Question

A solution of protein (extracted from crabs) was prepared by dissolving 0.75 g in $1.25c{m}^3$ of an aqueous solution. At $4^{o}C$ an osmotic pressure rise of 2.6 mm of the solution was observed. Then molecular weight of the protein is:

[Assume density of solution is $1.00g/c{m}^3$ ]

• A. $9.4\times {10}^5$
• B. $5.4\times {10}^5$
• C. $5.4\times {10}^{10}$
• D. $9.4\times {10}^{10}$

Answer 1

Answer: (B)

$5.4\times {10}^5$

$1atm=760mmHg$

$2\cdot 6mmHg=3\cdot 42\times {10}^{-3}atm$

$\pi =\underset{–}{M}RT$

$\pi =\frac{w}{M.wt}\times \frac{1000}{V\left(ml\right)}\times R\times T$

$M.wt=\frac{\cdot 75}{3.42\times {10}^{-3}}\times \frac{1000}{1.25}\times 0\cdot 0821\times 277$

$=5.4\times {10}^5$