Question 7.10: Calculate the osmolarity and the osmotic pressure that would...
Calculate the osmolarity and the osmotic pressure that would develop across a semipermeable membrane if the solutions of Example 7.9 were separated from pure water by the membrane. Assume a solution temperature of 27°C and use an R value of 62.4 L torr/K mol.
a. The molarity of the sugar solution was found to be 0.500 mol/L. Because sugar does not dissociate, n is equal to 1, and the osmolarity nM is also equal to 0.500 mol/L.
π = nMRT = (1) ( 0.500 \frac{\cancel{mol}}{\cancel{L}})(\frac{62.4 \cancel{L} torr}{\cancel{K} \cancel{mol}}) (300 \cancel{K} ) \\ = 9.36 × 10^{3} torrThus, this solution would develop a pressure sufficient to support a column of mer-cury 9.36 × 10^{3} mm high. This is equal to about 12.3 standard atmospheres of pres-sure (see Table 6.3 for pressure units).
| Table 6.3 Units of Pressure | ||
| Unit | Relationship to One Standard Atmosphere | Typical Application |
| Atmosphere | – | Gas laws |
| Torr | 760 torr = 1 atm | Gas laws |
| Millimeters of mercury | 760 mmHg = 1 atm | Gas laws |
| Pounds per square inch | 14.7 psi = 1 atm | Compressed gases |
| Bar | 1.01 bar 5 29.9 in. Hg = 1 atm | Meteorology |
| Kilopascal | 101 kPa = 1 atm | Gas laws |
b. The molarity of this solution was also found to be 0.500. But, because this solute dis-sociates into two ions, n = 2. This makes the osmolarity equal to (2)(0.500 mol/L) = 1.00 mol/L:
π = nMRT = (2) ( 0.500 \frac{\cancel{mol}}{\cancel{L}})(\frac{62.4 \cancel{L} torr}{\cancel{K} \cancel{mol}}) (300 \cancel{K} ) \\ = 1.87 × 10^{4} torr, or about 24.6 standard atmospheres