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Bookmark A solution containing 27.55 mg of an unknown protein per 25.0 mL of solution was found...

Bookmark A solution containing 27.55 mg of an unknown protein per 25.0 mL of solution was found to have an osmotic pressure of 3.22 torr at 25.0C. What is the molarmass of the protein?
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Answer #1
number of moles=(osmotic pressure*volume)/(R*T)
n=(3.22mmHg*.025L)/(62.36367*298K)
n=4.33*10^-6 mol
.02755g/4.33*10^-6 mol=6,360.25 g/mol
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answered by: Ling hoe
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