Question

An aqueous solution of a nonionic solute has a freezing point of -0.284 degrees Celsius. The solution contains 1.86 grams of solute dissolved in 77.7 grams of H₂O. What is the molar mass of the solute?

A) 134.6 g/mol
B) 78.9 g/mol
C) 46.1 g/mol
D) 92.5 g/mol

Answer 1

After calculating the molality (m) using the given freezing point depression (ΔTf) and the molal freezing-point depression constant (Kf), and then finding the moles of solute, we calculate the molar mass of the solute to be 156.8 g/mol.

Step-by-step explanation:

Calculating Molar Mass from Freezing Point Depression

To determine the molar mass of a solute from the freezing point depression of a solution, we can use the formula ΔTf = Kf x m, where ΔTf is the freezing point depression, Kf is the freezing point depression constant for the solvent (which is water in this case, with a value of -1.86°C/m), and m is the molality of the solution.

We are given that the freezing point depression (ΔTf) is -0.284°C and the mass of the solute is 1.86 grams, dissolved in 77.7 grams of water. First, we calculate the molality (m) as follows:

m = ΔTf / Kf = -0.284°C / -1.86°C/m = 0.1527 m

Next, using the molality definition, which is moles of solute/kg of solvent, we can solve for moles of solute:

moles of solute = molality (m) x kg of solvent = 0.1527 m x 0.0777 kg = 0.01186 moles

With the number of moles and the mass of the solute, we can now calculate the molar mass:

• Molar mass = grams of solute / moles of solute = 1.86 g / 0.01186 moles = 156.8 g/mol