What is the change in the freezing point of a solution made by dissolving 14.7 g of C6H1206 into 150 ml of water? Answer in 2 decimal.
The density of water is 1.00 g/ml and and Kf = 1.86 °C/m

The change in freezing point is 1.02°C

The change in the freezing point (ΔTf) is calculated from the following equation:

ΔTf = i x Kf x m

Given:

i = 1 (Van't Hoff factor, is equal to 1 because the solute is a nonelectrolyte)

Kf = 1.86°C/m (Freezing point constant of the solvent)

m = molality

We have to calculate the molality (m), which is equal to the moles of solute in 1 kg of solvent:

m = moles solute/ kg solvent

The solute is C₆H₁₂O₆. Thus, we calculate the moles of solute by dividing the mass into its molecular weight (Mw):

Mw(C₆H₁₂O) = (12 g/mol C x 6) + ( 1 g/mol H x 12) + (16 g/mol O x 6) = 180 g/mol

moles solute = mass C₆H₁₂O/Mw(C₆H₁₂O) = 14.7 g/(180 g/mol) = 0.082 mol

The solvent is water. Its mass (in kg) is obtained from the volume and the density, as follows:

kg solvent = V x density = 150 mL x 1.00 g/mL x 1 kg/1000 g = 0.15 kg

Now, we calculate the molality (m):

m = moles of solute/kg solvent = 0.082 mol/(0.15 kg) = 0.546 m

Finally, we calculate the change in freezing point:

ΔTf = i x Kf x m = 1 x 1.86°C/m x 0.546 m = 1.02°C

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