What mass of calcium chloride is required to prepare 3.20 L of 0.850 M of calcium chloride solution?

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.

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