Answer: The boiling point of a 3.70 m solution of phenol in benzene is
![89.5^0C](https://img.qammunity.org/2021/formulas/chemistry/high-school/gmsevndj626exe7qs33gsrwmgbi1p6o5b1.png)
Step-by-step explanation:
Elevation in boiling point:
![\Delta T_b=i* k_b* m](https://img.qammunity.org/2021/formulas/chemistry/high-school/d4pud7ut627113cp25asnozrbrh6lkz3i1.png)
where,
= change in boiling point
i= vant hoff factor = 1 (for benzene which is a non electrolyte )
= boiling point constant =
![2.53^0C/kgmol](https://img.qammunity.org/2021/formulas/chemistry/high-school/nab1a5equ5xbw865h866vf0vx76t2q3q4m.png)
m = molality = 3.70
![T_(solution)-T_(solvent)=i* k_b* m](https://img.qammunity.org/2021/formulas/chemistry/high-school/9plynv6yvis6p6t8dxdg5lfn6tlssea8m8.png)
![T_(solution)-80.1^0C=1* 2.53* 3.70](https://img.qammunity.org/2021/formulas/chemistry/high-school/mji7sqe6quscxwgmtvwzn6dmqpu6dgovvb.png)
![T_(solution)=89.5^0C](https://img.qammunity.org/2021/formulas/chemistry/high-school/h4fj6mq3p8veask7m3vawj14rqsni11emm.png)
Thus the boiling point of a 3.70 m solution of phenol in benzene is
![89.5^0C](https://img.qammunity.org/2021/formulas/chemistry/high-school/gmsevndj626exe7qs33gsrwmgbi1p6o5b1.png)