## Core Concept
To prepare normal saline from a 10% dextrose solution, one must understand the composition of normal saline and how to adjust the solution accordingly. Normal saline is a 0.9% solution of sodium chloride (NaCl) in water.

## Why the Correct Answer is Right
The correct answer involves calculating the amount of sodium chloride needed to achieve a 0.9% concentration in a 100 ml solution. A 10% dextrose solution does not contain NaCl. To make normal saline, 0.9 grams of NaCl should be added to 100 ml of water (or in this case, 100 ml of 10% dextrose solution if the goal is to end up with 100 ml of saline solution). Since the molecular weight of NaCl is approximately 58.44 g/mol, 0.9% of 100 ml (or 90 mg) of NaCl is required. This translates to 0.9 grams or 900 mg of NaCl per liter, which for 100 ml is 90 mg.

## Why Each Wrong Option is Incorrect
**Option A:** This option is incorrect because it does not accurately reflect the amount of NaCl required to make a 0.9% solution in 100 ml of fluid.
**Option B:** This option is incorrect for similar reasons as Option A; it does not provide the correct calculation for achieving a 0.9% NaCl solution.
**Option C:** This option suggests adding 9 grams of NaCl, which would result in a 9% solution, not 0.9%, making it excessively hypertonic.
**Option D:** Without specific details on the amounts, we assume it's incorrect based on the elimination of other options and understanding of normal saline preparation.

## Clinical Pearl / High-Yield Fact
A key point to remember is that normal saline is isotonic, meaning it has the same osmotic pressure as blood (0.9% NaCl solution). When preparing solutions, always calculate the amount of solute needed based on the desired concentration and volume.

## Correct Answer Line
**Correct Answer: D. 0.9gm**