How many liters of a 2.75 M NaBr solution can be made using 100.5 grams of NaBr?

To determine how many liters of a 2.75 M NaBr (sodium bromide) solution can be made using 100.5 grams of NaCl (sodium chloride), we will go through the following steps:

1. Calculate the moles of NaCl:
- First, we need to find out how many moles are present in 100.5 grams of NaCl. To do this, we use the molar mass of NaCl.
- The molar mass of NaCl is approximately 58.44 g/mol.
- Moles of NaCl = mass (in grams) / molar mass = 100.5 grams / 58.44 g/mol ≈ 1.72 moles of NaCl.

2. Convert moles of NaCl to moles of NaBr:
- In a stoichiometric reaction where NaCl is used to produce NaBr, 1 mole of NaCl is assumed to produce 1 mole of NaBr (stochiometric ratio = 1:1).
- Therefore, moles of NaBr = moles of NaCl = 1.72 moles.

3. Calculate the volume of the NaBr solution:
- We know the target molarity (concentration) of the NaBr solution is 2.75 M, which means there are 2.75 moles of NaBr per liter of solution.
- To find the volume of the solution that contains 1.72 moles of NaBr, use the formula:
[tex]\[
\text{Volume} = \frac{\text{Moles of NaBr}}{\text{Molarity of NaBr}}
\][/tex]

- So, Volume = 1.72 moles / 2.75 M = approximately 0.625 liters.

Therefore, you can make approximately 0.625 liters of a 2.75 M NaBr solution using 100.5 grams of NaCl.