Answer
a. 66 g of $NaNO_3$
b. 14 g of $NaNO_3$
Work Step by Step
1. Determine the solubility of $NaNO_3$ in 75 g of $H_2O$ at 20 $^oC$ and at $50 \space ^o C$.
At $20 ^o C$:
$$75 \space g \space H_2O \times \frac{88 \space g \space NaNO_3}{100 \space g \space H_2O} = 66 \space g \space NaNO_3$$
At $50 ^o C$:
$$75 \space g \space H_2O \times \frac{110 \space g \space NaNO_3}{100 \space g \space H_2O} = 83 \space g \space NaNO_3$$
a. Since 80 g > 66 g, the solution is saturated at 20 $^o C$, which means that only 66 g of $NaNO_3$ will remain in solution.
b. Subtract the amount of KCl in solution at 20 $^o $ C from the original amount of KCl in solution.
$$80 \space g - 66 \space g = 14 \space g$$
