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  • 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 2.57 x 10-3 bar. The molar mass of protein will be (R = 0.08

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 2.57 x 10-3 bar. The molar mass of protein will be (R = 0.083 L bar mol-1 K-1)

Option: 1

31011 g mol-1


Option: 2

61038 g mol-1


Option: 3

51022 g mol-1


Option: 4

122044 g mol-1


Answers (1)

best_answer

We know that 

Osmotic pressure,  \mathrm{\pi\:=\:CRT}

Where,

\pi = Representation of osmotic pressure.

\textup{C = concentration / molarity}

\mathrm{R =Gas\ constant= 0.0821\ \frac{atm-L}{mol-k}}

\textup{T=Temperature in Kelvin}

Now,

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

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

Now, Molarity = moles/volume(L)

Hence,

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

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

Option 2 is correct.

Posted by

Kuldeep Maurya

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