calculate the mass of barium hydroxide required to prepare 2.0L of a 0.500mol/L solution. I got 342.64 g.
I don't know how you did it but the answer you have is exactly twice what it should be.
How many mols do you want? That's M x L = ?
Then grams = mols x molar mass
To calculate the mass of barium hydroxide required to prepare a solution, you need to consider the molar mass and the desired concentration of the solution.
The molar mass of barium hydroxide (Ba(OH)2) can be calculated by summing the atomic masses of its constituent elements, which are:
- Barium (Ba) with an atomic mass of 137.327 g/mol
- Oxygen (O) with an atomic mass of 15.999 g/mol
- Hydrogen (H) with an atomic mass of 1.00784 g/mol
Molar mass of Ba(OH)2 = (1 * Ba) + (2 * O) + (2 * H)
Plugging in the values:
Molar mass of Ba(OH)2 = (1 * 137.327 g/mol) + (2 * 15.999 g/mol) + (2 * 1.00784 g/mol)
= 137.327 g/mol + 31.998 g/mol + 2.01568 g/mol
= 171.34 g/mol
Now, you need to calculate the mass of barium hydroxide required to prepare the solution.
First, determine the number of moles of barium hydroxide needed:
Number of moles = concentration (mol/L) * volume (L)
= 0.500 mol/L * 2.0 L
= 1.0 mol
Finally, calculate the mass using the equation:
Mass(g) = Number of moles * Molar mass(g/mol)
= 1.0 mol * 171.34 g/mol
= 171.34 g
Therefore, the correct mass of barium hydroxide required to prepare a 2.0L, 0.500 mol/L solution is 171.34 grams.