The vapor pressure of CCl₃F at 300 K is 856 torr. If 11.5 g of CCl₃F is enclosed in a 1.0-L container, will any liquid be present? If so, what mass of liquid?

Video transcript

Hey everyone, we're told that a three liter close vessel contains 2.75 g of acetone with a vapor pressure of 282 mm of mercury at 30°C determine if there will be any liquid present in the vessel. If so, calculate the mass of the liquid. To answer this question, we need to use our ideal gas law equation in order to determine the number of moles of acetone. So this is going to be our pressure times are volume, which is equal to the number of moles times, r gas constant times our temperature. Now, since we want to determine the number of moles of acetone, we have to rearrange this equation into the number of moles is equal to our pressure times are volume divided by our gas constant times our temperature. So let's go ahead and write out our values. So we have a pressure of 282 mm of mercury. Now we want to convert this into atmospheric pressure. Using dimensional analysis, we know that we have 760 of mercury per one atmospheric pressure. Now, when we calculate this out, we end up with the pressure of 0.3711 atmospheric pressure. Now for our volume, we were told that we had 3.00 l And our temperature is said to be 30°C. Now we want to convert this into Kelvin and we can do so by adding 273. And this will get us to 303.15 Kelvin. So now let's go ahead and solve for the number of moles of our acetone in its gas phase. So plugging in our values, we have a pressure of 0.3711 atmospheric pressure And we're going to multiply this buyer volume of 3.00 leaders. This will all be divided by our ideal gas constant which is 0.08-06 leaders times atmospheric pressure divided by Mole Times Kelvin. And we're going to multiply this value by our temperature of 303.15 Kelvin. Now, when we calculate this out and cancel out all of our units, We end up with a value of 0.0448 mol. Now let's go ahead and solve for the mass of acetone in our gas phase. To solve for the mass of our acetone In its gas phase, we're going to take the value we just calculated which was 0.0448 mol. And we're going to convert this into grams by using dimensional analysis Using acetone Mueller mass. We know that we have 58.79 g of acetone per one mole of acetone. This will get us to a mass of 2. g. Now we want to determine the mass of our acetone in its liquid phase and we can do so by taking our total mass and subtracting the mass of our acetone in its gas phase. So, plugging in these values, we have 2.75g -2.6019g, and this will get us to A value of 0.148 g, which is going to be our final answer. So to answer our question Yes, there is 0.148g of liquid acetone present in our vessel. Now, I hope this made sense and let us know if you have any questions.