Calculation of Mass of Solute in Aqueous Solution

Given:

Cryoscopic constant of water (Kf) = 1.86 K mol^-1 kg

Freezing point depression (ΔTf) = -0.372 °C

Molecular weight of cane sugar (C₁₂H₂₂O₁₁) = 342.3 g/mol


Formula:

ΔTf = Kf x molality

Molality (m) = moles of solute / mass of solvent (kg)

Moles of solute = Mass of solute / Molecular weight of solute


Calculation:

ΔTf = -0.372 °C = -0.372 K

Kf = 1.86 K mol^-1 kg

Molecular weight of cane sugar (C₁₂H₂₂O₁₁) = 342.3 g/mol

Let's assume the mass of the solvent (water) to be 1 kg.

Therefore, molality (m) = moles of solute / 1 kg


We can calculate the moles of solute as follows:

ΔTf = Kf x molality

molality = ΔTf / Kf = -0.372 / 1.86 = -0.2 mol/kg

moles of solute = molality x mass of solvent (in kg)

moles of solute = -0.2 x 1 = -0.2 mol


We have obtained a negative value for moles of solute, which is not feasible. This indicates that our assumption for the mass of solvent was incorrect. Let's assume the mass of the solvent to be 1000 g.

Therefore, molality (m) = moles of solute / 1 kg


We can calculate the moles of solute as follows:

ΔTf = Kf x molality

molality = ΔTf / Kf = -0.372 / 1.86 = -0.2 mol/kg

moles of solute = molality x mass of solvent (in kg)

moles of solute = -0.2 x 0.1 = -0.02 mol


Again, we have obtained a negative value for moles of solute. This indicates that the mass of solute is less than what we assumed earlier. Let's assume the mass of solute to be x g.

Therefore, molality (m) = moles of solute / 1 kg


We can calculate the moles of solute as follows:

ΔTf = Kf x molality

molality = ΔTf / Kf = -0.372 / 1.86 = -0.2 mol/kg

moles of solute = molality x mass of solvent (in kg)

moles of solute = -0.2 x 1 = -0.