Final answer:
We calculate the number of moles of barium sulfate from the given mass and use the molarity formula to determine that approximately 233.2 milliliters of water are required to make a 2.5 M solution.
Step-by-step explanation:
To find out how many liters of water are needed to make a 2.5 M solution using 136 grams of barium sulfate (BaSO4), we first need to calculate the number of moles of BaSO4 in 136 grams. The molar mass of BaSO4 is 233.4 g/mol. Therefore:
- Number of moles of BaSO4 = Mass (in g) ÷ Molar Mass (in g/mol) = 136 g ÷ 233.4 g/mol
- Number of moles of BaSO4 = 0.583 moles (approximately)
Since we want a 2.5 M solution, and we have 0.583 moles of BaSO4, we can then use the formula for molarity (M = moles of solute ÷ liters of solution) to find the volume of solution needed:
- 2.5 M = 0.583 moles ÷ liters of solution
- Liters of solution = 0.583 moles ÷ 2.5 M
- Liters of solution = 0.2332 liters or 233.2 milliliters
Therefore, 233.2 milliliters of water are needed to make a 2.5 M solution of barium sulfate.