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.

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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)(0.500molL)(62.4 L torrK mol)(300 K)=9.36×103 torrπ = 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}  torr

Thus, this solution would develop a pressure sufficient to support a column of mer-cury 9.36×1039.36 × 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)(0.500molL)(62.4 L torrK mol)(300 K)=1.87×104 torrπ = 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

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