Answer:
301.92 g of CaCl2.
Explanation:
The following data were obtained from the question:
Volume of CaCl2 solution = 3.20 L
Molarity of CaCl2 solution = 0.850 M
Mass of CaCl2 =?
Next, we shall determine the number of mole of CaCl2 in the solution. This can be obtained as illustrated below:
Volume of CaCl2 solution = 3.20 L
Molarity of CaCl2 solution = 0.850 M
Mole of CaCl2 =?
Molarity = mole /Volume
0.850 = mole of CaCl2 /3.20
Cross multiply
Mole of CaCl2 = 0.850 × 3.20
Mole of CaCl2 = 2.72 moles
Finally, we shall determine the mass of the CaCl2 needed to prepare the solution as follow:
Mole of CaCl2 = 2.72 moles
Molar mass of CaCl2 = 40 + (2×35.5)
=40 + 71
= 111 g/mol
Mass of CaCl2 =?
Mole = mass /Molar mass
2.72 = mass of CaCl2 /111
Cross multiply
Mass of CaCl2 = 2.72 × 111
Mass of CaCl2 = 301.92 g
Therefore, 301.92 g of CaCl2 is needed to prepare the solution.
