# 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.

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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.

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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

Using,

osmotic pressure = MRT

So,

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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