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200mL of an aqueus solution of a protien 1.26 g of the protein. The osmotic pressure of such a solution is found to be 2.7xx10^(-3) bar at 300 K. Calculate the molar mass of protein `(R=0.083 L bnar mol^(-1)K^(-1))` |
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Answer» Osmtic pressure `(pi)=(W_(B)xxRxxT)/(M_(B)xxV) or M_(B)=(W_(B)xxRxxT)/(pixxV)` `pi=2.7xx10^(-3) bar, V=200mL=200/1000=0.2L.` `W_(B)=1.26g, R=0.083 L bar mol^(-1)K^(-1), T=300K` `M_(B)=((1.26g)xx(0.083 L bar mol^(-1)K^(-1))xx(300K))/((2.7xx10^(-3)bar)xx(0.2L))=58,100g mol^(-1)` |
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