Chapter 13 - Solutions and Their Physical Properties - Example 13-8 - Calculating Osmotic Pressure - Page 579: Practice Example B

$8.3\times10^{-3}\,g$

$\Pi= M\times RT\implies M=\frac{\Pi}{RT}$ $=\frac{0.015\,atm}{0.08206\,L\,atm K^{-1}mol^{-1}\times298\,K}$ $=6.134\times 10^{-4}\,mol/L$ Moles of urea= $6.134\times10^{-4}\,mol/L\times0.225\,L$ $=1.38\times10^{-4}\,mol$ Mass of urea= $1.38\times10^{-4}\,mol\times60.06\,g/mol$ $=8.3\times10^{-3}\,g$