Question

Imagine two solutions with the same concentration and the same boiling point, but one has ethanol as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b ), and boiling point,Tb.Given:Ethanolnormal Boiling point: 78.4Kb: 1.22CCL4normal boiling point: 76.8Kb: 5.03

Answer 1

Answer:

$m=0.42m$

$Tb_{sol}=78.9^oC$

Explanation:

Hello,

In this case, we use the boiling point elevation colligative property equation for both ethanol and carbon tetrachloride as shown below:

$Tb_{sol}-Tb_{et}=Kb_{et}m\\Tb_{sol}-Tb_{CTC}=Kb_{CTC}m$

Now, since both the boiling point and the concentration of the solutions are equal, we solve a 2x2 system of lineal equations:

$\left \{ {{Tb_{sol}-1.22m=78.4} \atop {Tb_{sol}-5.03m=76.8}} \right.$

Subtracting the equations one obtains the molality (concentration):

$Tb_{sol}-Tb_{sol}-5.03m+1.22m=76.8-78.4\\-3.81m=-1.6\\m=0.42m$

Finally, solving for the boiling temperature of the solution one obtains:

$Tb_{sol}=Kb_{et}m+Tb_{et}\\Tb_{sol}=1.22^oC/m*0.42m+78.4^oC\\Tb_{sol}=78.9^oC$

Best regards.