Answer
There are necessary 55.5 g of $CaCl_2$ to prepare that solution.
Work Step by Step
1. Since the molarity of that solution is equal to $ 0.100 \space M \space CaCl_2$:
$1 \space L \space solution = 0.100 \space moles \space CaCl_2$
$\frac{1 \space L \space solution}{ 0.100 \space moles \space CaCl_2} $ and $\frac{ 0.100 \space moles \space CaCl_2}{1 \space L \space solution}$
2. Determine the molar mass of this compound ($CaCl_2$), and setup the conversion factors:
Molar mass :
$Ca: 40.08g $
$Cl: 35.45g * 2= 70.90g $
40.08g + 70.90g = 110.98g
$ \frac{1 \space mole \space (CaCl_2)}{ 110.98 \space g \space (CaCl_2)}$ and $ \frac{ 110.98 \space g \space (CaCl_2)}{1 \space mole \space (CaCl_2)}$
3. Use the conversion factors to calculate the mass of solute in $ 5.00$ L of that solution:
$ 5.00 \space L \space solution \times \frac{ 0.100 \space mole \space CaCl_2}{1 \space L \space solution} \times \frac{ 110.98 \space g \space CaCl_2}{1 \space mole \space CaCl_2} = 55.5 \space g \space CaCl_2$
