Using the data of 0.3 degrees Celsius freezing point of water-ice mixture, mass of solid unknown used 5.034g, freezing point of unknown liquid solution -3.0 degrees celsius for trial 1 and -2.5 degrees celsius for trial 2, mass of empty beaker 163.20g and full beaker 207.245 for trial one, and empty beaker with a mass of 163..25g and full beaker with a mass of 204.97g for trial two. Calculate the molality for each trial in mol/kg. The theoretical molar mass of the solid unknown is 110.98g/mol
To calculate the molality for each trial, we can use the formula:
Molality (m) = moles of solute / mass of solvent (in kg)
First, let's calculate the moles of solute for each trial.
For Trial 1:
Mass of solid unknown = 5.034g
Molar mass of solid unknown = 110.98g/mol
moles of solute = mass of solid unknown / molar mass of solid unknown
= 5.034g / 110.98g/mol
= 0.0453 mol
Now, let's calculate the mass of solvent (water-ice mixture) for Trial 1.
Mass of full beaker = 207.245g
Mass of empty beaker = 163.20g
Mass of solvent = Mass of full beaker - Mass of empty beaker
= 207.245g - 163.20g
= 44.045g
Next, let's calculate the molality for Trial 1.
Molality (m) = moles of solute / mass of solvent (in kg)
= 0.0453 mol / (44.045g / 1000g/kg)
= 1.03 mol/kg (rounded to two decimal places)
For Trial 2:
The same steps will be followed.
Mass of solid unknown = 5.034g
Molar mass of solid unknown = 110.98g/mol
moles of solute = mass of solid unknown / molar mass of solid unknown
= 5.034g / 110.98g/mol
= 0.0453 mol
Mass of solvent (water-ice mixture) for Trial 2:
Mass of full beaker = 204.97g
Mass of empty beaker = 163.25g
Mass of solvent = Mass of full beaker - Mass of empty beaker
= 204.97g - 163.25g
= 41.72g
Molality (m) = moles of solute / mass of solvent (in kg)
= 0.0453 mol / (41.72g / 1000g/kg)
= 1.09 mol/kg (rounded to two decimal places)
So, the molality for Trial 1 is 1.03 mol/kg and for Trial 2 is 1.09 mol/kg.