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9) Insulin is dissolved in a suitable solvent and the osmotic pressure $\left(\mathrm{\pi}\right)$ of solutions of various concentrations (g/cm^{3}) C is measured at 20^{0}C. The slope of a plot of $\mathrm{\pi}$ against C is found to be 4.65 x 10^{-3}. The molecular weight of the insulin is

(A) 3 x 10^{5} (B) 9 x 10^{5} (C) 4.5 x 10^{5} (D) 5.16 x 10^{5}

We need to use the concepts of colligative property to get the answer. As per given in the question, insulin is dissolved in a particular solvent. Since the graph was plotted and the slope was found to be 4.65x10

^{-3}

Using the equation for the osmotic pressure, one can write,

$\pi =CRTwhereCistheconcentrationofisulininthesolution,andTisthetemperatureinKandRistheuniversalgascons\mathrm{tan}t\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}equatingittoy=mx+c,heretheinterceptis0soweget\phantom{\rule{0ex}{0ex}}y=mx\phantom{\rule{0ex}{0ex}}nowonecanwrite,\phantom{\rule{0ex}{0ex}}\pi =CRT\phantom{\rule{0ex}{0ex}}C=\frac{numberofmoles}{volumeofsolution}\phantom{\rule{0ex}{0ex}}C=\frac{givenmass}{molarmass\times volumeofsolution}=\frac{concentration}{molarmass}\phantom{\rule{0ex}{0ex}}thus\pi =\frac{RT\times C}{molarm}\phantom{\rule{0ex}{0ex}}thusmolarmass=\frac{RT}{4.65\times {10}^{-3}}=\frac{8.314J/K/mol\times 293K}{4.65\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}THUSmolarmassis5.16\times {10}^{5}\phantom{\rule{0ex}{0ex}}$

Regards

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