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 molar mass of unknown solid for each trial. The theoretical molar mass of the solid unknown is 110.98g/mol
To calculate the molar mass of the unknown solid for each trial, we can use the formula:
Molar mass of unknown solid = (mass of solid unknown used / moles of solid unknown)
First, let's calculate the moles of solid unknown using the freezing point depression method:
ΔT = Kf * molality
Trial 1:
Given:
ΔT = 0.3 degrees Celsius
Kf = Freezing point depression constant (given as 1.86 degrees Celsius/m)
Mass of solid unknown used = 5.034g
Mass of empty beaker = 163.20g
Mass of full beaker = 207.245g
Mass of solvent = Mass of full beaker - Mass of empty beaker
= 207.245g - 163.20g
= 44.045g
Molality = moles of solute / mass of solvent (in kg)
= moles of solid unknown / 0.044045kg
ΔT = 1.86 * molality
0.3 = 1.86 * (moles of solid unknown / 0.044045)
moles of solid unknown = (0.3 * 0.044045) / 1.86
= 0.007113
Molar mass of solid unknown for Trial 1 = (5.034g / 0.007113)
= 706.86 g/mol
Trial 2:
Given:
ΔT = 3.0 degrees Celsius
Kf = Freezing point depression constant (given as 1.86 degrees Celsius/m)
Mass of solid unknown used = 5.034g
Mass of empty beaker = 163.25g
Mass of full beaker = 204.97g
Mass of solvent = Mass of full beaker - Mass of empty beaker
= 204.97g - 163.25g
= 41.72g
Molality = moles of solute / mass of solvent (in kg)
= moles of solid unknown / 0.04172kg
ΔT = 1.86 * molality
2.5 = 1.86 * (moles of solid unknown / 0.04172)
moles of solid unknown = (2.5 * 0.04172) / 1.86
= 0.055944
Molar mass of solid unknown for Trial 2 = (5.034g / 0.055944)
= 89.98 g/mol
Therefore, the molar mass of the unknown solid for Trial 1 is 706.86 g/mol, and for Trial 2 is 89.98 g/mol.