Question

Osmotic pressure measurements can be used to determine the molecular weights of large molecules such as proteins. For a solution of large molecules to qualify as "dilute," its molar concentration must be very low and hence the osmotic pressure can be too small to measure accurately. For this reason, the usual procedure is to measure the osmotic pressure at a variety of concentrations, then extrapolate the results to the limit of zero concentration. Here are some data for the protein hemoglobin dissolved in water at $3^{o}C$:

Concentration (grams/liter)	$∆h$ (cm)
5.6	2.0
16.6	6.5
32.5	12.8
43.4	17.6
54.0	22.6

The quantity $∆h$is the equilibrium difference in fluid level between the solution and the pure solvent,. From these measurements, determine the approximate molecular weight of hemoglobin (in grams per mole).

An experimental arrangement for measuring osmotic pressure. Solvent flows across the membrane from left to right until the difference in fluid level,$∆h$, is just enough to supply the osmotic pressure.

Short answer

The approximate molecular weight is $66.3kg/mol$.

Step-by-step solution

Step 1. Given Information

We are given a table,

Step 2. The osmotic pressure 

For dilute solutions, the osmotic pressure can be approximated as,

$\mathrm{\pi }=\frac{\mathrm{nRT}}{\mathrm{V}}$

$$\begin{gathered} \mathrm{\pi }=\frac{\mathrm{nRT}}{\mathrm{V}} \\ =\frac{\mathrm{cRT}}{\mathrm{M}} \end{gathered}$$

Therefore,

$M=\frac{cRT}{\mathrm{\pi }}$

Here the number density of the solute $\left(\frac{n}{V}\right)$in moles/liter is replaced by $\left(\frac{c}{M}\right)$, $c$is the concentration of the solute in (grams/liter) and $M$is its molecular weight (in grams/mole). But in this experiment the osmotic pressure is balanced by the difference in the fluid level $∆h$so for each solute concentration, one can calculate the osmotic pressure by the following formula,

$\mathrm{\pi }=\mathrm{\rho g}∆\mathrm{h}$

Substitute $\rho g∆h$for $\mathrm{\pi }$in the equation $M=\frac{cRT}{\mathrm{\pi }}$

$$\begin{gathered} \mathrm{\pi }=\frac{\mathrm{nRT}}{\mathrm{V}} \\ =\frac{\mathrm{cRT}}{\mathrm{M}} \end{gathered}$$

Use the density of water $\left(\rho =1g/c{m}^3\right)$. The table given below shows the value of the osmotic pressure for each solute condition and the corresponding estimated value of the molecular weight from the following relation,

Step 3. Consider the following table,

The table is as follows,

Concentration $grams/liter=kg/m^3$	$∆hcm$	$\mathrm{\pi }\mathrm{N}/{\mathrm{m}}^2$	$Mkg/mole$
5.6	2.0	196.2	65.5
16.6	0.5	637.6	59.7
32.5	12.8	1255.7	59.4
43.4	17.6	1726.6	57.7
54.0	22.6	2217.1	55.9

Step 4. Graph of concentration versus molecular mass

The graph of concentration versus molecular mass is as follows,

Step 5. Approximated Molecular Weight

The molecular weight estimated from each measurement is plotted versus the concentration. As the equation that relates osmotic pressure to the concentration of solutes is most accurate in the limit of very dilute solution, one should extrapolate the results to zero concentration. If apparently one bad point at $c=16.6g/L$is ignored, the molecular weight is $M=66.3kg/mol$. Therefore, the approximate molecular weight is$66.3kg/mol$