12. A solution contains 4.08 g of chloroform 1CHCl32 and 9.29 g of acetone 1CH3COCH32. The vapor pressures at 35 °C of pure chloroform and pure acetone are 295 torr and 332 torr, respectively. Assuming ideal behavior, calculate the vapor pressure above the solution. (LO 13.12) (a) 256 torr (b) 314 torr (c) 325 torr (d) 462 torr

Video transcript

Hello. In this problem we are asked what is the vapor pressure above the solution containing 1.47 g of vaccine and 2.18 g of methanol at four degrees Celsius. If the vapor pressure of pure hexane and methanol are 279 tour and 265. To respectfully, we are told that we can also assume ideal behavior when trying to find the vapor pressure above the solution. Recall that we can make use of rail slaw and in this case we have a two volatile component system made up of hexane and methanol according to rail slaw. Then the total vapor pressure above the solution is equal to the vapor pressure of component A. Plus that of component B. The vapor pressure of component A. Is equal to the mole fraction of component A. Times vapor pressure of puree and the vapor pressure of component B is equal to the mole fraction of component B. Times the vapor pressure of pure B. Again in this case are A and B. Are hexane and methanol. So to find the pressure above the solution, we have to find the mole fraction of a fraction of B, Which requires us. Then to convert our mass of hexane and methanol to moles. Beginning with hexane we have 1.47 g of hexane. You can make use of its molar mass to convert from master moles one mole of hexane has a mass of 86.175g, make sure to accept the mole master that grams, the numerator numerator cancels with grams in the denominator. And this works out to 0.01706 moles of Hexane. Similarly for Methanol we have 2.18 g of methanol. Making use of its molar mass as a conversion factor one mole of methanol As a massive .042g. Set up the mole mass. So that g and the numerator cancels with g and the denominator. This works out to 0.06804 moles of methanol. We can then find the total moles by combining molds of hexane and the molds of methanol. So our total moles is equal to 0. Plus 0.06804. So I told the moles works out to zero 08510 moles. You can now find the mole fraction of fixing. It's equal to the molds vaccines. 0.1706 divided by our total number of moles. This works out to 0.2005. And our mole fraction of methanol Is equal to our M.nolds. Methanol 0.06804. Right by our total number of moles. 0.08510. This works out to 0.7995. So our mole fraction of these two components should add up to one. So that's a check on our work. So .2005 and .7995 to add up to one. Going back to Real Saw. Then the vapor pressure above the solution will then be equal to our mole fraction of hexane, 2.2005. In terms of vapor pressure, pure hexane, which is 279 tour stated in the problem. Plus our mole fraction of methanol Times the vapor pressure of pure methanol, which is given as 265 tour. So the vapor pressure above the solution Then works out to 268 tour. So we can find the vapor pressure above the solution by finding the mole fractions of each of the components that makes up the solution and multiplying those by the vapor pressure for the peer components. Um in this case that works out to 268 tour for a mixture of hexane and methanol. Hope this helps. Thanks for watching.