Answer
a. 68 g of KCl remain in solution at $20 \space ^o C$
b. 12 g of KCl crystallized after cooling.
Work Step by Step
1. Determine the solubility of $KCl$ in 200. g of $H_2O$ at 20 $^oC$ and at $50 \space ^o C$.
At $20 ^o C$:
$$200. \space g \space H_2O \times \frac{34 \space g \space KCl}{100 \space g \space H_2O} = 68 \space g \space KCl$$
At $50 ^o C$:
$$200. \space g \space H_2O \times \frac{43 \space g \space KCl}{100 \space g \space H_2O} = 86 \space g \space KCl$$
a. Since 80 g > 68 g, the solution is saturated at 20 $^o C$, which means that only 68 g of $KCl$ will remain in solution.
b. Subtract the amount of KCl in solution at 20 $^o $ C from the original amount of KCl in solution.
$$80 \space g - 68 \space g = 12 \space g$$