How many grams of CaCl2 are required to prepare 2.00 liters of 7.00 M CaCl2?
To calculate the number of grams of CaCl2 required to prepare 2.00 liters of 7.00 M CaCl2 solution, we need to use the formula:
Moles = Molarity x Volume
First, we need to convert the volume from liters to milliliters:
2.00 liters = 2,000 milliliters
Next, we can use the equation to calculate the number of moles of CaCl2:
Moles = 7.00 M x 2,000 ml
Moles = 14,000 millimoles (mmol)
Since the molar mass of CaCl2 is 111 grams/mole, we can use this information to calculate the number of grams of CaCl2 required:
Grams = moles x molar mass
Grams = 14,000 mmol x 111 g/mol
Grams = 1,554,000 grams
Therefore, to prepare 2.00 liters of 7.00 M CaCl2, you would need 1,554,000 grams of CaCl2.
To find the number of grams of CaCl2 required to prepare 2.00 liters of a 7.00 M solution, we can use the equation:
moles of solute = molarity x volume of solution in liters
First, we need to convert the volume of the solution from liters to moles using the equation:
moles = Molarity x volume in liters
moles = 7.00 M x 2.00 L = 14.00 moles of CaCl2
Next, we need to convert moles of CaCl2 to grams. To do this, we need to know the molar mass of CaCl2.
CaCl2 consists of one calcium atom (Ca) with a molar mass of 40.08 grams/mole and two chlorine atoms (Cl) with a molar mass of 35.45 grams/mole each. Adding these values together gives us the molar mass of CaCl2:
Molar mass of CaCl2 = (1 x 40.08 g/mol) + (2 x 35.45 g/mol) = 110.98 g/mol
Finally, we can use the molar mass and moles to calculate the number of grams of CaCl2:
grams = moles x molar mass
grams = 14.00 moles x 110.98 g/mol = 1553.72 grams
Therefore, 1553.72 grams of CaCl2 are required to prepare 2.00 liters of a 7.00 M CaCl2 solution.