Example 1. Prepare 100 ml isotonic solution of 0.5 % (w/v) ephedrine sulfate (E value = 0.23).
Answer 1. Mass of ephedrine sulfate = concentration × volume = 0.5 (g/ 100 ml) × 100 ml = 0.5 g.
According to the E value each gram of ephedrine sulfate is equivalent to 0.23 g sodium chloride. Therefore, 0.5 g is equivalent to 0.23 × 0.5 = 0.115 g sodium chloride. The amount of sodium chloride needed to make the solution isotonic is equal to 0.9 - 0.115 = 0.785 g. The formula is written as the following:
Ephedrine sulfate 0.5 g
Sodium chloride 0.785 g
Water q.s. 100 ml
Example 2. What mass of sodium chloride is included in 100 ml of 5.0 % (w/v) dextrose monohydrate (E value = 0.16) isotonic solution.
Answer 2. Mass of dextrose monohyrate in 100 ml = concentration × volume = 5.0 (g/ 100 ml) × 100 ml = 5.0 g. Each gram of dextrose monohydrate is equivalent to 0.16 g sodium chloride (E value = 0.16). Therefore, 5 g dextrose monohydrate is equivalent to 0.8 g (5 × 0.16) sodium chloride. To make 100 ml of this solution isotonic some amount of sodium chloride needs to be included such that total solute effect is equivalent to that of 0.9 g sodium chloride. Thus, the formula for 100 ml is written as the following:
Dextrose monohydrate 5.0 g
Sodium chloride 0.1 g (0.9 - 0.8)
Water q.s. 100 ml
Example 3. Prepare an isotonic solution of ethanol (E value = 0.7).
Answer 3. Each gram of ethanol (anhydride) is equivalent to 0.7 g of sodium chloride. Thus, the amount needed for 100 ml solution in order to obtain an equivalent effect to that of 0.9 % sodium chloride is 1.29 g (0.9/ 0.7). Therefore, 1.29 % ethanol solution is isotonic.
Related posts:
Isotonic solutions in pharmaceutical preparations.
How to prepare isotonic solutions.
Answer 1. Mass of ephedrine sulfate = concentration × volume = 0.5 (g/ 100 ml) × 100 ml = 0.5 g.
According to the E value each gram of ephedrine sulfate is equivalent to 0.23 g sodium chloride. Therefore, 0.5 g is equivalent to 0.23 × 0.5 = 0.115 g sodium chloride. The amount of sodium chloride needed to make the solution isotonic is equal to 0.9 - 0.115 = 0.785 g. The formula is written as the following:
Ephedrine sulfate 0.5 g
Sodium chloride 0.785 g
Water q.s. 100 ml
Example 2. What mass of sodium chloride is included in 100 ml of 5.0 % (w/v) dextrose monohydrate (E value = 0.16) isotonic solution.
Answer 2. Mass of dextrose monohyrate in 100 ml = concentration × volume = 5.0 (g/ 100 ml) × 100 ml = 5.0 g. Each gram of dextrose monohydrate is equivalent to 0.16 g sodium chloride (E value = 0.16). Therefore, 5 g dextrose monohydrate is equivalent to 0.8 g (5 × 0.16) sodium chloride. To make 100 ml of this solution isotonic some amount of sodium chloride needs to be included such that total solute effect is equivalent to that of 0.9 g sodium chloride. Thus, the formula for 100 ml is written as the following:
Dextrose monohydrate 5.0 g
Sodium chloride 0.1 g (0.9 - 0.8)
Water q.s. 100 ml
Example 3. Prepare an isotonic solution of ethanol (E value = 0.7).
Answer 3. Each gram of ethanol (anhydride) is equivalent to 0.7 g of sodium chloride. Thus, the amount needed for 100 ml solution in order to obtain an equivalent effect to that of 0.9 % sodium chloride is 1.29 g (0.9/ 0.7). Therefore, 1.29 % ethanol solution is isotonic.
Related posts:
Isotonic solutions in pharmaceutical preparations.
How to prepare isotonic solutions.
Nice info, great
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