Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning
ISBN 10: 1285199049
ISBN 13: 978-1-28519-904-7

Chapter 13 - The Chemistry of Solutes and Solutions - Problem Solving Practice 13.13 - Page 595: a


$ 6.8\times10^{-4}\,g/mol $

Work Step by Step

Gas constant $R=0.0821\,L\,atm\,mol^{-1}K^{-1}$ Osmotic pressure $\Pi= 1.8\times10^{-3}\,atm$ Absolute temperature$T=(25+273)K=298\,K$ van't Hoff factor $i=1$ The relationship is $\Pi=cRTi$ where $c$ is the molarity of the solution. $\implies c=\frac{\Pi}{RTi}=\frac{1.8\times10^{-3}\,atm}{0.0821\,L\,atm\,K^{-1}mol^{-1}(298K)(1)}$ $=7.357\times10^{-5}\,mol/L$ But $c=\frac{moles\,of\,solute}{volume\,of\, solution\,in \,L}=\frac{moles\,of\,solute}{1.0\,L}$ (Volume of 5.0g solute is neglected as it is negligible compared to 1.0 L) Moles of solute= $7.357\times10^{-5}\,mol/L\times1.0\,L=7.357\times10^{-5}\,mol$ Molar mass of solute= molar mass of haemoglobin= $\frac{mass\,of \, haemoglobin}{moles\,of\, haemoglobin}=\frac{5.0\,g}{7.357\times10^{-5}\,mol}$ $= 6.8\times10^{-4}\,g/mol $
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