Molar Mass Lab
09/17/13
Che 102L
Lab # 4 Molar Mass by Freezing Point in Anti-Freeze. Objective: to determine the freezing point of pure naphthalene, determine the molar mass of an unknown by measuring the freezing point depression of a solution of the unknown in naphthalene.
Materials: Beakers 600 mL, 2-250 mL,Crystal Ice,20 g of NaCl,Stirring Rod,Thermometer,Foil, Water, 10 mL, Cylinder, 25 mL 100 mL Dropper, 2 Test Tubes.
Procedure:
Filled 600 mL beaker with ice, took temperature until it got to -10 c. The next step was to take 1 test tube with water placed in water bath, freezing point of distilled water -0.0 c.
Placed a 250 mL beaker on the balance weighed 113g. Added 10 mL of anti-freeze came to 126g. This is called solution 1. Placed …show more content…
Pure Biphenyl freezes at 71.0 c. A solution of 1.44 g of Anthracene (C14 H10 molar mass = 178.2 g/mol) and 12.87 g of biphenyl froze at 21.1 c. Calculate Kf of Biphenyl.
The solution has 1.44 g of anthracene in 12.87 g of solvent also have 1.44 g/ 178.2 g / mole of the solute = 0.00808 moles in 12.89 g of solvent moles of solute / kg of solvent = 0.008008 x 1000 / 12.89 = 0.62690 m
delta t = m x kf in this case
kf = ( 71.0 - 21.1 ) / 0.62690 = 79.60 C/m
4. Suppose that a student calculated Kf of Biphenyl to be 81.2 c.Kg/mol. calculate the molar mass of an unknown if a solution containing 1.64g of unknown in 18.22 g of Biphenyl froze at 39.9 c.
Delta t = M x Kf
M = (71 - 39.9) / 81.2 = 0.3830 moles of solute / kg of solvent
also have 1.64 g / 18.22 g of solvent = 1.64 x 1000 / 18.22 g / kg = 90.0 g
so in 1 kg we have .3830moles = 90.0 g
mass of 1 mole = 90.0 / .3830 = 235.0 g/ mole
References: General Chemistry Laboratory Manual for Alabama A&M
Che 102L
Lab # 4 Molar Mass by Freezing Point in Anti-Freeze. Objective: to determine the freezing point of pure naphthalene, determine the molar mass of an unknown by measuring the freezing point depression of a solution of the unknown in naphthalene.
Materials: Beakers 600 mL, 2-250 mL,Crystal Ice,20 g of NaCl,Stirring Rod,Thermometer,Foil, Water, 10 mL, Cylinder, 25 mL 100 mL Dropper, 2 Test Tubes.
Procedure:
Filled 600 mL beaker with ice, took temperature until it got to -10 c. The next step was to take 1 test tube with water placed in water bath, freezing point of distilled water -0.0 c.
Placed a 250 mL beaker on the balance weighed 113g. Added 10 mL of anti-freeze came to 126g. This is called solution 1. Placed …show more content…
Pure Biphenyl freezes at 71.0 c. A solution of 1.44 g of Anthracene (C14 H10 molar mass = 178.2 g/mol) and 12.87 g of biphenyl froze at 21.1 c. Calculate Kf of Biphenyl.
The solution has 1.44 g of anthracene in 12.87 g of solvent also have 1.44 g/ 178.2 g / mole of the solute = 0.00808 moles in 12.89 g of solvent moles of solute / kg of solvent = 0.008008 x 1000 / 12.89 = 0.62690 m
delta t = m x kf in this case
kf = ( 71.0 - 21.1 ) / 0.62690 = 79.60 C/m
4. Suppose that a student calculated Kf of Biphenyl to be 81.2 c.Kg/mol. calculate the molar mass of an unknown if a solution containing 1.64g of unknown in 18.22 g of Biphenyl froze at 39.9 c.
Delta t = M x Kf
M = (71 - 39.9) / 81.2 = 0.3830 moles of solute / kg of solvent
also have 1.64 g / 18.22 g of solvent = 1.64 x 1000 / 18.22 g / kg = 90.0 g
so in 1 kg we have .3830moles = 90.0 g
mass of 1 mole = 90.0 / .3830 = 235.0 g/ mole
References: General Chemistry Laboratory Manual for Alabama A&M