#### Answer

(a) 20.0% of $CaCl_2$ by mass.
(b) The molarity of $CaCl_2$ is equal to $2.13M$.

#### Work Step by Step

(a)
Percent by mass $CaCl_2 = \frac{mass(CaCl_2)}{Totalmass} \times 100\%= \frac{16.0}{16.0+64.0} \times 100\%= 20.0\%$
(b)
Use the conversion factors to calculate the molarity of the solution:
** Total mass = 16.0 g + 64.0 g = 80.0 g (Solution)
Molar mass $(CaCl_2) = 40.08* 1 + 35.45* 2 = 110.98g/mol$
80.0 g (Solution) $\times \frac{1mL}{1.180g} = 67.8mL(Solution)$
$16.0g (CaCl_2) \times \frac{1mol (CaCl_2)}{110.98g(CaCl_2)} \times \frac{1}{67.8mL (Solution)} \times \frac{1000mL}{1L} = 2.13 \frac{mol(CaCl_2)}{L(Solution)}$
= $2.13 M (CaCl_2)$