Answer:
[tex]m=0.42m[/tex]
[tex]Tb_{sol}=78.9^oC[/tex]
Explanation:
Hello,
In this case, we use the boiling point elevation colligative property equation for both ethanol and carbon tetrachloride as shown below:
[tex]Tb_{sol}-Tb_{et}=Kb_{et}m\\Tb_{sol}-Tb_{CTC}=Kb_{CTC}m[/tex]
Now, since both the boiling point and the concentration of the solutions are equal, we solve a 2x2 system of lineal equations:
[tex]\left \{ {{Tb_{sol}-1.22m=78.4} \atop {Tb_{sol}-5.03m=76.8}} \right.[/tex]
Subtracting the equations one obtains the molality (concentration):
[tex]Tb_{sol}-Tb_{sol}-5.03m+1.22m=76.8-78.4\\-3.81m=-1.6\\m=0.42m[/tex]
Finally, solving for the boiling temperature of the solution one obtains:
[tex]Tb_{sol}=Kb_{et}m+Tb_{et}\\Tb_{sol}=1.22^oC/m*0.42m+78.4^oC\\Tb_{sol}=78.9^oC[/tex]
Best regards.
