InterviewSolution
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InterviewSolution
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200 mL of an aqueous solution of a protein contains its 1.26g. The osmotic pressure of this solution at 300K is found to be `2.57 xx 10^(-3)` bar. The molar mass of protein will be `(R = 0.083 L bar mol^(-1)K^(-1))`A. `51022 g mol^(-1)`B. `122044 g mol^(-1)`C. `31011 g mol^(-1)`D. `61038 g mol^(-1)` |
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Answer» Correct Answer - D `pi = cRT = [(W_(2))/(M_(2)) xx (1)/(V)] RT` or `M_(2) = (W_(2)RT)/(piV)` `= (1.26 g xx 0.083 L "bar" mol^(-1)K^(-1) xx 300 K)/(2.57 xx 10^(3)"bar" xx 0.200L)` `= 61038 g mol^(-1)`. |
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