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, T b .

Imagine two solutions with the same concentration and the same boiling point.?

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 concentration and boiling point.

ethanol: 78.4 °C Kb=1.22 c/m

carbon tetrachloride: 76.8 °C Kb=5.03 c/m

The molal concentrations are 0.42 m.

Explanation:

This question seeks to explain the boiling point elevation as a colligative property of a mixture.

Colligative property of a mixture/solution is one that describes any property of that that depends only on the number of molecules present.

∆T=Kb.m

∆T - Change in boiling point

Kb- boiling point constant

m - molality

The change in boiling points are 1.6 degrees apart; i.e.,∆T= 78.4-76.8 = 1.6 so the CCl4 must increase 1.6 more than the ethanol.

Remember that

If you want the boiling point to be the same, that means you want

T (ethanol) = T (CCl4)

T (CCl4) =5.03*m

∆T = [1.22*m + 1.6]

[1.22*m + 1.6] = 5.03*m

If concentration is the same;

and solve for m

m = 0.42.

We can check that out from

∆T = Kb*m

∆T(ethanol)= 1.22*0.42 = 0.51

so boiling point. = 78.4 + 0.51 = 78.9

∆T(CCl4) = 5.03*0.42 = 2.1

so boiling point. = 76.8 + 2.1 = 78.9

and the boiling point are the same.

The molal concentrations are 0.42 m.

3 hours because it is a decent distance.