What volume of water would be required to dissolve 1 g of BaF2?
mass of BaF2 is 1.31 g and it dissolves in 1 L of water. The answer is 0.763. Can someone please explain how they got this?
One gallon of water dissolves 1.31 g. If you put in more BaF2 than that, the solution is saturated and the remainder will not dissolve.
1 g of BaF2 is 1/1.31 or 76.3% of the amount that will saturate one gallon of water.
If you use 76.3% of 1 gallon of water, you will have a saturated solution.
To calculate the volume of water required to dissolve 1 g of BaF2, we can use the given information:
Mass of BaF2: 1 g
Mass of BaF2 in the given information: 1.31 g
Volume of water required to dissolve 1.31 g of BaF2: 1 L
We can set up a proportion to find the volume of water required to dissolve 1 g of BaF2:
(1.31 g) / (1 L) = (1 g) / (x L)
Cross-multiplying and solving for x:
1.31 g * x L = 1 g * 1 L
1.31x = 1
x = 1 / 1.31
x ≈ 0.763 L
Therefore, approximately 0.763 L or 763 mL of water would be required to dissolve 1 g of BaF2.
To calculate the volume of water required to dissolve 1 g of BaF2, you can use the concept of molarity.
Molarity is defined as the number of moles of solute (in this case, BaF2) dissolved in one liter of solution (in this case, water).
First, you need to find the number of moles of BaF2 in 1.31 g using its molar mass.
The molar mass of BaF2 can be calculated by adding the atomic masses of barium (Ba) and fluorine (F):
Molar mass of BaF2 = (atomic mass of Ba) + 2 × (atomic mass of F)
The atomic mass of Ba is 137.33 g/mol, and the atomic mass of F is 18.998 g/mol.
So, the molar mass of BaF2 = 137.33 + 2 × 18.998 = 175.326 g/mol
Now, you can calculate the number of moles of BaF2:
Moles of BaF2 = Mass of BaF2 / Molar mass of BaF2
Moles of BaF2 = 1.31 g / 175.326 g/mol = 0.00747 mol
Since we know that 1 L of water can dissolve 1.31 g of BaF2, which is equivalent to 0.00747 mol, we can conclude that 1 g of BaF2 can be dissolved in:
Volume of water = (1 g BaF2 / 1.31 g BaF2) × 1 L
Volume of water = 0.763 L or 763 mL
Therefore, the volume of water required to dissolve 1 g of BaF2 is 0.763 liters (or 763 milliliters).