A protein solution is prepared by dissolving 1.26 g of protein to make 200 mL of solution. At 27°C, the osmotic pressure of the solution is \(2.57\times10^{-3}\ \text{bar}\).
Interpretation
This question uses osmotic pressure to determine the molar mass of a protein. The underlying idea is quite elegant: although protein molecules are too large to count directly, their presence in solution creates an osmotic pressure that depends on the number of molecules present. By measuring that pressure, we can work backwards and determine the molar mass.
For dilute solutions, osmotic pressure follows a relation analogous to the ideal gas equation:
\(\pi = CRT\)where \(\pi\) is the osmotic pressure, \(C\) is the molar concentration, \(R\) is the gas constant, and \(T\) is the absolute temperature.
Given Data
Mass of protein:
\(W = 1.26\ \text{g}\)Volume of solution:
\(V = 200\ \text{mL}\)Temperature:
\(T = 27^\circ\text{C} = 300\ \text{K}\)Osmotic pressure:
\(\pi = 2.57\times10^{-3}\ \text{bar}\)Gas constant:
\(R = 0.083\ \text{L·bar·mol}^{-1}\text{K}^{-1}\)Concept Application
The osmotic pressure equation contains concentration, but the quantity we need is the molar mass.
Since molar concentration is defined as
\(C=\frac{\text{moles of solute}}{\text{volume of solution}}\)and the number of moles is
\(n=\frac{W}{M}\)we can write
\(C=\frac{W}{MV}\)Substituting this into the osmotic pressure equation gives
\(\pi=\frac{WRT}{MV}\)This expression is especially useful because it directly connects the measurable quantities to the unknown molar mass.
Solution
Starting with
\(\pi = CRT\)we obtain
\(C=\frac{\pi}{RT}\)Substituting the given values:
\(C=
\frac{2.57\times10^{-3}}
{0.083\times300}
\) \(
C=
\frac{2.57\times10^{-3}}
{24.9}
\) \(
C=1.032\times10^{-4}\ \text{mol L}^{-1}
\)
Now, expressing concentration in terms of mass and molar mass,
\(C=\frac{W\times1000}{M\times V}
\)
Substituting the known values:
\[
\frac{1.26\times1000}{M\times200}
= 1.0321285\times10^{-4}
\]
\[
\frac{1260}{200M} = 1.0321285\times10^{-4}
\]
\frac{6.3}{M} = 1.0321285\times10^{-4}
\)
Solving for \(M\):
\(M=
\frac{6.3}{1.0321285\times10^{-4}}
\) \(
M=61038.9\ \text{g mol}^{-1}
\)
Therefore,
\(M = 61039\ \text{g mol}^{-1}
\)
Answer
\(Molar Mass = 61039\ \text{g mol}^{-1}
\)
Insight
A powerful result worth remembering is
\(\pi = \frac{WRT}{MV}
\)
This equation is simply the osmotic pressure equation rewritten in terms of the mass and molar mass of the solute. Whenever a problem provides mass of solute, volume of solution, temperature, and osmotic pressure, this relation should immediately come to mind.
Notice the physical meaning of the answer as well. The calculated molar mass is about \(6.1\times10^4\ \text{g mol}^{-1}\), which is very large compared with ordinary molecules. That is exactly what we expect for a protein, reinforcing that the calculation is chemically reasonable.
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