1. a)100ml solution has 0.5gm of tetracaine hydrochloride.
Then 15ml solution has 0.5/100×15= 0.075gm of tetracaine hydrochloride.
b) E- value of tetracaine hydrochloride = 0.18
0.075gm ×0.18=0.0135gm of NaCl is represented by tetracaine hydrochloride.
Volume of solution :-
15ml (total value)- 10ml (volume of epinephrine bitartrate) =5ml.
0.9gm of NaCl is present in 100ml of solution
Then amount of NaCl in 5ml = 0.9÷100 ×5= 0.045gm of NaCl is present in 5ml of isotonic solution.
The actual amount of NaCl is 0.045 gm - 0.0135gm =0.0315gm of NaCl only.
0.9gm of NaCl is present in 100ml
So, 0.0315gm of NaCl is present in 0.9÷100 × 0.0315 = 3.5 ml
3.5 ml of Sodium chloride is required to make this isotonic solution .
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