Question

A solution is prepared by dissolving 315 grams of methanol, CH3OH, in 1000 grams of water. What is the freezing point of this solution? [The freezing point depression constant for water is 1.86°C/mole solute in 1000g of water]

Answer 1

Answer:

(1) Tcs = - 17,44 °C

Explanation:

When a solute is dissolved in a solvent (like water), propierties of the pure solvent change. It prevents solidification at the characteristic temperature and an even more drastic decrease in energy is needed to achieve the ordering of the molecules. Said decrease in the freezing point of the solvent is proportional to the number of dissolved particles. These propierties are called “colligative”.  

To solve this problem you must make following calculations:

Data:

-m CH3OH: solute mass = 315 g

-PM CH3OH: solute molecular weight = 32,04 g/mol

-m H2O: solvent mass = 1000 g or 1 kg

- Kc: freezing point depression constant = 1,86 °C. solvent Kg / solute moles

- Tcw: freezing point pure water to 1 atm = 0 °C

- Tcs: freezing point solution =? (unknown)

Calculation

(1) ΔTc = Tcw - Tcs ------------ clearing the unknown: Tcs = Tcw - ΔTc

(2) ΔTc = Kc. m

Where:

- ΔTc: freezing point variation [°C]  

- (3) m: molality = solute moles / solvent kg ------------ n CH3OH / m H2O  

- (4) n CH3OH: solute moles = solute moles / solute PM

Calculating:

(4) n CH3OH = solute moles / solute PM = 315 g / 32,04 g/mol = 9,83 mol

(3) m = n CH3OH / m H2O = 9,83 mol solute / 1 kg water = 9,38 mol/kg

(2) ΔTc = Kc. m = 1,86 °C. Kg / moles * 9,38 mol/ kg = 17,44 °C

(1) Tcs = Tcw - ΔTc = 0°C - 17,44 °C = - 17,44 °C