What mass of Ca(OH)2must be dissolved in 0.504 L of a solution to make a 0.56 M solution?
0.28 g
21 g
1.1 g
0.90 g
1.1 g??
To calculate the mass of Ca(OH)2 required to prepare a 0.56 M solution in 0.504 L, you can use the formula:
Molarity (M) = moles/volume (L)
First, let's rearrange this equation to solve for moles:
moles = Molarity x volume
moles = 0.56 M x 0.504 L
moles = 0.28224 mol
Next, we can calculate the molar mass of Ca(OH)2:
Ca = 1 x 40.08 g/mol = 40.08 g/mol
O = 2 x 16.00 g/mol = 32.00 g/mol
H = 2 x 1.008 g/mol = 2.016 g/mol
Molar mass of Ca(OH)2 = 40.08 + 32.00 + 2.016 = 74.096 g/mol
Finally, we can calculate the mass of Ca(OH)2:
mass = moles x molar mass
mass = 0.28224 mol x 74.096 g/mol
mass ≈ 20.93 g
So, the correct answer is approximately 21 g, which is the closest option to the calculated mass of Ca(OH)2 required for the solution.
To solve this problem, you'll need to use the equation for molarity, which is:
Molarity (M) = moles of solute / volume of solution (in liters)
Given that the molarity (M) is 0.56 M and the volume of the solution is 0.504 L, you can rearrange the equation to solve for moles of solute:
moles of solute = Molarity (M) x volume of solution (in liters)
moles of solute = 0.56 M x 0.504 L = 0.28224 moles
Now, you'll need to find the molar mass of Ca(OH)2, which is:
Ca = 1 x 40.08 g/mol = 40.08 g/mol
O = 2 x 16.00 g/mol = 32.00 g/mol
H = 2 x 1.01 g/mol = 2.02 g/mol
Adding these up, you get:
Molar mass of Ca(OH)2 = 40.08 g/mol + 32.00 g/mol + 2.02 g/mol = 74.10 g/mol
Finally, to calculate the mass of Ca(OH)2 needed, you can use the mole-mass conversion:
mass of Ca(OH)2 = moles of solute x molar mass of Ca(OH)2
mass of Ca(OH)2 = 0.28224 moles x 74.10 g/mol = 20.91 g
Therefore, the closest answer would be 21 g, so the correct answer is 21 g.
mols Ca(OH)2 needed = M x L = 0.56 M x 0.504 L = ?
Then grams Ca(OH)2 = mols Ca(OH)2 x molar mass Ca(OH)2 = ? .