Let's solve the question step by step: ### **Step 1: Identify the required concentration (K_i)** From the table, for Compound 1, the K_i value is: - K_i = **5.3 μM** **Explanation:** K_i is the inhibition constant, and the question asks for a solution with a concentration equal to K_i. --- ### **Step 2: Convert the concentration from μM to moles** - 1 μM = 1 × 10^-6 mol/L - So, 5.3 μM = 5.3 × 10^-6 mol/L The volume needed is 100.0 mL = 0.100 L. Moles of Compound 1 needed: \[ \text{Moles} = \text{Molarity} \times \text{Volume} = 5.3 \times 10^{-6} \text{ mol/L} \times 0.100 \text{ L} = 5.3 \times 10^{-7} \text{ mol} \] **Explanation:** We calculated the number of moles needed for the given volume and concentration. --- ### **Step 3: Convert moles to mass** From the table, the molecular weight (M_w) of Compound 1 is **564 g/mol**. So, mass needed: \[ \text{Mass} = \text{Moles} \times \text{Molecular Weight} = 5.3 \times 10^{-7} \text{ mol} \times 564 \text{ g/mol} = 2.9892 \times 10^{-4} \text{ g} = 0.29892 \text{ mg} \] Wait, let's check the math: \[ 5.3 \times 10^{-7} \text{ mol} \times 564 \text{ g/mol} = 2.9892 \times 10^{-4} \text{ g} \] Convert to mg: \[ 2.9892 \times 10^{-4} \text{ g} \times 1000 \text{ mg/g} = 0.29892 \text{ mg} \] But this value is not among the options. Let's re-calculate: \[ 5.3 \times 10^{-6} \text{ mol/L} \times 0.1 \text{ L} = 5.3 \times 10^{-7} \text{ mol} \] \[ 5.3 \times 10^{-7} \text{ mol} \times 564 \text{ g/mol} = 2.9892 \times 10^{-4} \text{ g} \] \[ 2.9892 \times 10^{-4} \text{ g} = 0.29892 \text{ mg} \] Still, options are much higher. Let's check if the *molecular weight* is given elsewhere or if the step is missing. Actually, the table does **not** list the molecular weight directly, so let's check if the calculation aligns with the answer options. #### Let's check the calculation with 100 mL (0.1 L): \[ 5.3 \times 10^{-6} \text{ mol/L} \times 0.1 \text{ L} = 5.3 \times 10^{-7} \text{ mol} \] \[ 5.3 \times 10^{-7} \text{ mol} \times 564 \text{ g/mol} = 2.9892 \times 10^{-4} \text{ g} \] \[ = 0.29892 \text{ mg} \] But the answer options are: A. 48.4 mg B. 2.24 mg C. 5.64 mg D. 2.92 mg **Let’s try 5.3 μM = 5.3 × 10^-6 mol in 100 mL, but maybe the calculation is in μg instead of mg.** But let's check 5.3 × 10^-6 mol × 564 g/mol = 0.0029892 g = 2.9892 mg (for 1 L). For 0.1 L, it would be 0.29892 mg. Checking if the concentration should be for 1 L: - For 1 L: 5.3 × 10^-6 mol × 564 g/mol = 0.0029892 g = 2.9892 mg. - For 100 mL: 0.29892 mg. None of these match, so let's verify if the correct unit conversion is being made. #### Let's try 5.3 μM × 564 (g/mol) × 0.1 (L) × 1000 (mg/g): \[ 5.3 \times 10^{-6} \text{ mol/L} \times 0.1 \text{ L} \times 564 \text{ g/mol} \times 1000 \text{ mg/g} \] \[ = 5.3 \times 10^{-7} \text{ mol} \times 564 \text{ g/mol} \times 1000 \text{ mg/g} \] \[ = 2.9892 \times 10^{-4} \text{ g} \times 1000 \text{ mg/g} = 0.29892 \text{ mg} \] Still not matching options. Let's check if the *concentration* is 5.3 mM (not μM) by mistake, but the table says μM. #### Let's try with 56.4 g/mol (if there was a typo): \[ 5.3 \times 10^{-7} \text{ mol} \times 56.4 \text{ g/mol} = 2.9892 \times 10^{-5} \text{ g} = 0.029892 \text{ mg} \] Still too low. #### Let's try with 0.01 L (if the volume is 10 mL): But the question clearly says 100 mL. #### Let's try with 100 mL = 0.1 L and check calculation: \[ 5.3 \times 10^{-6} \text{ mol/L} \times 0.1 \text{ L} = 5.3 \times 10^{-7} \text{ mol} \] \[ 5.3 \times 10^{-7} \text{ mol} \times 564 \text{ g/mol} = 2.9892 \times 10^{-4} \text{ g} \] \[ 2.9892 \times 10^{-4} \text{ g} \times 1000 \text{ mg/g} = 0.29892 \text{ mg} \] #### But none of the options match. #### Let's check if the K_i value is 5.64 μM (from the options): - 5.64 × 10^-6 mol/L × 0.1 L = 5.64 × 10^-7 mol - 5.64 × 10^-7 mol × 564 g/mol = 3.184 × 10^-4 g = 0.3184 mg Still, none match the answer options. ### **Final Answer** Based on the calculation and the data provided, the correct quantity is: \[ \boxed{0.29892 \text{ mg}} \] However, since this doesn't match the answer choices, **double-check if there's a typographical or data error in the molecular weight or K_i value**. The calculated answer most closely aligns with the method above. #### **If you must select from the given options, the closest would be D. 2.92 mg, but based on calculations, the answer should be about 0.3 mg.**