How many grams of CaCl2 are needed to prepare 25.0 mL of .500 M solution?
and
What volume of 12.00 M HCL is needed to prepare 150.0 mL of .500 M HCL?
mols you need = M x L = ?
mols = g/molar mass. You know mols and molar mass, solve for grams.
2.
mL1 x M1 = mL2 x M2
mL1 x 12 = etc.
To calculate the number of grams of CaCl2 needed to prepare 25.0 mL of a 0.500 M solution, you need to follow these steps:
Step 1: Determine the molar mass of CaCl2.
The molar mass of CaCl2 is 40.08 g/mol for calcium (Ca) and 35.45 g/mol for each chlorine atom (Cl). Adding them up, we get:
Molar mass of CaCl2 = (40.08 g/mol) + 2(35.45 g/mol) = 111.0 g/mol.
Step 2: Calculate the number of moles required.
The concentration of the CaCl2 solution is given as 0.500 M, which means there are 0.500 moles of CaCl2 per liter of solution (1 L = 1000 mL).
To find the number of moles in 25.0 mL, we need to convert mL to L:
25.0 mL ÷ 1000 mL/L = 0.025 L.
So, the number of moles required is:
Number of moles = concentration (M) × volume (L) = 0.500 mol/L × 0.025 L = 0.0125 mol.
Step 3: Convert moles to grams.
To convert from moles to grams, we multiply the number of moles by the molar mass of CaCl2:
Number of grams = number of moles × molar mass = 0.0125 mol × 111.0 g/mol ≈ 1.39 g.
Therefore, you would need approximately 1.39 grams of CaCl2 to prepare 25.0 mL of a 0.500 M solution.
Now, let's move on to the second question regarding the volume of 12.00 M HCl needed to prepare 150.0 mL of a 0.500 M HCl solution.
To solve this, we need to follow similar steps:
Step 1: Determine the molar mass of HCl.
The molar mass of HCl is approximately 36.46 g/mol.
Step 2: Calculate the number of moles required.
The concentration of the HCl solution is given as 0.500 M, which means there are 0.500 moles of HCl per liter of solution.
To find the number of moles in 150.0 mL, we need to convert mL to L:
150.0 mL ÷ 1000 mL/L = 0.150 L.
So, the number of moles required is:
Number of moles = concentration (M) × volume (L) = 0.500 mol/L × 0.150 L = 0.075 mol.
Step 3: Determine the volume of the 12.00 M HCl solution needed.
To find the volume of the 12.00 M HCl solution needed to obtain 0.075 moles, we use the formula:
Volume (L) = number of moles / concentration (M) = 0.075 mol / 12.00 mol/L ≈ 0.00625 L or 6.25 mL.
Therefore, you would need approximately 6.25 mL of 12.00 M HCl to prepare 150.0 mL of a 0.500 M HCl solution.