Final answer:
To make a 410 mL solution of 0.45% NaCl isotonic, you would need 4.28 mL of a 50% dextrose solution, using the E-value to equate the osmotic pressure to that of a 0.9% NaCl solution.
Step-by-step explanation:
To calculate the volume of a 50% dextrose solution needed to add to 410 mL of 0.45% sodium chloride to make it isotonic, we need to consider the E-value of dextrose, which is given as 0.16. The E-value represents the amount of substance that provides the same osmotic pressure as 1 g of NaCl. Since we want an isotonic solution, we want the osmotic pressure of the dextrose solution to equate to that of a 0.9% NaCl solution, which is considered isotonic.
Firstly, calculate the grams of NaCl that would make the 410 mL solution isotonic. For a 0.9% NaCl solution, we would need 0.9 grams of NaCl for every 100 mL. Therefore, for 410 mL, we need:
0.9 g/100 mL × 410 mL = 3.69 g NaCl.
Since the current 410 mL of 0.45% NaCl solution already contains some amount of NaCl (1.845 g), we need to find the additional amount needed to make it isotonic:
3.69 g (needed for isotonicity) - 1.845 g (already present) = 1.845 g NaCl.
Next, we convert this required mass of NaCl to an equivalent mass of dextrose using the E-value:
1.845 g × (1 + E-value) = 1.845 g × (1 + 0.16) = 1.845 g × 1.16 = 2.14 g dextrose.
Since the dextrose solution is 50% by volume, to get 2.14 g of dextrose we need:
2.14 g ÷ (0.5 g/mL) = 4.28 mL of 50% dextrose solution.