Question
Asked Nov 21, 2019
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When 2.63 g of a polypeptide is dissolved in 101 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.)

 

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Expert Answer

Step 1

To calculate the molecular weight first it required to calculate the number of moles and to calculate number of moles, it is required to calculate the concentration of the substance which can be calculated as,

Osmotic pressure (II) CRT
Where, C is concentration, R is universal gas constant and T is temperature in
Kelvin
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Osmotic pressure (II) CRT Where, C is concentration, R is universal gas constant and T is temperature in Kelvin

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Step 2

Substituting the values to calculate the concentration.

II 0.125 atm, R 0.0821 L-atm mol1 K- T
37 + 273 = 310 K
II CRT
0.125 C x 0.0821 x 310
C 0.0049 mol/L
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II 0.125 atm, R 0.0821 L-atm mol1 K- T 37 + 273 = 310 K II CRT 0.125 C x 0.0821 x 310 C 0.0049 mol/L

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Step 3

Now since the volume is given as 101 mL (0.101 L...

Number of moles = Concentration x Volume
Number of moles = 0.0049 x 0.101
Number of moles = 4.96 x 104 mols
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Number of moles = Concentration x Volume Number of moles = 0.0049 x 0.101 Number of moles = 4.96 x 104 mols

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