How mush CaCl2, in grams, is needed to make 2.0 L of a 3.5M solution?
g CaCl2 = (2.0L) *( 3.5 mol Cl-/1L) * (1mol CaCl2/ 2mol Cl-)
*( 110.98 g CaCl2/ 1 mol CaCl2) = 5.549 g CaCl2
i don't think i did any of this right
You made two errors. The rest is ok.
You want grams CaCl2 and not Cl so omit Cl.
That's mols needed = M x L = 2.00L x 3.5M = 7.00 mols.
Then grams = mols x molar mass = 7.00 x 110.98 = about 777 g.
How you obtained 5.55 grams I don't know. YOu must have punched the wrong numbers into the calculator.
To determine the mass of CaCl2 needed to make a 3.5M solution in 2.0L, we can use the formula:
Mass (g) = Volume (L) × Molarity (mol/L) × Molar mass (g/mol)
First, we need to calculate the moles of Cl- ions in the solution. Since CaCl2 dissociates into 2 Cl- ions, the moles of Cl- ions would be 2 × Molarity (3.5M) × Volume (2.0L).
Moles of Cl- ions = 2 × 3.5 mol/L × 2.0 L = 14 moles
Next, we convert the moles of Cl- ions to moles of CaCl2 using the ratio of moles of Cl- ions to moles of CaCl2. Since the ratio is 2:1 (2 moles of Cl- ions per 1 mole of CaCl2), the moles of CaCl2 would be half of the moles of Cl- ions.
Moles of CaCl2 = 14 moles ÷ 2 = 7.0 moles
Now, we multiply the moles of CaCl2 by its molar mass to get the mass of CaCl2 in grams. The molar mass of CaCl2 is 110.98 g/mol.
Mass (g) = 7.0 moles × 110.98 g/mol = 776.86 g ≈ 776.9 g
Therefore, approximately 776.9 grams of CaCl2 is needed to make a 2.0 L 3.5M solution.
To calculate the amount of CaCl2 needed to make a 3.5M solution in 2.0L, you can follow these steps:
Step 1: Determine the molar mass of CaCl2
The molar mass of CaCl2 is 40.08 g/mol (Ca) + 35.45 g/mol (Cl) + 35.45 g/mol (Cl) = 110.98 g/mol.
Step 2: Use the formula for molarity to find the number of moles of CaCl2:
Molarity (M) = moles of solute / volume of solution in liters
Rearrange the formula to solve for moles of solute:
moles of solute = Molarity (M) * volume of solution (L)
moles of solute = 3.5 M * 2.0 L = 7 moles of CaCl2
Step 3: Convert moles of CaCl2 to grams:
grams of CaCl2 = moles of CaCl2 * molar mass of CaCl2
grams of CaCl2 = 7 moles * 110.98 g/mol = 776.86 grams
Therefore, you would need approximately 776.86 grams of CaCl2 to make a 3.5M solution in 2.0 liters.