A 90 g sample of potassium nitrate holds 12 g of sodium chloride as an impurity…

Step 1. The water. $90\ \text{g}$ dissolves in $\dfrac{90}{169} \times 100 = 53.3\ \text{g}$ of water at $80\ ^\circ\text{C}$.

Step 2. The impurity when hot. At $80\ ^\circ\text{C}$, $53.3\ \text{g}$ of water can hold $0.380 \times 53.3 = 20.2\ \text{g}$ of sodium chloride. The sample has only $12\ \text{g}$, so it all dissolves.

Step 3. The crystals. At $20\ ^\circ\text{C}$ the water holds $0.316 \times 53.3 = 16.8\ \text{g}$ of potassium nitrate. So the crystals are $90 - 16.8 = 73.2\ \text{g}$.

Step 4. The impurity when cold. At $20\ ^\circ\text{C}$ the water can hold $0.359 \times 53.3 = 19.1\ \text{g}$ of sodium chloride. It has only $12\ \text{g}$, so the sodium chloride stays dissolved and the crystals are pure.

Step 5. The yield. $\dfrac{73.2}{90} \times 100 = 81.3\ \%$.