Let's solve the problem step by step: **Given:** - Volume of solution = 400 mL = 0.400 L - Desired concentration = 0.0500 mol/L - Solute: KNO₃·8H₂O (potassium nitrate octahydrate) --- ### 1. Calculate moles of solute needed \[ \text{Moles} = \text{Molarity} \times \text{Volume (in L)} \] \[ \text{Moles} = 0.0500\, \text{mol/L} \times 0.400\, \text{L} = 0.0200\, \text{mol} \] --- ### 2. Find the molar mass of KNO₃·8H₂O - K: 39.10 g/mol - N: 14.01 g/mol - O₃: 16.00 × 3 = 48.00 g/mol - 8H₂O: 8 × (2.02 + 16.00) = 8 × 18.02 = 144.16 g/mol \[ \text{Molar mass} = 39.10 + 14.01 + 48.00 + 144.16 = 245.27\, \text{g/mol} \] --- ### 3. Calculate the mass required \[ \text{Mass} = \text{Moles} \times \text{Molar mass} \] \[ \text{Mass} = 0.0200\, \text{mol} \times 245.27\, \text{g/mol} = 4.905\, \text{g} \] --- ### 4. Report with correct significant figures The limiting sig figs are 3 (from 0.0500 and 0.400), so: \[ \boxed{4.91\, \text{g}} \] --- ## **Final Answer** **4.91 g of KNO₃·8H₂O is needed to make 400 mL of a 0.0500 mol/L solution.**