Notice: A non well formed numeric value encountered in /home/jshzub/web/rxcal.org/public_html/practice/practice/7320181004063653-94-rxcal-53.php on line 35

Notice: A non well formed numeric value encountered in /home/jshzub/web/rxcal.org/public_html/practice/practice/7320181004063653-94-rxcal-53.php on line 36

Notice: A non well formed numeric value encountered in /home/jshzub/web/rxcal.org/public_html/practice/practice/7320181004063653-94-rxcal-53.php on line 37

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: 1.05 %
      Drug XYZ: 0.85 %
      NaCl qs
      Aqua dist q.s. a.d. 50 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 0.076.


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 0.397 g kg

    Answer: 0.397 g grams of NaCl will be needed.
    • Volume of the solution: 50 mL
    • Amount of NaCl needed to prepare an isotonic solution of this volume: 0.9/100 × 50 = 0.45 g
    • Concentration of the PQR drug: 1.05 %.
    • Quantity of the PQR drug present in the solution: 1.05 /100 × 50 = 0.525 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 (0.525) and its E value ( 0.04 ), which is: 0.525 × 0.04 = 0.021 g
    • Quantity of the XYZ drug present in the solution: 0.85 /100 × 50 = 0.425 g
    • E value of the XYZ drug (provided): 0.076
    • Now, the NaCl-equivalent tonicity contributed by the XYZ drug will be the product of drug's weight in g (0.425) and its E value ( 0.076 ), which is: 0.425 × 0.076 = 0.0323 g
    • Therefore, the quantity of additional amount of NaCl required to make the product isotonic: 0.45 g - 0.021 g - 0.0323 g = 0.3967 g, rounded to 0.397 g g





Notice: Undefined index: TOTALTRY in /home/jshzub/web/rxcal.org/public_html/practice/practice/includes/body.php on line 675