NaCl needed to make isotonic, One drug
8. Buffer and Isotonic Solution 8.2) Isotonic Solution 8.2.2) NaCl needed, One drug
8.2.2.1) NaCl needed, One drug Easy 1
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Calculate the gram amount of NaCl required to make the following sterile product isotonic.
Rx:
Drug ABC: 1.15 %
NaCl qs
Aqua dist q.s. a.d. 100 mL
M.ft. isotonic solution
Provided, the E value of the drug is 0.53 .
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lb equals 0.291 g kg
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Answer: 0.2905 grams of NaCl will be needed.
- Volume of the product: 100 mL
- Concentration of the drug: 1.15 %.
- Amount of NaCl needed to prepare an isotonic solution of this volume: 0.9/100 × 100 = 0.9 g
- Quantity of the drug present in the solution: 1.15 /100 × 100 = 1.15 g
- E value of the drug (provided): 0.53
- Now, the NaCl-equivalent tonicity contributed by the drug will be the product of drug's weight in g (1.15) and its E value ( 0.53 ), which is: 1.15 × 0.53 = 0.6095 g
- Therefore, the quantity of additional amount of NaCl required to make the product isotonic: 0.9 g - 0.6095 g = 0.2905 g