Question

A 6.0 gram sample of an unknown compound is dissolved in 125 g of water. The solutions freezing point is lowered to 1.38 C° below the freezing point of pure water. What is the molar mass of the sample? K for water is 1.86 C°/m.

Answer 1

Using the freezing point depression constant (Kf) for water and the observed freezing point depression, the molality of the solution was calculated. This was then used to determine the number of moles of solute present. The molar mass of the unknown compound was found to be 64.72 g/mol.

Step-by-step explanation:

To calculate the molar mass of the unknown compound we will use the phenomenon known as freezing point depression, which is a colligative property. This property is related to the number of moles of solute dissolved in the solvent. Given the freezing point depression (ΔTf = 1.38°C), the mass of the sample (6.0 g), and the mass of the water (125 g), along with water's freezing point depression constant (Kf = 1.86°C/m), we can find the molality of the solution and then the molar mass of the compound.

First, calculate the molality (m), which is moles of solute per kilogram of solvent:
m = ΔTf / Kf = 1.38°C / 1.86°C/m = 0.7419 m

Then, use the molality to find the number of moles of solute (n):
n = molality × mass of solvent (kg) = 0.7419 m × 0.125 kg = 0.0927 mol

Finally, calculate the molar mass (M) of the solute:
M = mass of solute (g) / number of moles of solute = 6.0 g / 0.0927 mol = 64.72 g/mol

Answer 2

Change in the freezing point = Kf * m

m = molality
Kf = 1.86 °C/m

m = Change in the freezing point / k = 1.38°C / 1.86 °C/m =0.742 m

m = #mol of solute / Kg of solvent

Kg of solvent = 125g/1000 g/kg = 0.125 kg.

#mol of solute = m* Kg of solvent = 0.742m * 0.125 kg =0.09275mol

Molar mass = 6. 0 g / 0.09275 mol = 64.7 g/mol