NaCl needed to make isotonic, Two drugs
8. Buffer and Isotonic Solution 8.2) Isotonic Solution 8.2.3) NaCl needed, Two drugs
8.2.3.1) NaCl needed, Two drugs Easy 1
Total tried: Correct: Wrong:
Calculate the gram amount of NaCl required to make the following sterile product isotonic.
Rx:
Drug PQR: 0.85 %
Drug XYZ: 0.25 %
NaCl qs
Aqua dist q.s. a.d. 20 mL
M.ft. isotonic solution
Provided, the E value of the drug PQR is 0.04, and the E value of the drug XYZ is 0.456.
Click on the button below to see the answer and explanations
Notice: Undefined variable: input_value1 in /home/jshzub/web/rxcal.org/public_html/practice/practice/includes/body.php on line 220
lb equals 0.150 g kg
-
Answer: 0.150 g grams of NaCl will be needed.
- Volume of the solution: 20 mL
- Amount of NaCl needed to prepare an isotonic solution of this volume: 0.9/100 × 20 = 0.18 g
- Concentration of the PQR drug: 0.85 %.
- Quantity of the PQR drug present in the solution: 0.85 /100 × 20 = 0.17 g
- E value of the PQR drug (provided): 0.04
- Now, the NaCl-equivalent tonicity contributed by the PQR drug will be the product of drug's weight in g (0.17) and its E value ( 0.04 ), which is: 0.17 × 0.04 = 0.0068 g
- Quantity of the XYZ drug present in the solution: 0.25 /100 × 20 = 0.05 g
- E value of the XYZ drug (provided): 0.456
- Now, the NaCl-equivalent tonicity contributed by the XYZ drug will be the product of drug's weight in g (0.05) and its E value ( 0.456 ), which is: 0.05 × 0.456 = 0.0228 g
- Therefore, the quantity of additional amount of NaCl required to make the product isotonic: 0.18 g - 0.0068 g - 0.0228 g = 0.1504 g, rounded to 0.150 g g