Question

0.0500 g of insulin was dissolved in 50mL of water. The osmotic pressure of the resulting solution is 45.108mm water at 25 degrees C.
Density of water 1.00 g/mL
Desnity of Hg 13.9 g/mL
What is the molar mass of insulin?

Answer 1

I'll try to walk you through this step by step since we're going to need to do a few conversions. For an ideal solution, the osmotic pressure can be calculated using the following equation:

\[\Pi=iMRT\]
Where Π is the osmotic pressure, i is the van't Hoff factor (the number of particles each molecules of solute produces when dissolved), M is the molarity, R is the gas constant, and T is the absolute temperature. Eventually we're going to want the molar mass of insulin, and we can use molarity to find it. Remember, M = n/V:

\[\Pi=i\frac{n}{V}RT\]
Where n is the moles of solute and V is the volume of solution. Now, n = m/MM (where m is the actual mass of the compound and MM is its molar mass). That means we can update our equation once more:

\[\Pi=i\frac{m}{MM \times V}RT\]
Now we can solve this question in just one step! I'll just rearrange the equation to isolate MM:

\[MM=i\frac{m}{\Pi \times V}RT\]
Alright...now let's look at each variable:

van't Hoff factor (i): Insulin does not split into more than one particle after dissolving, and since one molecules of insulin produces 1 molecule of dissolved solute, i = 1.

mass (m): It's given in the question: 0.05 g

gas constant (R): 0.08206 L atm/K/mol

absolute temperature (T): This is temperature in Kelvin. Just add 273 to the 25C given to get 298.

volume (V): Given in the question: 50 mL or 0.05 L

osmotic pressure (Π): This one's tricky. We need to express it in atm to match the units of R, so we need to convert what's given in the question. It's easy to convert mmHg to atm (just divide by 760), but we're given pressure in mmH2O. Luckily, we can use the densities of H2O and Hg to convert mmH2O to mmHg, then mmHg to atm. Hg is 13.9 times more dense than H2O. Therefore, it will occupy 13.9 times LESS volume at the same pressure. This means we have 45.108/13.9 = 3.245 mmHg, which means we have 3.245/760 = 0.00427 atm of pressure. Whew!

Now we can plug all that nonsense in to our equation to find the molar mass of insulin:

\[MM=1 \times \frac{0.05}{0.00427 \times 0.05}\times 0.08206 \times 298=5726.9 \space g/mol\]

So the molar mass of insulin is 5726.9 g/mol. Wikipedia tells me the molar mass is 5808 g/mol, so we're not too far off! Blame the difference on rounding.

Anyways...I know that may have been a lot to take in. If you have any questions about any of this please let me know!

Answer 2

Wow! Thanks, that was really helpful and easy to follow. The only thing I don't understand is the osmotic pressure part. I know we need it in atm, but I don't understand why it makes sense to divide 45.108 by 13.9.

Answer 3

I also have an additional question regarding another problem, if you don't mind. So, for one question, I'm asked to find the concetration of NaCl to raise the boiling point temperature of water by 10 degrees C. I did this. But following that question is another asking me to find the concentration of NaCl to LOWER the freezing point by 10 degrees C, would delta T be -10 degrees C then?

Answer 4

Regarding your question about pressure, we're actually dividing by the specific gravity of Hg. Specific gravity is a ratio of densities: the density of a substance of interest (in this case Hg) to the density of a reference substance (in this case H2O). Since the density of H2O is 1 g/mL, the specific gravity of whatever substance we look at will usually be the same as its actual density, which makes things easy.

Say now you have 100 g of H2O and 100 g of Hg. Because Hg is much denser than water (in fact, 13.9x denser), it will occupy 13.9 times less volume than the same mass of water. So, if you have a pressure-measuring device such as a manometer, and you replace the H2O in it with Hg, the height difference (measured in mm) between the arms of the manometer will be 13.9 times less than it was with H2O.

As for your second question, the K(f) and K(b) will have different values, even for the same solvent. So although your change in temperature might be 10 degrees for boiling point elevation, this may not be true for freezing point depression as well.