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 need to use the formula:
molality (m) = moles of solute / mass of solvent (in kg)
First, let's calculate the moles of solute for each trial:
Trial 1:
Mass of solid unknown = 5.034g
Molar mass of unknown = 110.98g/mol
moles of solute = mass / molar mass
moles of solute = 5.034g / 110.98g/mol
moles of solute = 0.04534 mol
Trial 2:
Mass of solid unknown = 5.034g
Molar mass of unknown = 110.98g/mol
moles of solute = mass / molar mass
moles of solute = 5.034g / 110.98g/mol
moles of solute = 0.04534 mol
Next, let's calculate the mass of the solvent for each trial:
Trial 1:
Mass of empty beaker = 163.20g
Mass of full beaker = 207.245g
mass of solvent = mass of full beaker - mass of empty beaker
mass of solvent = 207.245g - 163.20g
mass of solvent = 44.045g = 0.044045 kg
Trial 2:
Mass of empty beaker = 163.25g
Mass of full beaker = 204.97g
mass of solvent = mass of full beaker - mass of empty beaker
mass of solvent = 204.97g - 163.25g
mass of solvent = 41.72g = 0.04172 kg
Finally, let's calculate the molality for each trial:
Trial 1:
molality (m) = 0.04534 mol / 0.044045 kg
molality (m) ≈ 1.03 mol/kg
Trial 2:
molality (m) = 0.04534 mol / 0.04172 kg
molality (m) ≈ 1.09 mol/kg
Therefore, the molality for Trial 1 is approximately 1.03 mol/kg, and the molality for Trial 2 is approximately 1.09 mol/kg.