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  • Answer please! 200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57x10-3 bar. The molar mass of protein will be (R=0.083 L bar mol-1 K-1)

200 mL of an aqueous solution of a protein contains its 1.26 g. The osmotic pressure of this solution at 300 K is found to be 2.57x10-3 bar. The molar mass of protein will be (R=0.083 L bar mol-1 K-1)

  • Option 1)

    51022 g mol-1

  • Option 2)

    122044 g mol-1

  • Option 3)

    31011 g mol-1

  • Option 4)

    61038 g mol-1

 

Answers (2)

As we learned in concept

Mathematical Expression for Osmotic Pressure -

\pi =CRT
 

- wherein

\pi = Representation of osmotic pressure.

C =constant\: in \: moles

R =Gas \: constant= 0.0821\frac{atm.lit}{mol.k}

T=Temp. \: in\: kelvin

 

 Osmotic pressure, \pi\:=\:CRT

2.57\times 10^{-3}=C\times 0.083\times 300

C = 10^{-4}\:mol\:L^{-1}\times 1.0321

\frac{1.26}{M\times 0.2}\:=\:10^{-4}

M = 61038\:g\:mol^{-1}

 

 


Option 1)

51022 g mol-1

This option is incorrect.

Option 2)

122044 g mol-1

This option is incorrect.

Option 3)

31011 g mol-1

This option is incorrect.

Option 4)

61038 g mol-1

This option is correct.

Posted by

Vakul

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W2=1.26g T=300k π=2.57x10*-3  R=0.083

V=0.2

We know that M2=W2 R T/ πV

1.26x0.083x300/2.57x10*-3x 0.2

31.374/0.514x10*-3=61038g mol*-1

Posted by

Yeshas

View full answer

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