Peter Jeschofnig, Ph.D. Version 42-0149-00-01
Lab Report Assistant
This document is not meant to be a substitute for a formal laboratory report. The Lab Report Assistant is simply a summary of the experiment’s questions, diagrams if needed, and data tables that should be addressed in a formal lab report. The intent is to facilitate student’s writing of lab reports by providing this information in an editable file which can be sent to an instructor
Observations & Questions for Part 1
Record your observations and your time and …show more content…
C. Historically certain colligative properties – freezing point depression, boiling point elevation, and osmotic pressure – have been used to determine molecular mass. (Now there are instrumental methods to determine this.) Of these three, osmotic pressure is the most sensitive and gives the best results. Molecular mass can be found according the following equation:
Π = MRT
Where:Π = osmotic pressure,
M = molarity of solution,
R = the ideal gas constant (0.0821 L×atm/mol×K), and
T = Kelvin temperature.
0.125 grams of a starch is dissolved in 100 mL of water at 25oC and has an osmotic pressure of
5.15 mmHg. What is its molecular mass?
Since the gas constant, R, requires atmospheres as pressure units, we have to convert 5.15 mmHg to atmospheres:
5.15 mm Hg * 1 atm/760 mm Hg = 0.00678 atm.
0.00678 atm = M (0.0821) (273 + 25 = 298 K); We solve for M = 0.00678/(0.0821*298);
M (molarity) = 2.77 x 10 –4
Molarity = moles/L: 2.77 x 10–4 = moles/.1 = 2.77 x 10-5 moles of starch; If 2.77 x 10-5 moles = 0.125 grams,
Then 1 mole = 0.125 g/2.77 x 10-5 = 4512 g/mole = molar mass of starch.
Problem for Lab