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Q.200 cm³ of an aqueous solution of a protein contains 1.26 gm of the Protein. The osmotic pressure of such a solution at 300 K is found to be 2.57×10⁻³ atm. Calculate the molar mass of the protein.

Uttarakhand UbseUttarakhand Board Intermediate (Class 12) 2019Subjective· 3mImportance★★★★★
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M=wRTπV=1.26×0.0821×3002.57×10−3×0.2≈6.04×104 g mol−1M = \dfrac{wRT}{\pi V} = \dfrac{1.26\times0.0821\times300}{2.57\times10^{-3}\times0.2} \approx 6.04\times10^4\ g\,mol^{-1}. (OR — Henry's law, mole fraction, molarity below.)

Concept. Osmotic pressure (π\pi) is a colligative property given by the van't Hoff equation:

πV=nRT=wMRT  ⇒  M=wRTπV\pi V = nRT = \frac{w}{M}RT \;\Rightarrow\; M = \frac{wRT}{\pi V}

Given data. w=1.26 gw = 1.26\,g, V=200 cm3=0.2 LV = 200\,cm^3 = 0.2\,L, T=300 KT = 300\,K, π=2.57×10−3 atm\pi = 2.57\times10^{-3}\,atm, R=0.0821 L atm K−1mol−1R = 0.0821\,L\,atm\,K^{-1}mol^{-1}.

Calculation.

M=1.26×0.0821×3002.57×10−3×0.2=31.035.14×10−4≈6.04×104 g mol−1M = \frac{1.26 \times 0.0821 \times 300}{2.57\times10^{-3} \times 0.2} = \frac{31.03}{5.14\times10^{-4}} \approx 6.04\times10^4\ g\,mol^{-1}

OR — Definitions.

  • Henry's law: at a constant temperature, the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas over the solution, p=KH xp = K_H\,x (where xx is the mole fraction of the gas and KHK_H the Henry constant). …

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