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To prepare a 400 mL (0.400 L) solution of 1.2 M NaCl, we need to calculate the mass of NaCl required.

### Steps for calculation:
1. **Molarity formula**:
   \[
   M = \frac{n}{V}
   \]
   Where:
   - \( M \) is the molarity (1.2 M in this case),
   - \( n \) is the number of moles of solute (NaCl),
   - \( V \) is the volume of the solution in liters (0.400 L).

   Rearranging the formula to solve for \( n \):
   \[
   n = M \times V
   \]

2. **Calculate moles of NaCl**:
   \[
   n = 1.2 \, \text{M} \times 0.400 \, \text{L} = 0.48 \, \text{moles of NaCl}
   \]

3. **Convert moles of NaCl to grams**:
   The molar mass of NaCl is approximately 58.44 g/mol.

   Now, calculate the mass of NaCl required:
   \[
   \text{mass} = 0.48 \, \text{moles} \times 58.44 \, \frac{\text{g}}{\text{mol}} = 28.05 \, \text{g}
   \]

### Procedure for preparation:
1. Measure out 28.05 grams of NaCl.
2. Dissolve the NaCl in a small amount of distilled water in a beaker.
3. Transfer the solution to a 400 mL volumetric flask.
4. Add distilled water to the flask until the total volume reaches exactly 400 mL.
5. Mix the solution thoroughly to ensure complete dissolution.

This will give you a 400 mL solution of 1.2 M NaCl.

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Would you like further details or have any questions? Here are some related questions:

1. How do you calculate the molarity of a solution given mass and volume?
2. What are the steps to dilute a solution from a higher concentration?
3. How does temperature affect the solubility of NaCl in water?
4. Can molarity change if the solution evaporates?
5. How would you prepare a 1.2 M NaCl solution if only a 6 M stock solution is available?

**Tip**: Always ensure that the final solution volume, not the added water, matches the desired volume in molarity preparations.

How would you prepare 400 mL of 1.2 M solution of NaCl?

Learn how to prepare a 400 mL solution of 1.2 M NaCl. This process includes calculating the moles of NaCl needed, converting moles to grams using the molar mass, and detailed steps for mixing the solution. Ideal for high school chemistry students.

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