At 60 °C, compound X has a vapor pressure of 96 mm Hg, benzene 1C6H62 has a vapor pressure of 395 mm Hg, and a 50:50 mixture by mass of benzene and X has a vapor pres- sure of 299 mm Hg. What is the molar mass of X?

Question

Accepted Answer

Hello. In this problem we are told vapor pressure of compound X. Halloween and a 50 50 mixture by mass of compound X. And valuing at 50 degrees Celsius are 141 millimeters of mercury, 93 millimeters of mercury and 100 24 millimeters of mercury. Respectfully, we are asked to determine the molar mass of X. Recall when dealing with the vapor pressure of a solution, we can make use of rails law. In this case we have a 22 volatile component system which is made up of compound X and Halloween rails Law says that the vapor pressure above the solution then is equal to the vapor pressure of component A plus that of component B. The vapor pressure of component A. Is found by taking the mole fraction of a time is the vapor pressure of puree. And the pressure of B is found by taking the mole fraction of B. Times of vapor pressure of peer B. So in this case we have now one equation with two unknowns. We need to get this to be one unknown with one equation. So we make use of the fact that the mole fraction of a plus that of B will be equal to one. And we will write then the mole fraction of B. In terms of A. So the mole fraction of B then is equal to one minus the mole fraction of a. Rewriting rails above. We have the pressure of vapor above the solution then is equal to the mole fraction of a. Times of a pressure puree plus then substituting in for mole fraction of B. We have one minus mole fraction of a times the vapor pressure of Peer be applying the foil method. We have then one times the vapor pressure Pier B. Which would be equal to then the pressure Pier B minus. Then the mole fraction of a times the vapor pressure of Peer B. We are gonna move the vapor pressure pier B to the left hand side by subtracting it from both sides of the equation. So we have then the pressure for the solution minus the pressure for Pure B is equal to the mole fraction of a. Times of a pressure. Puree minus the mole fraction of a. Times of vapor pressure. For Peer B will factor out the mole fraction of a more fraction of a. Then we have times of a pressure Pier a minus the vapor pressure of Peer B. Isolating the mole fraction of a bible size. Then by the pressure puree minus that of Pure B. You get the pressure above the solution minus the vapor pressure of Peer B divided by the pressure. Pure a minus the vapor pressure of Pure B. So we have again two components, we will call a in this case to Halloween. We are told in the problem that pure Halloween has a favorite pressure of 93 mm of mercury and we will call component B. X. We are told then that X. has a pressure of a higher and 41 mm of mercury. That's for pure X. So we can plug those into our equation one more value. Um We also need then again the vapor pressure above the solution which we were told was 124 millimeters of mercury. So the mole fraction then of Halloween Will be equal to the vapor pressure above the solution. So 124 mm of Mercury minus the vapor pressure peer B. Which we're calling X, which is 100 41 millimeters of mercury all over the very pressure of puree which is Halloween. It was 93 mm of Mercury minus the vapor pressure Pier B which is X. And was 100 41 millimeters of mercury. This thing gives us a mole fraction for valuing of 0.3542. again the mole fraction of Halloween plus out of mole fraction of X. Z equal to one. So that means then the mole fraction of X is equal to one - at Halloween. So 1 -0.3542. Which works out to 0.6458. So now we are going to assume that we just have one more total. We have one more total. That means then we have 0. moles Of X. and we have 0.3542 moles of Halloween. So we'll make use of our moles of Halloween and determine. Then the massive Halloween. So we have 0.3542 moles of Halloween. Make use of its molar mass convert from moles to mass one more of Halloween has a mass of 92.14 g. We set up the Mueller mass so that our molds of talia and cancel this works out to 32.63 g. Halloween. We were told that the solution was made up of a 50 mixture by mass of compound X. And Halloween. So that means if we have 32.63 g of Halloween, it's 50 50 mixture. We also have 32.63 g of X. So again based on our assumption Of one mole hole we found then the moles of X to be .6453 Oh sorry, 5 8. And so we take our mass of X 62.32 point 63 g of X. Divided by our moles of X, which is 0.6458 moles of X. And this works out to then 50.5 g per mole. So our molar mass of X is equal to 50.5 g per mole. So we need to remember always that when we're dealing with the vapor pressure of a solution, we can make use of real slaw. We can take rail slaw and simplify it. Keeping in mind that the mole fractions of our two components have to sum to one. We can rewrite rail slaw so that we have one equation and one unknown. We can then rewrite the equation in terms of our one unknown in this case the mole fraction of a. Which we term Halloween. Once we then calculate the mole fraction of Halloween, we can determine the mole fraction of our unknown compound X. We need to know the number of moles of each of these. We can convert it to mass so we can find the molar mass. Can make a simple relationship based on assuming that we have one more total. If we have one more total, we have .6458 moles of X and .3542 moles of valuing. We determine then the massive valuing. Given it's a 50 50 mixture. We also then know the mass of our unknown X. We then take the mass of our unknown X. Divided by the molds of X. To determine then the molar mass which worked out to again 50. g per mole. Thanks for watching. Hope this help

At 60 °C, compound X has a vapor pressure of 96 mm Hg, benzene 1C6H62 has a vapor pressure of 395 mm Hg, and a 50:50 mixture by mass of benzene and X has a vapor pres- sure of 299 mm Hg. What is the molar mass of X?

Verified Solution

Key Concepts

Vapor Pressure

Raoult's Law

Molar Mass Calculation