The boiling point of a solution increases directly as a function of the number of moles of solute present in a given mass of solution. This relationship is expressed mathematically by the following equation.

ΔTb = Kb·m

ΔTb is the boiling-point elevation of a solution, m is the molality of solute particles in solution, and Kb is the molal boiling-point-elevation constant.

Determine the number of moles of sucrose present in a 11.70 g sample of water that has a boiling point of 102.89°C. The boiling point for pure water is 100.00°C and the molal boiling-point-elevation constant for water is 0.51°C/m

To determine the number of moles of sucrose present in the sample, we need to use the equation ΔTb = Kb·m, where ΔTb is the boiling-point elevation, Kb is the molal boiling-point-elevation constant, and m is the molality of solute particles in solution.

Given:
Sample mass of water (solvent) = 11.70 g
Boiling point of the solution = 102.89°C
Boiling point of pure water = 100.00°C
Molal boiling-point-elevation constant for water (Kb) = 0.51°C/m

First, we need to calculate the boiling-point elevation (ΔTb) of the solution by subtracting the boiling point of pure water from the boiling point of the solution:

ΔTb = 102.89°C - 100.00°C
ΔTb = 2.89°C

Next, we can use the equation ΔTb = Kb·m to find the molality (m). Rearranging the equation, we have:

m = ΔTb / Kb

Substituting the given values:

m = 2.89°C / 0.51°C/m
m = 5.6618 m

Now we can determine the number of moles of sucrose present in the sample using the molality and the mass of the solvent. First, we need to convert the mass of water from grams to kilograms:

Mass of water = 11.70 g = 0.0117 kg

The formula to calculate the number of moles (n) is:

n = m × mass of solvent (in kg)

Substituting the values:

n = 5.6618 m × 0.0117 kg
n = 0.06613 mol

Therefore, the number of moles of sucrose present in the 11.70 g sample is approximately 0.06613 mol.

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