Answer to Question #341663 in General Chemistry for sue

Question

A. Calculate the molality, freezing point, and boiling point for each of the following water solutions of nonionizing solutes: 1. 144 g of C6H12O6 dissolved in 1000 g of H2O 2. 48 g of CH3OH dissolved in 200 g of H2O 3. 184 g of C2H5OH dissolved in 400 g of H2O 4. 600 g of C3H7OH dissolved in 600 g of H2O 5. 100 g of of C2H6O2 dissolved in 200 g of H2O

Answer 1

1.m = n(C₆H₁₂O₆) / m(H₂O)_kg = (144 g / 180g/mol)/ 1 kg = 0.8 mol/kg

ΔT = K_fm = 1.86 °C/m * 0.8 mol/kg = 1.496 °C

Freezing point of water = 0 °C

The freezing point of the solution = 0 - 1.496 = - 1.496 °C

ΔT_b​ = K_bm = 0.512 ^oC/m * 0.8 mol/kg = 0.41^oC

The boiling pint of pure water is 100 ^oC.

The boiling point of aqueous solution = 100 + 0.41 = 100.41 ^oC

2.m = n(CH₃OH) / m(H₂O)_kg = (48 g / 32 g/mol)/ 0.2 kg = 7.5 mol/kg

ΔT = K_fm = 1.86 °C/m * 7.5 mol/kg = 13.95 °C

The freezing point of the solution = 0 - 13.95 = - 13.95 °C

ΔT_b​ = K_bm = 0.512 ^oC/m * 7,5 mol/kg = 3.84 ^oC

The boiling point of aqueous solution = 100 + 3.84 = 103.84 ^oC

3.m = n(C₂H₅OH) / m(H₂O)_kg = (184 g / 46 g/mol)/ 0.4 kg = 10 mol/kg

ΔT = K_fm = 1.86 °C/m * 10 mol/kg = 18.6 °C

The freezing point of the solution = 0 - 18.6 = - 18.6 °C

ΔT_b​ = K_bm = 0.512 ^oC/m * 10 mol/kg = 5.12 ^oC

The boiling point of aqueous solution = 100 + 5.12 = 105.12 ^oC

4.m = n(C₃H₇OH) / m(H₂O)_kg = (600 g / 60 g/mol)/ 0.6 kg = 16.67 mol/kg

ΔT = K_fm = 1.86 °C/m * 16.67 mol/kg = 31.01 °C

The freezing point of the solution = 0 - 31.01 = - 31.01 °C

ΔT_b​ = K_bm = 0.512 ^oC/m * 16.67 mol/kg = 8.54 ^oC

The boiling point of aqueous solution = 100 + 8.54 = 108.54 ^oC

5.m = n(C₂H₆O₂) / m(H₂O)_kg = (100 g / 62 g/mol)/ 0.2 kg = 8.06 mol/kg

ΔT = K_fm = 1.86 °C/m * 8.06 mol/kg = 14.99 °C

The freezing point of the solution = 0 - 14.99 = - 14.99 °C

ΔT_b​ = K_bm = 0.512 ^oC/m * 8.06 mol/kg = 4.13 ^oC

The boiling point of aqueous solution = 100 + 4.13 = 104.13 ^oC