When 2.63 g of a polypeptide is dissolved in 445 mL of water, theresulting solution is found to have an osmotic pressure of 0.125 atm at37.0 °C. What is the molar mass of the polypeptide? (Assume thevolume doesn't change when the polypeptide is added.)

Answered: When 2.63 g of a polypeptide is…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Question

When 2.63 g of a polypeptide is dissolved in 445 mL of water, the
resulting solution is found to have an osmotic pressure of 0.125 atm at
37.0 °C. What is the molar mass of the polypeptide? (Assume the
volume doesn't change when the polypeptide is added.)

Expert Solution

Step 1

Equation for Osmotic pressure:

$π$ = i * M * R * T

where,

$π$ = Osmotic pressure = 0.125 atm

i = van't Hoff factor = 1 for a polypeptide since it does not ionize or dissociate

M = Molarity

R = Universal gas constant = 0.0821 Latm/Kmol

T = Temperature = 37.0 ⁰C = 310.15 K (Temperature in K = Temperature in ⁰C + 273.15)

Putting values we get,

$0.125 a t m = 1 \times M \times 0.0821 L a t m / K m o l \times 310.15 K M = \frac{0.125 a t m}{0.0821 L a t m / K m o l \times 310.15 K} M = 0.00491 m o l / L$

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