Question

A solution of crab hemocyanin, a pigmented protein extracted from crabs, was prepared by dissolving 0.750 g in 125 cm${}^3$ of an aqueous medium. At $4^{\circ }C$ an osmotic pressure rise of 2.6 mm of the solution was observed. The solution had a density of 1.00 g/cm${}^3.$ Determine the molecular weight of the protein.

• A. $2.7\times {10}^{5}g/mol$
• B. $5.4\times {10}^{5}g/mol$
• C. $6.3\times {10}^{6}g/mol$
• D. $12.6\times {10}^{6}g/mol$

Answer 1

The correct option is B $5.4\times {10}^{5}g/mol$

In terms of osmotic pressure, the molecular weight of solute is defined as,

$M=\frac{w\times R\times T}{O.P\times V}$

$O.P=\frac{2.6}{13.6\times 760}atm$

Now,

$M=\frac{0.750\times 277\times 0.0821\times 760\times 13.6}{2.6\times 125\times {10}^{-3}}$

$M=5.4\times {10}^{5}g/mol$

Hence, option B is correct