Question

For serum bovine albumin, the following data at 25 °C and 0.15 M NaCl were collected.

                                 at pH = 5.3                  

C (g/L)	8.8	17.5	27.5	56.3
p/c (torr L/g)	.28	.284	.308	.348

                                 at pH = 7.00

C (g/L)	16.9	29.4	50.6	56.9
p/c (torr L/g)	.312	.336	.384	.399

At pH = 5.3, the polymer is uncharged. At 7.00, it has a charge of 12.

Calculate M for each pH. Why are they the same or different?

Answer 1

Sol Grpen (18 L) & Ple (for Lgt) fat pH= 5.39 pH 7.00 for sexum berine albumm , aut 250¢ ¢ 0.159 wal l first we have to plet the graph between a (get) & Ple (for Lgt) at both pH=5. 39 pH = 7.00.1 a we know that the van't Hoff reduced osmatic pressure ' B'ha eam is P = m + 2 Ret 8 let B2 M2 RTC = = = TB)C+ mamcleuetat weight in the form of Here L = reduce e reduced osmo #2 pressule : The plot between / rs c is .. y=mate .“. Intercept = { - RI = = ma Slope = B = Le

0.45 0.42 • pH = 5.3 pH = 7 0.39 p/c (torr Lg) TTTTTTT 0.271 0 10 20 50 60 70 30 40 C (g L"")

from the abuee plot atprt. 53; the mtercept = 0.26377 (torr 297 Intercept = € → Me BT RT Plc mtercent M= 62.3635 *tor kt mott x 2984 R 0.26377 terry g1 : 18,584.323 mouth 0.26377 97 Barit 2 70457.54 gmail M 2 70.456 kgmot at pH= 5:3 ART RT from the plot, at pH = 7 intercept = = 629363 teor Lg : - Ma Mdevelar weight - - Plc intercept = 62.3635 %Tro KT molt x 2988 0.27363 tefrigt att = 18,584,323 mort 0.27363 gt 367,9177.7 g most. song 2 67.9 kg most molecular weight mat pH=9.3: 70.456 kgmoll P4-7 67.91x9 mo? pH-7

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