Answer
There are necessary 59.6 g of $KCl$ to prepare that solution.
Work Step by Step
1. Since the molarity of that solution is equal to $ 0.200 \space M \space KCl$:
$1 \space L \space solution = 0.200 \space moles \space KCl$
$\frac{1 \space L \space solution}{ 0.200 \space moles \space KCl} $ and $\frac{ 0.200 \space moles \space KCl}{1 \space L \space solution}$
2. Determine the molar mass of this compound (KCl), and setup the conversion factors:
Molar mass :
$K: 39.10g $
$Cl: 35.45g$
39.10 g + 35.45 g = 74.55 g
$ \frac{1 \space mole \space (KCl)}{ 74.55 \space g \space (KCl)}$ and $ \frac{ 74.55 \space g \space (KCl)}{1 \space mole \space (KCl)}$
3. Use the conversion factors to calculate the mass of solute in $ 4.00$ L of that solution:
$ 4.00 \space L \space solution \times \frac{ 0.200 \space moles \space KCl}{1 \space L \space solution} \times \frac{ 74.55 \space g \space KCl}{1 \space mole \space KCl} = 59.6 \space g \space KCl$
