Final answer:
To make the drug solution isotonic with body fluids, 460 mg of sodium chloride is needed for a 200 mL solution, based on the freezing point depression of the drug and the known colligative properties of sodium chloride solutions.
Step-by-step explanation:
The student is asking how much sodium chloride is necessary to create an isotonic solution that has the same osmotic pressure as body fluids. The question mentions a freezing point depression of a drug solution and intends to use this information to find out the amount of sodium chloride needed to reach isotonicity. An isotonic solution usually has a freezing point depression of -0.52 degrees Celsius, which is equivalent to a 0.9% sodium chloride solution. Given the colligative properties of solutions, which relate the concentration of solute particles to the freezing point depression, we can calculate the amount of sodium chloride needed. Considering the freezing point depression (1% solution of the drug was -0.12 degrees Celsius), we can use the fact that a 1% NaCl solution would have a freezing point depression of -0.58 degrees Celsius to find that about half of the 1% NaCl is needed to make the drug solution isotonic with body fluids.
Since the question requires practical isotonicity, we will use 0.9% NaCl solution (which has a freezing point of -0.52 degrees Celsius) rather than the theoretical 1%. Thus, for a 0.52/0.12 ratio, one would need roughly 0.52/0.12 or 4.33 times less NaCl than in a 1% solution to make the drug solution isotonic. Since a 1% NaCl solution contains 1 g of NaCl per 100 mL, we'd need 1g/4.33 per 100 mL for isotonicity. Therefore, for a 200 mL solution, we need 2g/4.33, which equals 0.461 grams or 461 mg of sodium chloride. The answer is approximately 460 mg (option d).