Homework 8 altering product strength part A; potency key.pdf - Homework 8 – Altering product strength part A potency KEY 1 If a gallon of a 30 w/v

# Homework 8 altering product strength part A; potency...

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1 Homework 8 – Altering product strength part A; potency - KEY 1. If a gallon of a 30% w/v solution is to be evaporated so that the solution will have a strength of 50% w/v, what will be its volume in milliliters? Answer Q1 x C1 = Q2 x C2 1 gallon = 3,785 mL 3,785 mL x 30% = V mL x 50% V mL = 3,785 mL x 30 mL x 100 mL = 2,271 mL, answer 100 mL 50 mL 2. How many milliliters of a 15% (w/v) ethanol solution should be diluted with water to yield 750 mL of a 1:50 solution? Answer Let’s write down all the information that we have in the form of the QC equation. The ratio strength will be w/v because the percentage is also w/v. V mL x 15% w/v = 750 mL x 1:50 w/v Let’s express the concentrations as fractions using their definitions V mL x 15 g = 750 mL x 1 g 100 mL 50 mL Let’s use algebra to solve this equation V mL = 750 mL x 1 g x 100 mL = 100 mL, answer 50 mL 15 g 3. How many milliliters of a diluent should be added to 25 mL of a 5% w/v drug solution in order to prepare a solution with a final concentration of 1g/100 mL? Answer Since this is a practice problem of dilution, let’s first write the QC equation 25 mL x 5% w/v = V mL x 1 g 100 mL Let’s express the percentage as a fraction 2 25 mL x 5 g = V mL x 1 g 100 mL 100 mL Let’s solve the equation using algebra V = 25 mL x 5 g = 125 mL 1 g Since the final volume is 125 mL and we start with 25 mL, volume of diluent is 125 mL – 25 mL = 100 mL, answer 4. How many milliliters of a 1:2,500 w/v solution of the phenylmercuric acetate can be made from 125mL of a 0.2% solution? Answer The qualifier of the percent will be w/v because this is the qualifier of the ratio strength Let’s write the information in the form of the QC equation 125 mL x 0.2% w/v = V mL x 1:2,500 w/v Let’s express concentrations as fractions 125 mL x 0.2 g = V mL x 1 g 100 mL 2,500 mL Let’s use algebra to solve the equation V mL = 125 mL x 0.2 g x 2,500 mL 100 mL 1 g V = 625 mL, answer 5. How many milliliters of a 50% w/v stock solution of Atropine are needed to prepare 250 mL of a solution containing 25 mg of drug per milliliter? Answer Let’s write the information in the form of the QC equation. V mL x 50% w/v = 250 mL x 25 mg/mL We will now express the percent as a fraction, while changing the mg to g 1 g = W g W = 0.025 g 3 1,000 mg 25 mg V mL x 50 g = 250 mL x 0.025 g 100 mL 1 mL Arranging the equation V = 250 mL x 0.025 g x 100 mL = 12.5 mL, answer 50 g 6. How many milliliters of a diluent should be added to 10 mL of a 5% w/v antibiotic solution in order to prepare a solution with a final concentration of 2.5 g/100 mL? [Please provide the answer in mL and not ml] Answer Since this is a practice problem of dilution, let’s first write the QC equation 10 mL x 5% w/v = V mL x 2.5 g 100 mL Let’s use the definition of w/v as gram in 100 mL, to re-write the equation to 10 mL x 5 g = V mL x 2.5 g 100 mL 100 mL V mL = 10 mL x 5 g = 20 mL 2.5 g Since the final volume is 20 mL and we start with 10 mL 20 mL – 10 mL = 10 mL We will need to add 10 mL of diluent, answer 7. How many milliliters of normal saline should be added to 75 mL of a solution containing 1,500 mg of atropine sulfate to make a 0.05% (w/v) solution?  #### You've reached the end of your free preview.

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