Let's break down the problem step by step: ### **Step 1: Identify the value of \( K_i \)** From the table: - \( K_i \) (μM) = 6.0 ### **Step 2: Volume of solution to be prepared** - Volume = 100.0 mL = 0.100 L ### **Step 3: Calculate moles of Compound 1 needed** We want a concentration of 6.0 μM: \[ \text{Moles needed} = \text{Concentration} \times \text{Volume} \] Convert μM to M: \[ 6.0\, \mu\text{M} = 6.0 \times 10^{-6}\, \text{mol/L} \] So: \[ \text{Moles needed} = 6.0 \times 10^{-6}\, \text{mol/L} \times 0.100\, \text{L} = 6.0 \times 10^{-7}\, \text{mol} \] ### **Step 4: Calculate the mass of Compound 1 needed** From the table, the molecular weight (\( M_w \)) of Compound 1 is 564 g/mol. \[ \text{Mass} = \text{moles} \times \text{molecular weight} \] \[ \text{Mass} = 6.0 \times 10^{-7}\, \text{mol} \times 564\, \text{g/mol} = 3.384 \times 10^{-4}\, \text{g} \] Convert grams to milligrams: \[ 3.384 \times 10^{-4}\, \text{g} = 0.3384\, \text{mg} \] ### **Step 5: Compare with answer choices** - A. 48.4 mg - B. 2.42 mg - C. 5.64 mg - D. 2.92 mg ### **Step 6: Double-check calculations** **Wait:** Let's check the molecular weight. The table shows a value "564", which is likely the molecular weight (since it's not a kinetic parameter). If so: \[ \text{Mass} = 6.0 \times 10^{-7}\, \text{mol} \times 564\, \text{g/mol} = 3.384 \times 10^{-4}\, \text{g} = 0.3384\, \text{mg} \] None of the answer choices match this. Let's check if the concentration is actually **6.0 μM** or **60.0 μM** (check the table and question). **Review the table:** - The highest concentration in the table for Compound 1 is 60.0 μM. - The question says: "What quantity of Compound 1 must be provided to prepare 100.0 mL of solution **with a concentration equal to \( K_i \)?**" So, the correct value is **\( K_i = 6.0\, \mu M \)**. ### **Step 7: Check for calculation errors** Let's recompute: \[ 6.0\, \mu\text{M} = 6.0 \times 10^{-6}\, \text{mol/L} \] \[ 0.100\, \text{L} \times 6.0 \times 10^{-6}\, \text{mol/L} = 6.0 \times 10^{-7}\, \text{mol} \] \[ 6.0 \times 10^{-7}\, \text{mol} \times 564\, \text{g/mol} = 3.384 \times 10^{-4}\, \text{g} \] \[ 3.384 \times 10^{-4}\, \text{g} \times 1000 = 0.3384\, \text{mg} \] ### **Step 8: Re-evaluate the answer choices** None of the answer choices match. There may be a typo in the question or a misinterpretation. #### **Let's try with 60.0 μM, just in case** \[ 60.0\, \mu\text{M} = 60.0 \times 10^{-6}\, \text{mol/L} \] \[ 0.100\, \text{L} \times 60.0 \times 10^{-6}\, \text{mol/L} = 6.0 \times 10^{-6}\, \text{mol} \] \[ 6.0 \times 10^{-6}\, \text{mol} \times 564\, \text{g/mol} = 0.003384\, \text{g} = 3.384\, \text{mg} \] Still, not matching. Let's try 5.64 mg: \[ \frac{5.64\, \text{mg}}{564\, \text{g/mol}} = \frac{5.64 \times 10^{-3}\, \text{g}}{564\, \text{g/mol}} = 1 \times 10^{-5}\, \text{mol} \] \[ \text{Concentration} = \frac{1 \times 10^{-5}\, \text{mol}}{0.1\, \text{L}} = 1 \times 10^{-4}\, \text{mol/L} = 100\, \mu\text{M} \] ### **FINAL ANSWER** **Based strictly on calculation and the provided \( K_i = 6.0\, \mu M \):** \[ \boxed{0.338\, \text{mg}} \] But this value does **not** match any answer choices. **If you are to select from the options, based on calculation shown, none is exactly correct.** If you believe there is a typo and the intended answer was for 5.64 mg, the corresponding concentration would be 100 μM, not 6 μM. **If you have to pick the closest, pick the lowest value. Otherwise, the calculation above is correct for 6.0 μM.** --- **If you need clarification, let me know!**