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 13.12 - Page 591: a



Work Step by Step

$T_{f}(solution)= 5.15^{\circ} C$ (given) $T_{f}(solvent)=5.5^{\circ} C$ (freezing point of benzene) $\Delta T_{f}=T_{f}(solution)-T_{f}(solvent)$ $=5.15^{\circ} C-5.5^{\circ} C= -0.35^{\circ} C$ $K_{f}$ for benzene=$-5.10^{\circ} C\,kg/mol$ But $\Delta T_{f}= K_{f}\times m_{solute}$ $\implies molality\,\,m_{solute}=\frac{\Delta T_{f}}{K_{f}}$ $=\frac{-0.35^{\circ} C}{-5.10^{\circ} C\,kg/mol}=0.0686\,mol/kg$ As molality=$\frac{moles\,of\,solute}{mass\,of\, solvent\,in\,kg}$, we have Moles of solute= $m_{solute}\times$mass of solvent in kg $=0.0686\,mol/kg\times0.050\,kg=0.00343\,mol$ Molar mass of solute=$\frac{mass\,of\, solute}{number\, of\, moles}=\frac{0.180\,g}{0.00343\,mol}=52.5\,g/mol$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.