## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 13 Solutions and Their Behavior - Study Questions - Page 505d: 66

#### Answer

$C_{10}H_{12}O_2$

#### Work Step by Step

B.p. elevation: $61.82-61.70=0.12^{\circ}C$ Molality: $\Delta T_b=K_b,m\rightarrow m=0.12^{\circ}C\div3.63^{\circ}C/m=0.033\ mol/kg$ Number of mols: $0.033\ mol/kg\times(25/1000)kg=8.26\times10^{-4}\ mol$ Molar mass: $0.135\ g\div 8.26\times10^{-4}\ mol=163.35\ g/mol$ Molar mass of empirical formula: $82.10\ g/mol$ Ratio of the molar masses: $163.35\div 82.10=1.99\approx 2$ Molecular formula: $C_{10}H_{12}O_2$

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.