Final answer:
Using the formula ΔT = i * Kb * m and given data, it is calculated that 124.22 g of sucrose would be required to raise the boiling point of 500 g of water from 99.63°C to 100°C at 750 mm Hg. None of the options provided are sufficient to achieve this boiling point elevation.
Step-by-step explanation:
The boiling point of a solution can be increased by adding a nonvolatile solute such as sucrose. This is known as the boiling point elevation. It is a colligative property, meaning it's dependent on the number of particles of solute present in the solution, not their identity. The molal boiling point elevation constant (Kb) for water is 0.51°C/m.
To find out how much sucrose needs to be added to 500 g of water to achieve a boiling point of 100°C, we use the formula:
ΔT = i * Kb * m
Where ΔT is the boiling point elevation, i is the van't Hoff factor (which is 1 for non-ionic compounds like sucrose), Kb is the boiling point elevation constant, and m is the molality of the solution.
Given that the boiling point of water at 750 mm Hg is 99.63°C, to raise the boiling point by 0.37°C (from 99.63°C to 100°C), we can calculate the required molality:
m = ΔT / (i * Kb) = 0.37°C / (1 * 0.51°C/m) = 0.7255 m
The molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol. To calculate the grams of sucrose, we use the formula:
mass = molality * mass of solvent (in kg) * molar mass of solute
mass = 0.7255 m * 0.500 kg * 342.3 g/mol
mass = 124.21575 g
However, the initial question asks for a range from 10 g to 40 g, none of which would be enough to achieve the desired boiling point elevation. Therefore, based on the calculations, the correct choice would be none of the given options.