1. **State the problem:** We need to find how many grams of water are formed when 85.0 mL of 0.750 M KOH reacts completely with HNO3. 2. **Write the balanced chemical equation:** $$\text{KOH (aq)} + \text{HNO}_3 \text{(aq)} \rightarrow \text{KNO}_3 \text{(aq)} + \text{H}_2\text{O (l)}$$ 3. **Identify the mole ratio:** From the equation, 1 mole of KOH produces 1 mole of H2O. 4. **Calculate moles of KOH:** $$\text{Moles KOH} = Molarity \times Volume = 0.750 \frac{mol}{L} \times 0.0850 L = 0.06375 \text{ mol}$$ 5. **Calculate moles of water formed:** Since the ratio is 1:1, $$\text{Moles H}_2\text{O} = 0.06375 \text{ mol}$$ 6. **Calculate mass of water:** Molar mass of water is approximately 18.015 g/mol. $$\text{Mass H}_2\text{O} = 0.06375 \text{ mol} \times 18.015 \frac{g}{mol} = 1.148 \text{ g}$$ 7. **Round to appropriate significant figures:** $$1.15 \text{ g}$$ **Final answer:** 1.15 g of water is formed.