Quinine (C20H24N2O2), is a key medication for treating malaria. Its molecular weight is 324.42 g/mol. a) A lab protocol requires you to prepare 250.0 mL of a 0.0150 mol dm−3 quinine solution for a cell culture experiment. Calculate the mass of quinine (in grams) you must dissolve

a) To calculate the mass of quinine needed, first convert the volume to dm³, then calculate the moles, and finally the mass.

Step 1: Convert volume from mL to dm³. $V = 250.0 mL \times \frac{1 dm^3}{1000 mL} = 0.2500 dm^3$

Step 2: Calculate the moles of quinine. $\text{Moles} = Concentration \times \text{Volume}$ $\text{Moles} = 0.0150 mol dm^{-3} \times 0.2500 dm^3 = 0.00375 \text{ mol}$

Step 3: Calculate the mass of quinine. $\text{Mass} = Moles \times \text{Molecular Weight}$ $\text{Mass} = 0.00375 mol \times 324.42 g/mol = 1.216575 \text{ g}$ Rounding to three significant figures based on the given concentration: $\text{Mass} = 1.22 \text{ g}$ The mass of quinine you must dissolve is $\boxed{1.22 g}$.

b) To calculate the volume of stock solution needed, use the dilution formula $M_1V_1 = M_2V_2$.

Step 1: Identify the knowns and unknowns. $M_1 = 2.50 mol dm^{-3}$ $V_1 = ?$ $M_2 = 0.0150 mol dm^{-3}$ $V_2 = 500.0 \text{ mL}$

Step 2: Apply the dilution formula and solve for $V_1$. $M_1V_1 = M_2V_2$ $(2.50 mol dm^{-3}) \times V_1 = (0.0150 mol dm^{-3}) \times (500.0 mL)$ $V_1 = \frac{(0.0150 mol dm^{-3}) \times (500.0 mL)}{2.50 mol dm^{-3}}$ $V_1 = \frac{7.50 mol mL}{2.50 mol}$ $V_1 = 3.00 \text{ mL}$ The volume of stock solution needed is $\boxed{3.00 mL}$.

c) To calculate the percent by mass of Bromine, divide the atomic mass of Bromine by the molecular mass of the derivative and multiply by 100%.

Step 1: Identify the mass of Bromine in one mole of C₂₀H₂₃BrN₂O₂. $\text{Mass of Br} = 79.90 \text{ g/mol}$

Step 2: Calculate the percent by mass of Bromine. $\text{Percent by mass of Br} = \frac{Mass of Br}{Molecular Weight of C_{20}H_{23}BrN_{2}O_{2}} \times 100\%$ $\text{Percent by mass of Br} = \frac{79.90 g/mol}{404.32 g/mol} \times 100\%$ $\text{Percent by mass of Br} = 0.19760096 \times 100\%$ $\text{Percent by mass of Br} = 19.760096\%$ Rounding to two decimal places: $\text{Percent by mass of Br} = 19.76\%$ The percent by mass of Bromine in the monobrominated derivative is $\boxed{19.76\%}$.

d) To find the moles of quinine alkaloid, first convert the dose to grams, then calculate moles of quinine sulfate, and finally use the mole ratio to find moles of quinine alkaloid.

Step 1: Convert the dose of quinine sulfate from mg to g. $\text{Mass of quinine sulfate} = 650 mg \times \frac{1 g}{1000 mg} = 0.650 \text{ g}$

Step 2: Calculate the moles of quinine sulfate. $\text{Moles of quinine sulfate} = \frac{Mass}{Molecular Weight}$ $\text{Moles of quinine sulfate} = \frac{0.650 g}{782.96 g/mol} = 0.00083015 \text{ mol}$

Step 3: Use the mole ratio to find moles of quinine alkaloid. The hint states that each mole of quinine sulfate salt contains 2 moles of quinine alkaloid. $\text{Moles of quinine alkaloid} = Moles of quinine sulfate \times 2$ $\text{Moles of quinine alkaloid} = 0.00083015 mol \times 2 = 0.0016603 \text{ mol}$ Rounding to three significant figures: $\text{Moles of quinine alkaloid} = 0.00166 \text{ mol}$ There are $\boxed{0.00166 mol}$ of actual quinine alkaloid in a single 650 mg dose.