How do you sole this problem: how many grams of CaCl2 are needed to prepare 7.25 L of .015M solution?
Molarity=moles/volume (L)=0.015= x moles/7.25 L
Solve for moles,
0.015*7.25L= moles of CaCl2 in solution
moles of CaCl2 in solution*(110.98 g/mol)= mass of CaCl2 in g.
To solve this problem, you need to use the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
First, let's calculate the moles of CaCl2 needed. We need to rearrange the formula to find the moles:
Moles of solute = Molarity × volume of solution
Moles of CaCl2 = 0.015 M × 7.25 L
Next, we need to convert moles into grams using the molar mass of CaCl2. The molar mass of CaCl2 is found by adding up the atomic masses of each element:
Ca: 40.08 g/mol
Cl: 2 × 35.45 g/mol = 70.90 g/mol
Molar mass of CaCl2 = 40.08 g/mol + 70.90 g/mol = 111.98 g/mol
Finally, we can calculate the grams of CaCl2 needed:
Grams of CaCl2 = Moles of CaCl2 × Molar mass of CaCl2
Now you can plug in the values:
Grams of CaCl2 = (0.015 M × 7.25 L) × 111.98 g/mol
Calculate the expression in the parentheses:
Grams of CaCl2 = 0.2175 moles × 111.98 g/mol
Now multiply the values:
Grams of CaCl2 = 24.32745 g
Therefore, you would need approximately 24.33 grams of CaCl2 to prepare a 7.25 L solution with a concentration of 0.015 M.