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A solution is prepared by dissolving `1.08 g` of human serum albumin, a protein obtained from blood plasma, in `50 cm^(3)` of aqueous solution. The solution has an osmotic pressure of `5.85 mm Hg` at `298 K`. a. What is the molar mass of albumin ? b. What is the height of water column placed in solution ? `d_((H_(2)O)) =1 g cm^(-3)` |
Answer» a. The molar massof albumin can be calculated using the following relation `Mw_(B)=(W_(B)xxRT)/(piV)` ….(i) Given,`W_(B)=1.08 g`,`R=0.0821 L atm K^(-1) mol^(-1)` `T=298 K , pi=(5.85)/(760) atm,V=(50)/(1000) = 0.005 L` Substituting these values in Eq.(i), we get `Mw_(B)=(1.08xx0.0821xx298)/((5.85//760)xx0.05)=68655 g mol^(-1)` b. `pi=hdg` `(5.85)/(760)xx101325=hxx1xx10^(-3)xx9.8` [1 atm = 101325 Pa] `:. h=7.958 xx10^(-2) m =7.958 cm` |
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