Describe (with calculations) how you prepare 1.00 L of 0.750 M barium hydroxide solution starting with Solid barium hydroxide
Molarity has the definition of mols/liter of solution.
You want 0.750 M; therefore, weigh 0.750 mol x (molar mass/1 mol) Ba(OH)2, dissolve it in some water and make the final volume to 1.00 liter.(Note I did NOT say to add 1.00 L of water. Adding 1 L of water to Ba(OH)2 results in a volume that is more than 1.00 liter of solution; hence, we dissolve the solute in a little solvent and make the final volume to the desired level.
Look up the molar mass of Ba(OH)2, or calculate it.
I have 171.342 for the molar mass of Ba(OH)2 but I don't know the reliability of those values I added together. The numbers you used may be more up to date. Anyway, then 0.750 x 171.342 = 128.506 which would round to 128 to three significant figures (0.750 has three s.f.). But I think the important point your prof is trying to impress on the class with this problem is to give directions for preparing. So the directions would go something like this. Weigh on an analytical balance 128.5 g Ba(OH)2, dissolve in water, and make the final volume to 1.00 liter to prepare 1.00 L of a 0.750 M Ba(OH)2 solution.
my answer is 128.495 is that right Dr.Bob?
thank you Dr. Bob
Well, isn't that a chemical delight! Let's crunch some numbers and concoct this solution for you.
First, we need to determine how many moles of barium hydroxide (Ba(OH)2) are required to make a 0.750 M solution. The concentration of 0.750 M means that there are 0.750 moles of Ba(OH)2 per liter of solution.
So, for 1.00 L of solution, we would need 1.00 x 0.750 moles of Ba(OH)2. That gives us 0.750 moles of Ba(OH)2.
Now, let's figure out the molar mass of Ba(OH)2. Ba has a molar mass of 137.33 g/mol, O has a molar mass of 16.00 g/mol, and H has a molar mass of 1.01 g/mol.
Thus, the molar mass of Ba(OH)2 is:
137.33 g/mol (Ba) + (2 * 1.01 g/mol (H)) + (2 * 16.00 g/mol (O))
= 137.33 g/mol + 2.02 g/mol + 32.00 g/mol
= 171.35 g/mol
To determine how much solid barium hydroxide we'll need, we can use the equation:
moles = mass / molar mass
Rearranging the equation, we'll find the mass needed:
mass = moles * molar mass
For 0.750 moles of Ba(OH)2, the required mass will be:
0.750 moles * 171.35 g/mol = 128.51 g
So, you'll need approximately 128.51 grams of solid barium hydroxide to prepare 1.00 L of 0.750 M solution.
Just remember to handle those chemicals with care and keep that colorful chemistry hat on!
To prepare 1.00 L of a 0.750 M barium hydroxide (Ba(OH)2) solution starting with solid barium hydroxide, you need to know the molar mass of Ba(OH)2 and the desired concentration.
The molar mass of Ba(OH)2 can be calculated by adding up the atomic masses of each element. Barium (Ba) has a molar mass of 137.33 g/mol, oxygen (O) has a molar mass of 16.00 g/mol, and hydrogen (H) has a molar mass of 1.01 g/mol. Since there are 2 hydrogen atoms in the formula, the molar mass of Ba(OH)2 is:
(1 × 137.33 g/mol) + (2 × 16.00 g/mol) + (2 × 1.01 g/mol) = 171.34 g/mol.
Now, let's calculate the amount of solid barium hydroxide needed to make the solution. To do this, we will use the formula:
moles = (volume × concentration) / molar mass.
We want to prepare 1.00 L of a 0.750 M solution, so the desired volume is 1.00 L and the desired concentration is 0.750 M. Plugging these values into the formula:
moles = (1.00 L × 0.750 M) / 171.34 g/mol = 0.004384 moles.
To convert moles into grams, we use the formula:
grams = moles × molar mass.
Therefore,
grams = 0.004384 moles × 171.34 g/mol = 0.75 grams (rounded to two decimal places).
So, starting with solid barium hydroxide, you would need approximately 0.75 grams to prepare 1.00 L of a 0.750 M barium hydroxide solution.