Question

200 cm³ of an aqueous solution of protein contains 1.26 g of the protein. The osmotic pressure of this solution at 300K is found to be 2.57 × 10-³ bar. Calculate the molar mass of the protein.​

Answer 1

Answer :

• Molar mass of protein = $\bold{61039\:g\:mol^{-1}}$

Step-by-step explanation :

Given that,

• Volume of the solution (V) = 200 cm³ = 0.200 L.

• Mass of the solute (protein), $\bold{w_2}$ = 1.26 g.

• Osmotic pressure, $\bold{\pi=2.57\times{}10^{-3}\:bar}$

• Temperature, T = 300 K.

• Universal gas constant, R = $\bold{0.083\:L\:atm\:K^{-1}\:mol^{-1}}$

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We know, $\bold{M_2=\dfrac{w_2RT}{\pi\:V}}$

Substituting the given values in the formula, we get,

$\implies\bold{M_2=\dfrac{1.26\:g\times{}0.083\:L\:atm\:K^{-1}\:mol^{-1}\times{}300\:K}{2.57\times{}10^{-3}\:bar\times{}0.200\:L}}$

$\implies\bold{61039\:g\:mol^{-1}}$