Chapter 13 - The Chemistry of Solutes and Solutions - Questions for Review and Thought - Topical Questions - Page 605d: 80

$1.2\times10^{4}\, g/mol$

$Osmotic\,pressure\,\Pi=7.6\,mmHg= 0.01013bar$ (As 760 mmHg= 1.013 bar) Grams of solute $w_{2}=5.0\,g$ Volume of the solution V= 1.0 L Temperature T= 298.15 K Gas constant $R= 0.083 L\,bar\,mol^{-1}K^{-1}$ Molarity mass of the polymer= $\frac{w_{2}RT}{\Pi V}=\frac{5.0g\times0.083 L\,bar\,mol^{-1}K^{-1}\times298.15K}{0.01013\,bar\times1.0L}=$ $1.2\times10^{4}\,g/mol$