A 1.50 M solution containing 25 Grams of Al(OH)3 would occupy what volume in liters?
1.5 moles/L of solution.
How many moles do we have in 25 g.
25/molar mass Al(OH)3.
M = moles/L
You know M (1.5) and you know moles. Solve for L.
To determine the volume of a solution, we can use the formula:
Volume (L) = mass (g) / concentration (mol/L) / molar mass (g/mol)
1. Calculate the molar mass of Al(OH)3:
- The molar mass of Al is 26.98 g/mol.
- The molar mass of O is 16.00 g/mol.
- The molar mass of H is 1.01 g/mol.
So, the molar mass of Al(OH)3 is:
(1 * 26.98 g/mol) + (3 * (16.00 g/mol + 1.01 g/mol)) = 78.00 g/mol
2. Convert the grams of Al(OH)3 to moles:
25 g Al(OH)3 * (1 mol Al(OH)3 / 78.00 g Al(OH)3) = 0.32 mol Al(OH)3
3. Use the given concentration to find the volume:
Concentration = 1.50 mol/L
0.32 mol Al(OH)3 / 1.50 mol/L = 0.21 L
Therefore, a 1.50 M solution containing 25 grams of Al(OH)3 would occupy 0.21 liters.
To calculate the volume of a solution, we need to use the formula:
Volume (in liters) = Mass (in grams) / Concentration (in moles per liter)
First, let's calculate the number of moles of Al(OH)3 in the solution:
1. Convert the mass of Al(OH)3 from grams to moles. Since the molar mass of Al(OH)3 is 78 g/mol, we can use the formula:
Number of moles = Mass / Molar mass = 25 g / 78 g/mol ≈ 0.32 mol.
Now we can use this information along with the given concentration to find the volume:
0.32 mol / 1.50 mol/L = 0.213 L.
Therefore, a 1.50 M solution containing 25 grams of Al(OH)3 would occupy a volume of approximately 0.213 liters.