Question

Answer questions 53-54 using the data given for the two solutions below assuming the two solutions are separated by a dialysis membrane with a molecular weight cutoff of 50 (concentrations are in mM).

                                    Solution A                                 Solution B

                                    1.0 glucose                               1.5 glucose

                                    1.0 KCl                                      2.0 NaCl

                                    1.5 CaCl₂                                   3.0 Albumin

                                    4.0 triglyceride

53.       What is the magnitude of the osmotic pressure difference between solutions A and B (in mmHg)? (Constants: R=0.082 L atm/mol K; T=310 K)

            a) 0.5;     b) 9.7;     c) 58.0;     d) 96.5;     e) none of the above

when I tried to solve this equation, I got answer C, but the correct answer is B according to the answer key.

Answer 1

Through a dialysis membrane has a cutoff molecular weight of 50 which means only ions or compounds with molecular weight lower than the cutoff value can pass through the membrane.

Now, in aqueous solutions, glucose, albumin, triglycerides stay as neutral molecules, while NaCl, KCl, CaCl2 dissociate into ions. For the latter, if the molar mass of the ions is lesser than the membrane cutoff value as stated earlier, they can pass through to the other side.

Now, molar mass of the constituents is as follows:

Constituent	Molar mass (molecular or ionic mass)
Glucose	180
Potassium ion	39
Calcium ion	40
Sodium ion	23
Chloride ion	35.5
Triglyceride	>50 (since glycol molecular weight is itself 62 and then the fatty acid molecular weight will add on)
Albumin	66.5 kilo Dalton

So, only the ions K+, Cl-, Na+, and Ca+2 can pass through the membrane.

These ions will pass through the membrane so that their concentration on both the sides become equal. For, K+, Ca+2, and Na+, since they are only present in one side of the membrane, their movement is unidirectional; K+ and Ca+2 moves from solution A to solution B while Na+ moves from solution B to solution A till each of their concentrations is the same in both solutions. Thus, at equilibrium the concentration of these ions in each solution is half of their original concentration in the solution.

Constituent	Concentration at equilibrium (solution A) mM	Concentration at equilibrium (solution B) mM
Potassium ion (initial conc. 1 mM in solution A)	0.5	0.5
Calcium ion (initial conc. 1.5 mM in solution A)	0.75	0.75
Sodium ion (initial conc. 2 mM in solution B)	1	1

Cl- ion is present in both the solutions. 1 mole of NaCl or KCl provides 1 mol of Cl- while 1 mol of CaCl2 provides 2 moles of Cl-. Thus, the movement of Cl- occurs both ways till the concentration of Cl- in solution A and in solution B is the same and the total concentration of Cl- is the sum of Cl- concentration at the beginning in solution A and B combined (i.e.,6 mM total = 3 mM from CaCl2 (1.5 mM of CaCl2 will provide 3.0 mM of Cl-) + 1 mM from KCl+ 2 mM from NaCl). So, at equilibrium, the conc. of Cl- in each solution will be half of 6, i.e. 3 mM. Rest of the constituents cannot pass the membrane. So, at equilibrium the concentration of each constituent will be as shown below:

Constituent	Concentration at equilibrium (solution A) mM	Concentration at equilibrium (solution B) mM
Glucose	1	1.5
Potassium ion	0.5	0.5
Calcium ion	0.75	0.75
Sodium ion	1	1
Chloride ion	3	3
TRiglyceride	4	0
Albumin	0	3

Since the ions have equal concentration in both the solution, they will not exert any osmotic pressure. The constituents which have a concentration differential between the solution will lead to the exertion of osmotic pressure. Those constituent and their conc. differential between solution A and B is shown below.

Constituent	Concentration at equilibrium (solution A) mM	Concentration at equilibrium (solution B) mM	Concentration differential (conc. In A – conc. In B) mM
Glucose	1	1.5	-0.5
TRiglyceride	4	0	4
Albumin	0	3	-3

The equation for osmotic pressure deltaP (between A and B) is:

deltaP = (deltac_glucoseRT) + (deltac_triglycerideRT) +(deltac_albuminRT)

where, deltac_component = concentration differential of component between A and B

Substituting values from the table into the equation

deltaP = (-0.5 x10^-3)*RT + (4x10^-3)* RT + (-3x10^-3)*RT = (-0.5x10^-3)*RT [Here 10^-3 is mu;tiplied to convert mM units to Molar, i.e. mol/L so that the unit is compatible for calulation with R as 0.082 L.atm/mol.K]

Replacing value of R = 0.082 L.atm/K.mol and T = 310 K into the above equation, we get

deltaP = 0.01271 atm

Now, 1 atm = 760 mmHg

So, 0.01271 atm = 760 * 0.01271 mm Hg = 9.7 mm Hg

So, deltaP = 9.7 mmHg