Answer:
See explanation
Explanation:
(a)
From the data provided;
Mass of methanol =11.0 g
Mass of solvent (H2O) =100 g = 0.1 kg
Molar mass of CH3OH =32.04 g/mol
Number of moles of solute = 0.343 mol
molality(m) of methanol = 3.43 m
Also,
Mass of ethanol =22.0 g
Mass of solvent (H2O) =200 g = 0.2 kg
Molar mass of C2H5OH =46.068 g/mol
Number of moles of solute = 0.477 mol
molality(m) of ethanol= 2.38 m
The greater the concentration of a solution, the higher the freezing point of solvent will be depressed. Therefore, higher molality, leader to a greater decrease of the freezing point.
Therefore, the freezing point of methanol CH3OH/H2O is lower than the freezing point of ethanol.
b)
Mass of solvent = 1 kg
Mass of water = 20g or 0.02 Kg
Molar mass of water = 18 gmol-1
Number of moles of solute = 20g/18 gmol-1 = 1.11 moles
Molality= 1.11 moles/1 kg = 1.11 m
Also;
Mass of ethanol = 20 g = 0.02 kg
Mass of solvent = 1.00 Kg
Molar mass of solute = 46 gmol-1
Number of moles of solute = 20g/ 46 gmol-1 = 0.43 moles
Molality= 0.43 moles/ 1 kg = 0.43 m
The greater the concentration of a solution, the higher the freezing point of solvent will be depressed. Therefore, higher molality, leader to a greater decrease of the freezing point.
Therefore, the freezing point of water H2O/CH3OH is lower than the freezing point of ethanol.
