$$\Delta T_b = K_b \times m$$
where ΔTb is the change in boiling point, Kb is the boiling point elevation constant of the solvent, and m is the molality of the solution.
We are given that ΔTb = 100.680 °C - 100.000 °C = 0.680 °C, and that the solvent is water, which has a boiling point elevation constant of Kb = 0.512 °C/m.
Substituting these values into the equation, we get:
$$0.680 °C = 0.512 °C/m \times m$$
Solving for m, we get:
$$m = 1.33 m$$
This means that the solution contains 1.33 moles of solute per kilogram of water.
To calculate the molar mass of the solute, we can use the following equation:
$$Molarity = \frac{Moles\text{ of Solute}}{Liters\text{ of Solution}}$$
We know that the solution contains 1.33 moles of solute, and we can calculate the liters of solution using the density of water (1 g/mL):
$$Liters\text{ of Solution} = \frac{3.90 \times 10^{2} g}{1 g/mL} = 390 mL$$
Now we can use molar mass formula:
$$Molarity = \frac{1.33\text{ mol}}{0.390 \text{ L}}$$
Molarity becomes:
$$Molarity = 3.41$$
Finally, we use the following equation to calculate the molar mass of the solute:
$$Molar\text{ Mass} = \frac{Grams\text{ of Solute}}{Moles\text{ of Solute}}$$
Substituting the values we know, we get:
$$Molar\text{ Mass} = \frac{64.3 g}{1.33 mol}$$
$$Molar\text{ Mass} = 48.3\text{ g/mol}$$
Therefore, the molar mass of the solute is 48.3 g/mol.
