Question

A solution of crab hemocyanin, a pigmented protein extracted from crabs, was prepared by dissolving $0.750g$ in $125c{m^3}$ of an aqueous medium. At $4^\circ C$, an osmotic pressure rise of $2.6mm$ of the solution was observed. The solution had a density of $1.00g/c{m^3}$. Determine the molecular weight of the protein.
A. $2.7 \times {10^5}g/mol$
B. $5.4 \times {10^5}g/mol$
C. $6.3 \times {10^5}g/mol$
D. $12.6 \times {10^5}g/mol$

Answer 1

We are given the osmotic pressure rise in the solution. We must first convert this into the osmotic pressure in terms of $atm$. As we are given rest of the values directly and we need to just substitute these into the equation for molecular mass when osmotic pressure is known. The rise in osmotic pressure is inversely proportional to the molecular mass of the compound.
Formulas used: ${h_{Hg}}{\rho _{Hg}}g = h\rho g$
Where $h$ is the increase in height, $\rho$ is the density, $g$ is the acceleration due to gravity and the subscript $Hg$ indicates the above parameters in terms of mercury.
$M = \dfrac{{w \times R \times T}}{{\pi \times V}}$
Where $M$ is the molecular mass of the protein, $w$ is the given mass, $R$ is the universal gas constant, $T$ is the absolute temperature, $\pi$ is the osmotic pressure and $V$ is the volume of the solution.

Complete step by step solution:
As we are given the increase in height due to the osmotic pressure as 2.6mm, we must first convert this into units of pressure. We can do this by expressing the height increase in terms of $mm$ of $Hg$. To do so, we equate the equation of pressure to mercury:
${h_{Hg}}{\rho _{Hg}}g = h\rho g$
Where $h$ is the increase in height, $\rho$ is the density, $g$ is the acceleration due to gravity and the subscript $Hg$ indicates the above parameters in terms of mercury.
Density of mercury is $13.6g/c{m^3}$ and density of the solution is given as $1.00g/c{m^3}$. Cancelling the acceleration due to gravity term present on both sides and substituting the values, we get:
${h_{Hg}} \times 13.6 = 2.6mm \times 1g/c{m^3}$
$\Rightarrow {h_{Hg}} = \dfrac{{2.6}}{{13.6}}mmHg$
We know that $760mm$ of $Hg$ is equivalent to $1atm$ pressure. Thus, the given osmotic pressure in terms of atmospheric pressure can be written as:

Pressure in $mm$ of $Hg$	Pressure in $atm$
$760$	$1$
$\dfrac{{2.6}}{{13.6}}$	$\pi$

Using the cross-multiplication rule, we have:
$\pi \times 760 = \dfrac{{2.6}}{{13.6}} \times 1$
$\Rightarrow \pi = \dfrac{{2.6}}{{13.6 \times 760}}atm$
The equation for molar mass is given by:
$M = \dfrac{{w \times R \times T}}{{\pi \times V}}$
Where $M$ is the molecular mass of the protein, $w$ is the given mass, $R$ is the universal gas constant, $T$ is the absolute temperature, $\pi$ is the osmotic pressure and $V$ is the volume of the solution.
The given mass $w = 0.75g$, $R = 0.0821L.atm/mol.K$
The temperature in Kelvin scale is: $4 + 273 = 277K$ and the volume, converted to litres is:
$125c{m^3} = 0.125L$ (since $1c{m^3} = 1mL = 0.001L$)
Substituting these values, we get:
$M = \dfrac{{0.75 \times 0.0821 \times 277}}{{\dfrac{{2.6}}{{13.6 \times 760}} \times 0.125}}$
$\Rightarrow M = \dfrac{{0.75 \times 0.0821 \times 277 \times 13.6 \times 760}}{{2.6 \times 0.125}}$’
On solving this, we get:
$M = 5.4 \times {10^5}g/mol$
Hence, the correct option to be marked is B.

Note:
Osmotic pressure is the pressure that must be applied to a solution to halt the flow of solvent molecules through a semipermeable membrane (osmosis). Note that when a pressure greater than the osmotic pressure is applied, the solvent molecules move from the region of lower concentration to the region of higher concentration, and is known as reverse osmosis.