NAMA : NENY SANDRAWATI

BP : 1111012043

KELAS : REGULER I

SOAL DAN PEMBAHASAN MATEMATIKA FARMASI BAB III

“ Using Percentages in Dosage Calculations “

1. How many grams of salicylic acid and benzoic acid should be used in preparing 100 grams of an ointment containing 12% and 6% of each, respectively?Answer: salicylic acid = 12 %

12100

= x100

= 100 x =1200 x = 12 g

Benzoic acid = 6%

6100

= x100

= 100 x = 600 x= 6 g

So, 12 g salicylic acid and 6 g benzoic acid should be used in preparing 100 grams of an ointment containing 12% and 6% of each respectively.

2. How many grams of boric acid should be used in compounding the following prescription:R

Boric acid 2: 100Aqua add to 90 mL

Answer: 2100

= x90

x = 2x 90100

x = 1.8So, boric acid should be used in compounding the prescription is 1.8 g

3. What is the amount of potassium permanganate (in g) that should be used in compounding the following prescription?R

potassium permanganate 0.03%Aqua ad 600 mL

Answer : 0.03100

= x600

x = 0.03 x600100

x = 0.18 so, the amount of potassium permanganate (in g) that should be used in compounding the prescription is 0.18 g

4. What is the percentage strength of boric acid if 50 grams are dissolved in 350 mL of deionized distilled water?

Answer : 50350

= x100

x = 5x 100350

x = 14.3 so, the percentage strength of boric acid is 14.3%

5. In preparing 500 mL of a topical formulation, a community pharmacist used 8 mL of liquid phenol. What is the percentage strength (v/v) of the liquefied phenol in this topical formulation?

Answer: 8500

= x100

x = 8 x100500

x = 1.6so, percentage strength liquefied phenol in this topical formulation

(v/v) is 1.6%

6. In preparation of 100 mL mouthwash, a pharmacist used 250 mg of menthol. What is the percentage (w/v) of menthol in this mouthwash?

Answer: 100250

= x100

x = 100x 100250

x = 0.25so, the percentage (w/v) of menthol in this mouthwash is 0.25%

7. What is the amount of codeine sulfate to be used in compounding the following prescription?R

Codeine sulfate 2%Aqua add to 150 mL

Answer: 2100

= x150

x = 2x 150100

x = 3so, the amount of codeine sulfate is 3 g

8. What is the amount of methyl salicylate in (mL) that should be used in compounding the following prescription?R

Methyl salicylate 6%Isopropanol 150 mL

Answer: 6100

= x150

x = 6 x150100

x = 9so, the amount of methyl salicylate in (mL) is 9 mL

9. What is the amount of peppermint oil (in mL) that should be used in compounding aromatic water in the following prescription?R

Peppermint oil 2%Water to 220 mL

Answer: 2100

= x220

x = 2x 220100

x = 4.4so, the amount of peppermint oil (in mL) is 4.4 mL

10. How many grams of drug A should be dissolved in 140 mL of distilled water to prepare 8% (w/v) solution?

Answer : 8100

= x140

x = 8 x140100

x = 11.2so, drug A should be dissolved in 140 mL of distilled water to prepare 8%

(w/v) solutionis 11.2 g

11. How many grams of drug substance are needed to prepare 200 mL of a 10% (w/w) solution in water?

Answer: 10100

= x200

x = 10x 200100

x = 20 so, drug substance are needed to prepare 200 mL of a 10% (w/w) solution

in water is 20 g

12. How many grams of a dorzolamide are needed to prepare 60 mL of 2% (w/w)ophthalmic solution in isotonicthe solution?

Answer: 2100

= x60

x = 2x 60100

x = 1.2so, a dorzolamide are needed to prepare 60 mL of 2% (w/w)ophthalmic

solution in isotonicthe solution is 1.2 g

13. Express each of following concentrations as a ratio strength:a. 2 g of active ingredient in 50 mL of the solutionb. 0.2 mg of active ingredient in 2 mL of the solutionc. 25 mg of active ingredient in 5 mL of the solutiond. 1 mg of active ingredient in 1 mL of the solution

Answer : a. 250

= x100

x = 2x 10050

x = 4

∴ concentrations is 4%

b. 0.2 mg = 0.0002 g

0.00022

= x100

x = 100x 0.0002

2

x = 0.01

∴ concentration is 0.01%

c . 25 mg = 0.025 g

0.0255

= x100

x = 100x 0.025

5x = 0.5

∴ concentration is 0.5%

d d . 1 mg = 0.001 g

0.0011

= x100

x = 100x 0.001

1x = 0.1

∴ concentration is 0.1%

14. How many milligrams of a benzalkonium chloride should be used in preparing 5 liters of a 0.001% solution?

Answer: 0.001100

= x5000

x = 0.001x 5000

100x = 50so, milligrams of a benzalkonium chloride is 50 mg

15. A liquid medication contains 2.5 mg of fluoxetine hydrochloride per 75 milliliters. Express the milligram percent (mg%) concentration of this preparation.

Answer: 2.575

= x100

x = 2.5x 10075

x = 3.3so, milligram percent (mg%) concentration of this preparation is 3.3 mg

%

16. An injection contains 50 mg pentobarbital sodium in each mL of solution. What is he percentage strength (w/v) of the solution?Answer: 50 mg = 0.05 g

0.051

= x100

x = 0.05 x100

1 x = 5so, the percentage strength (w/v) of the solution is 5%

17. If 500 g of dextrose monohydrate are dissolved in enough water to make 20 L, what is the percentage strength (w/v) of the solution?

Answer: 50020000

= x100

x = 500x 10020000

x = 2.5so, the percentage strength (w/v) of the solution is 2.5%

18. how many milliliters of a 0.9% (w/v) of sodium chloride can be prepared from

11.25 g of sodium chloride?

Answer: 0.9100

= 11.25x

x = 100x 11.25

0.9x = 1250 mLso, a 0.9% (w/v) of sodium chloride can be prepared from 11.25 g of

sodium chloride is 1250 mL

19. If 50 g magnesium citrate are dissolved in enough water to make 2.5 L, what is the percentage strength (w/v) of the solution?

Answer: 502500

= x100

x = 50x 1002500

x = 2so, the percentage strength (w/v) of the solution is 2%

20. What is the percentage strength (w/w) of a solution made by dissolving 250 g of potassium chloride in 750 mL of water?

Answer: 250750

= x100

x = 250x 100750

x = 33.33 so, the percentage strength (w/w) of a solution is 33.33%

21. If 100 mg of docusate sodium are dissolved in enough water and the volume is made up to 10 ml what is percentage strength (w/w) of the solution?

Answer : 100mgxmg

= 10ml100ml

x=100 x10010

=1000mg%atau 1gr%

So, the percent (w/w) strength of a solution is 1%

22. a formula for a mouth rinse contains 1/10 % (w/v) of zinc chloride. How many grams of zinc chloride should be used in preparing 10 liters of the mouth rinse?

Answer: 1/10% = 0,1%= 0,1 gr in 100 ml

0,1100

= x10.000

x=0,1x 10.000100

=10 g

So, grams of zinc chloride is 10 g

23. what is the percentage strength (w/w) of a solution made by dissolving 0,3 g of cimetidine hydrochloride in 90 g of water?

Answer: 0,3gr90gr

= 100%x%

x= 0,390 x100

=0.33%

So, the percent (w/w) strength of a solution is 0.33%

24. how many milliliters of resorcinol monoacetate liquid should be used to prepare a liter of a 15% (v/v) lotion?

Answer: 15100

= x ml1000ml

x=15 x1000100

=150ml

So, resorcinol monoacetate liquid should be used to prepare a liter of a 15% (v/v) lotion is 150 mL

25. how many milligrams of sodium fluoride must be dissolved in 500ml of water to make 0,05% (w/w) solution?

Answer: 5mg

10000ml= xmg500ml

x=5 x50010000

=0,25mg= 250 g

So, sodium fluoride must be dissolved in 500ml of water to make 0,05% (w/w) solution is 250 g

26. if one liter of a solution contains 250 g of active drug, what is its percentage strength expressed as w/v?

Answer: 250gx

=1000ml100ml

x=250 x1000100

= 25 %

So, the percent (w/v) strength of a solution is 25%

27. how many grams of calcium carbonate should be usedin preparing 100 ml of 10% calcium carbonate suspension?

Answer: 10100

= x gr100

x=10 x100100

=10 g

So, calcium carbonate should be usedin preparing 100 ml of 10% calcium carbonate suspension is 10 g

28. if a patient is instructed to take 250 ml of 16% w/v of an oral solution, how many grams of the drug will he consume?

Answer: 16100

= x gr250ml

x=16 x250100

=40g

So, grams of the drug will consume is 40 g

29. how many grams of iodine are represented in a 20 ml of 20% w/v iodine solution?

Answer: 20100

= x20ml

x=20 x20100

=4 g

So, grams of iodine is 4 g

30. if 230 ml of a solution contain 46 mg of drug, what is the strength of this solution solution in percent (w/v)?

Answer: 46 grx

=230ml100ml

x=46 x 100230

=¿ 20 mg% = 0.02 gr%

So, the percent (w/v) strength of a solution is 0.02%

31. if 250 ml of a 0,1% solution is mixed with 150 ml of a 0,3% solution, what will be the strength (percent w/v) of the final solution?

Answer: 0.1100

= x250

x = 0.1x 250100

x = 0.25%

0.3100

= y150

y = 0.3 x150100

y = 0.45%

z = x + y = 0.25% + 0.45% = 0.75%

so, the percent (w/v) strength of a solution is 0.75%

32. what is the percent (w/v) strength of a solution that has 500 mg of drug in 200 ml?

Answer: 500mgxmg

=200ml100ml

x=500 x100200

=250mg%∨0.25 gr%

So, the percent (w/v) strength of a solution is 250 mg% or 0.25%

33. what is the percentage strength (in w/v) of a 5 mg in 5 ml solution?

Answer: 5mgx

= 5ml100ml

x=5 x1005

=100mg%atau 0,1gr%

So, the percentage strength is 100mg%∨0,1gr%

34. while preparing 250 ml of 2% (w/v) solution, a pharmacist mistakenly used only 4,5 g of drug. How many more milligrams of this drug are needed to get the correct strength?

Answer: 2100

= x250ml

x=2x 250100

=5 gr=5000mg

4,5 gr= 4500 mg

So, drug needed to get the correct strength is 5000 mg – 4500 mg= 500 mg