College

Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ How many liters of a 2.75 M NaBr solution can be made using 100.5 grams of NaBr?

Answer :

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.