Answer:
[tex]Tb=63.8\°C[/tex]
Explanation:
Hello,
In this case, for the boiling point elevation, we have:
[tex]\Delta T=i*m*Kb[/tex]
Whereas the van't Hoff factor for the given substance is 1 since is not ionizing. Moreover, the molality is computed by:
[tex]m=\frac{mol_{solute}}{m_{solvent}}=\frac{175.3mg*\frac{1g}{1000mg}*\frac{1nol}{164g} }{1.50g*\frac{1kg}{1000g} } =0.713m[/tex]
In such a way, since the boiling point of pure chloroform is 61.2 °C, the boiling point of the solution is:
[tex]Tb=61.2\°C+1*0.713m*3.63\frac{\°C}{m} \\\\Tb=63.8\°C[/tex]
Regards.
