The vapor pressure of carbon tetrachloride, CCl₄, is 0.354 atm, and the vapor pressure of chloroform, CHCl₃, is 0.526 atm at 316 K. A solution is prepared from equal masses of these two compounds at this temperature. Calculate the mole fraction of the chloroform in the vapor above the solution. If the vapor above the original solution is condensed and isolated into a separate flask, what will the vapor pressure of chloroform be above this new solution?

Question

Accepted Answer

Hi everyone today, we have a question telling us that a solution is made by mixing equal masses of ethanol and acetone at 25.0°C, calculate the mole fraction of acetone in the vapor above the solution, the vapor pressures of pure acetone and pure ethanol at 25.0°C are 0.34040 atmosphere and 0.05872 atmosphere respectively. Assume that all the vapor above this solution is condensed and transferred to a new flask. What will acetone vapor pressure be above the new solution in the new flask? So first off, we're going to assume that we have 100 g of each of the two liquids. So acetone Mueller mass is 58.08 g per mole ethanol's moller mass Is 46.07 g per mole. The moles of acetone equals 100 0.0 g times one mole of acetone over 58 . g. So our grams are canceling out here And that equals 1.722 Molds of a system. Our moles of ethanol Equals 100 g times one mole of ethanol over .07 g. So our g are canceling out here, leaving us with 2.17 one moles of ethanol. So now to calculate our mole fractions, our mole fraction of acetone is 1. moles Divided by 1.72, 2 moles Plus 2.171 moles. And that equals 0.4423. Our mole fraction of ethanol Equals 2.171 moles Divided by 1.72, 2 moles Plus 2.171 moles, Which equals 0.5577. Now to calculate our vapor pressures above the solution. So our pressure of acetone Equals 0.442, three Times 0. Atmospheres, Which equals 0. atmospheres. Our pressure of ethanol Equals 0.557, seven Times 0. Atmospheres Equals 0. Atmospheres. So our total pressure Equals 0.13 for five atmospheres Plus 0. Atmospheres Equals 0. Atmospheres. Now to calculate the mole fraction of acetone in the vapor above the solution, the mole fraction of acetone will equal 0. atmospheres, Divided by 0.1672 Atmospheres, Which equals 0.8044. And the vapor pressure of Acetone. In the second flask, we have the mole fraction of Acetone in the new flask, which is 0.8044. So the vapor pressure of pure acetone is 0.3040 atmospheres. The vapor pressure in the new flask is going to be 0. Times, atmospheres. And that's going to equal 0. five atmospheres. So those are our two final answers. Thank you for watching. Bye.

Verified Solution

Key Concepts

Vapor Pressure

Raoult's Law

Mole Fraction