Question

Calculate the molality and the boiling point of a solution made by dissolving 155 g acetone (MM= 58.08 g/mol ) in 235 g of water. The molal boiling point constant for water is 0.512∘C/m.

Water normally boils at 100∘C
∆Tb = mKb
Molarity =m

Answer 1

The molality of the solution is 11.36 m, calculated by dividing the moles of acetone by the kilograms of water. The solution's boiling point is raised to 105.82°C, calculated by multiplying the molality with the boiling point elevation constant of water and adding it to water's normal boiling point.

Step-by-step explanation:

The student is interested in calculating the molality of a solution and determining its new boiling point after acetone is dissolved in water. To find the molality, we use the formula:

Molality (m) = moles of solute / kilograms of solvent

First, we calculate the moles of acetone:

moles of acetone = mass of acetone / molar mass of acetone = 155 g / 58.08 g/mol = 2.67 mol

Then we calculate the mass of the water in kilograms:

mass of water = 235 g = 0.235 kg

Now we can find the molality:

molality = 2.67 mol / 0.235 kg = 11.36 m

To determine the new boiling point, we use the formula:

∆Tb = Kb x m

Where Kb is the molal boiling point constant for water, and m is the molality of the solution.

∆Tb = 0.512°C/m x 11.36 m = 5.82°C

Therefore, the new boiling point of the solution is:

new boiling point = normal boiling point of water + ∆Tb = 100°C + 5.82°C = 105.82°C