Osmotic pressure can be very useful for estimating the molecular weights of macromolecules for whom a molecular formula may not be available as it can provide greater precision in the measurement. (a) 1.364 g of a protein with molecular mass 1.78 × 104 g/mol is dissolved in enough water to create 100.0 mL of solution at 298 K. What would be the expected osmotic pressure (in bars) of this solution? (b) If 1.853 g of a protein are dissolved in enough water to make up 100.0 mL of solution at 298 K, then the osmotic pressure is observed to be 2.521 torr (3.361 × 10-³ bar). What is the molar mass (g/mol) of the protein?

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Osmotic pressure can be very useful for estimating the molecular weights of
macromolecules for whom a molecular formula may not be available as it can provide
greater precision in the measurement.
(a) 1.364 g of a protein with molecular mass 1.78 × 104 g/mol is dissolved in enough water
to create 100.0 mL of solution at 298 K. What would be the expected osmotic pressure
(in bars) of this solution?
(b) If 1.853 g of a protein are dissolved in enough water to make up 100.0 mL of solution at
298 K, then the osmotic pressure is observed to be 2.521 torr (3.361 × 10-³ bar). What is
the molar mass (g/mol) of the protein?
Transcribed Image Text:Osmotic pressure can be very useful for estimating the molecular weights of macromolecules for whom a molecular formula may not be available as it can provide greater precision in the measurement. (a) 1.364 g of a protein with molecular mass 1.78 × 104 g/mol is dissolved in enough water to create 100.0 mL of solution at 298 K. What would be the expected osmotic pressure (in bars) of this solution? (b) If 1.853 g of a protein are dissolved in enough water to make up 100.0 mL of solution at 298 K, then the osmotic pressure is observed to be 2.521 torr (3.361 × 10-³ bar). What is the molar mass (g/mol) of the protein?
Expert Solution
Step 1

For part (a):

The mass of protein = 1.364 g

The molar mass of the protein = 1.78 × 104 g/mol

The volume of the solution = 100.0 mL

Temperature = 298 K

We have to find the osmotic pressure.

For part (b):

The mass of the protein = 1.853 g

The volume of the solution = 100.0 mL

The osmotic pressure = 3.361 × 10-3 bar

Temperature = 298 K

We have to find the molar mass of the protein.

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