Question

We typically measure osmotic pressure indirectly by determining the pressure required to stop osmosis. Osmotic pressure can be calculated using the following equation: π = iRT(ΔC), where

π = osmotic pressure (mm H2O)

i = number of osmotically active particles resulting from dissociation of each molecule in solution. =1

1 R = gas constant (848 L-mm H2O/mol-degree) 0.0821

T = temperature (K) 296

ΔC = concentration difference of solute on the two sides of the membrane (Cinside – Coutside) 1-0=1

Under ideal conditions (that is, conditions where the hydrostatic pressure is negligible, there are no wall effects, the solution is not diluted by water influx, etc.), how high would the fluid rise in the tube? (0.25 pt) (solve equation to find value)

How does the calculated value above compare to the height your solution actually rose during the time allotted in the lab? Explain the discrepancy. (0.25 pt) ?

f. How high would the fluid rise in the tube if the solution in the bag were changed to 0.5 M NaCl? How does this compare to the calculated height of the solution found in question 2d? Explain. (0.75 pt) (solve equation to find value)?

Answer 1

We know that

π= i c R T

osmotic pressure =1 x 1 X 0.0821 X 296 = 24.302 atm = 2462400.15 Pascals

Now this pressure = height X density X acceleration due to gravity

2462400.15 = height X 1000kg/m^2 * 9.8m/s^2

Height = 251.26 meters

b) we want experimental data

Pressure = i C R T = 2 X 0.5 X 0.0821 X 296 = 24.302 atm = 2462400.15 Pascals

Same values

Now this pressure = height X density X acceleration due to gravity

2462400.15 = height X 1000kg/m^2 * 9.8m/s^2

Height = 251.26 meters