Using the data of 0.3 degrees Celsius freezing point of water-ice mixture, mass of liquid unknown used 7.882g, freezing point of unknown liquid solution -3.5 degrees celsius for trial 1 and -2.9 degrees celsius for trial 2, mass of empty beaker 163.098g and full beaker 233.84 for trial one, and empty beaker with a mass of 163.098g and full beaker with a mass of 238.712g for trial two. Mass of unknown per kg water is 11.14g/kg for trial one and 10.4 g/kg for trial 2. Calculate the molar mass of the unknown liquid for both trials
To calculate the molar mass of the unknown liquid for both trials, we can use the formula:
Molar mass = (mass of unknown liquid / moles of solute) × (1 / mass of solvent)
First, let's calculate the moles of solute for both trials using the given mass of the unknown liquid per kg of water.
For trial 1:
Mass of unknown liquid = 7.882g
Mass of solvent (water) = 233.84g - 163.098g = 70.742g
Mass of solute (unknown liquid) = 7.882g
Mass of solvent in kg = 70.742g / 1000 = 0.070742kg
Moles of solute = Mass of solute / Molar mass of solute
Moles of solute = 7.882g / 11.14g/kg = 0.707 moles
For trial 2:
Mass of unknown liquid = 7.882g
Mass of solvent (water) = 238.712g - 163.098g = 75.614g
Mass of solute (unknown liquid) = 7.882g
Mass of solvent in kg = 75.614g / 1000 = 0.075614kg
Moles of solute = Mass of solute / Molar mass of solute
Moles of solute = 7.882g / 10.4g/kg = 0.758 moles
Now, let's calculate the molar mass of the unknown liquid for both trials.
For trial 1:
Molar mass = (mass of unknown liquid / moles of solute) × (1 / mass of solvent)
Molar mass = (7.882g / 0.707 moles) × (1 / 0.070742kg)
Molar mass = 13.604g/mol
For trial 2:
Molar mass = (mass of unknown liquid / moles of solute) × (1 / mass of solvent)
Molar mass = (7.882g / 0.758 moles) × (1 / 0.075614kg)
Molar mass = 10.434g/mol
Therefore, the molar mass of the unknown liquid is 13.604g/mol for trial 1 and 10.434g/mol for trial 2.