Asked Jan 29, 2019

When 0.713 grams of a protein were dissolved in 72.5 mL of solution at 28.5 degrees C, the osmotic pressure was found to be 35.6 torr. Calculate the molar mass of the protein.


Expert Answer

Step 1

Osmotic pressure is an colligative property meaning it depends upon the amount of solute or ions present in solution.

Step 2

Given here,

0.713 grams of a protein

dissolved in 72.5 mL (0.0725 L) of solution

[1 ml = 0.001 L]

at Temperature (T) = 28.5 degrees C + 273 = 301.5 K

the osmotic pressure found = 35.6 torr

convert torr to atm

conversion factor, 1 atm = 760 torr

So, osmotic pressure = 35.6 torr x (1 atm/760 torr) = 0.047 atm

R = 0.0821 L.atm/K.mol

M = molarity of solution


osmotic pressure = MRT


0.047 = M x 0.0821 x 301.5

Molarity of solution = 0.002 M

With, molarity = moles/Liter of solution


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