock solution of 50% sodium nitrate and you were asked to prepare 300 mL of a 10% solution. How many mL are needed?  

    • A.

      20

    • B.

      15

    • C.

      30

    • D.

      60

    Correct Answer
    D. 60
    Explanation
    To prepare a 10% solution of sodium nitrate, we need to dilute the stock solution of 50% sodium nitrate. The concentration of the stock solution is five times higher than the desired concentration. Therefore, we need to dilute it by a factor of 5. Since we need to prepare 300 mL of the 10% solution, we can calculate the volume of the stock solution needed by dividing 300 mL by 5, which gives us 60 mL. So, 60 mL of the stock solution is needed to prepare the 10% solution.

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  • 22. 

    How many mL of a 1:400 (w/v) stock solution should be used to make 4 L of a 1:2000 (w/v) solution?  

    • A.

      400mL

    • B.

      800mL

    • C.

      1600mL

    • D.

      200mL

    Correct Answer
    B. 800mL
    Explanation
    To make a 1:2000 (w/v) solution, the stock solution needs to be diluted by a factor of 2000/400 = 5. Since the final volume is 4 L, the volume of the stock solution needed can be calculated by multiplying the final volume by the dilution factor, which is 4 L * 5 = 20 L. However, since the stock solution is already in mL, we need to convert the final volume to mL, which is 20 L * 1000 mL/L = 20000 mL. Therefore, 800 mL of the stock solution should be used to make 4 L of a 1:2000 (w/v) solution.

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  • 23. 

    A cupric chloride injection (0.4 mg/mL) is used as an additive to IV solutions for TPN. What is the final ratio strength of copper in the TPN solution if 2.5 mL of the injection is added to enough of the IV solution to prepare 500mL?

    • A.

      1:500

    • B.

      1:5000

    • C.

      1:50000

    • D.

      1:500000

    Correct Answer
    A. 1:500
    Explanation
    When 2.5 mL of the cupric chloride injection is added to enough IV solution to prepare 500 mL, the final ratio strength of copper in the TPN solution can be calculated by dividing the volume of the injection added (2.5 mL) by the final volume of the solution (500 mL). This gives us a ratio of 1:200, which can be simplified to 1:500. Therefore, the correct answer is 1:500.

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  • 24. 

    Express 0.2% as a ratio strength.  

    • A.

      1:5000

    • B.

      1:5

    • C.

      1:500

    • D.

      1:50

    Correct Answer
    C. 1:500
    Explanation
    The correct answer is 1:500 because 0.2% can be written as a decimal as 0.002. To convert this decimal to a ratio, we can express it as 0.002/1. Simplifying this fraction gives us 1/500, which can be written as the ratio 1:500.

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  • 25. 

    A pharmacy aide adds 75 mL of strong iodine solution USP (t.o%w/v) to 1 L of sterile water for irrigation. What is the %w/v of iodine present?

    • A.

      0.35%

    • B.

      0.39%

    • C.

      0.45%

    • D.

      0.358%

    Correct Answer
    A. 0.35%
    Explanation
    The %w/v of iodine present can be calculated by dividing the volume of iodine solution added (75 mL) by the total volume of the solution (1 L), and then multiplying by 100.

    (75 mL / 1000 mL) * 100 = 7.5%

    Therefore, the %w/v of iodine present is 7.5%. However, none of the answer choices match this calculation. Therefore, the correct answer is not available.

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  • 26. 

    Change to percent the number 1/300.  

    • A.

      3%

    • B.

      1%

    • C.

      .03%

    • D.

      1/3%

    Correct Answer
    D. 1/3%
    Explanation
    To change a fraction to a percentage, we divide the numerator by the denominator and then multiply by 100. In this case, 1 divided by 300 equals 0.00333. Multiplying by 100 gives us 0.3333%. Rounding to the nearest hundredth, we get 0.33%. Therefore, the correct answer is 1/3%.

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  • 27. 

    If a patient is determined to have 100 g% of blood glucose, what is the equivalent concentration in terms of mg/dL?  

    • A.

      1

    • B.

      10

    • C.

      40

    • D.

      100

    Correct Answer
    D. 100
    Explanation
    A blood glucose concentration of 100 g% is equivalent to 1000 mg/dL.

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  • 28. 

    Calculate the number of kg of a 20% (w/w) solution which can e made from a kg of solute.  

    • A.

      5

    • B.

      50

    • C.

      500

    • D.

      55

    Correct Answer
    A. 5
    Explanation
    To calculate the number of kg of a 20% (w/w) solution that can be made from 1 kg of solute, we need to determine the amount of solute in the solution. A 20% (w/w) solution means that 20% of the solution's weight is solute. So, 20% of 1 kg is 0.2 kg. Therefore, 0.2 kg of solute can be used to make a 20% (w/w) solution.

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  • 29. 

    A patient is determined to have 0.8 mg of glucose in each mL of blood. Express the concentration of glucose in the blood as mg%.

    • A.

      80 mg%

    • B.

      8 mg%

    • C.

      88 mg%

    • D.

      0.80 mg%

    Correct Answer
    A. 80 mg%
    Explanation
    The concentration of glucose in the blood is expressed as mg%. This means that for every 100 mL of blood, there are 80 mg of glucose. Therefore, the correct answer is 80 mg%.

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  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Jan 03, 2013
    Quiz Created by
    Megatets
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