Find the freezing point of the solution containing 0.520g glucose dissolved in 80.2g of water. Kf for water is 1.86 k/molal.?

Question

Answer

Calculation of molality:
Molality is defined as the number of moles of solute per kilogram of solvent.

Molar mass of glucose (C6H12O6) = 6(12.01) + 12(1.01) + 6(16.00) = 180.18 g/mol
Number of moles of glucose = 0.520 g / 180.18 g/mol = 0.00289 mol
Mass of water = 80.2 g
Molality of the solution = 0.00289 mol / 0.0802 kg = 0.0360 mol/kg

Calculation of change in freezing point:
The change in freezing point is given by the formula ∆Tf = Kf x molality.

Kf for water = 1.86 K/molal
∆Tf = 1.86 K/molal x 0.0360 mol/kg = 0.067 K

Calculation of freezing point of the solution:
The freezing point of the solution is given by the formula Tf = Tf° - ∆Tf, where Tf° is the freezing point of the solvent (water).

For water, Tf° = 0°C
Tf = 0°C - 0.067°C = -0.067°C

Conclusion:
The freezing point of the solution containing 0.520 g glucose dissolved in 80.2 g of water is -0.067°C. This means that the solution will freeze at a temperature 0.067°C lower than the freezing point of pure water.

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