A laboratory assistant needs to prepare 325 mL of 0.110 M CaCl2 solution. How many grams of calcium chloride will she need?
To determine the number of grams of calcium chloride needed to prepare a 0.110 M solution, we can use the formula:
Molarity (M) = moles of solute / liters of solution
First, let's calculate the moles of calcium chloride needed:
Moles = Molarity × Volume (in liters)
Given:
Molarity (M) = 0.110 M
Volume = 325 mL = 325/1000 = 0.325 L
Moles = 0.110 M × 0.325 L
Moles = 0.03575
Now, we need to calculate the molar mass of calcium chloride (CaCl2):
1 calcium (Ca) atom = 40.08 g/mol
2 chlorine (Cl) atoms = 2 × 35.45 g/mol = 70.90 g/mol
Molar mass of calcium chloride (CaCl2) = 40.08 g/mol + 70.90 g/mol
Molar mass of calcium chloride (CaCl2) = 110.98 g/mol
Finally, we can calculate the grams of calcium chloride needed:
Grams = Moles × Molar mass
Grams = 0.03575 mol × 110.98 g/mol
Grams = 3.9731 g
Therefore, the laboratory assistant will need approximately 3.9731 grams of calcium chloride to prepare 325 mL of 0.110 M CaCl2 solution.
To find out how many grams of calcium chloride are needed to prepare the solution, you need to use the formula:
Amount of solute = Concentration × Volume
First, let's convert the volume from milliliters (mL) to liters (L):
325 mL = 325/1000 = 0.325 L
The concentration is given as 0.110 M (moles per liter).
Now, we can calculate the amount of solute (calcium chloride) using the formula:
Amount of solute = Concentration × Volume
Amount of solute = 0.110 mol/L × 0.325 L = 0.03575 mol
To convert moles (mol) to grams (g), we need to know the molar mass of calcium chloride (CaCl2). The molar mass can be calculated by adding up the atomic masses of each element:
Molar mass of CaCl2 = (atomic mass of Ca) + 2 × (atomic mass of Cl)
Using the periodic table, we find that the atomic mass of calcium (Ca) is 40.08 g/mol, and the atomic mass of chlorine (Cl) is 35.45 g/mol.
Molar mass of CaCl2 = 40.08 g/mol + 2 × 35.45 g/mol = 110.98 g/mol
Now, we can calculate the mass of calcium chloride using the formula:
Mass = Amount of solute × Molar mass
Mass = 0.03575 mol × 110.98 g/mol ≈ 3.979 g
Therefore, the laboratory assistant will need approximately 3.979 grams of calcium chloride to prepare the 325 mL of 0.110 M solution.
moles needed = M x L = ??
Solve for moles.
then M = grams/molar mass.
Solve for grams.