Answer:1.4 moles of calcium chloride are required.
Explanation:
Molarity of the calcium chloride solution = 0.700 M = 0.700 mol/L
Volume of the solution = 2 L
Moles of the calcium chloride = x
\(Molarity=\frac{\text{moles of the substance}}{\text{volume of the Solution (L)}}\)
\(0.700 mol/L=\frac{x}{2 L}\)
x = 1.4 moles of calcium chloride
1.4 moles of calcium chloride are required.
