(b) Calculate the molar mass of a vapor that has a density of 7.135 g>L at 12 °C and 99.06 kPa.

Video transcript

Hey everyone in this example, we're told that a sample of gas is found to have a density of 1.457 g per liter at 34 degrees Celsius and 8 35 Tour. We need to determine the molar mass of this gas. So we're going to recall our density form of our ideal gas equation and we should recall that that is going to be equal to our density symbol, which is rho equal to our pressure of our gas, multiplied by its molar mass and then divided by the gas constant R. Times temperature. Now, in terms of this question, we need to solve for Mueller mass. So we're going to reorganize this to solve for Mueller mass so that we have Mueller mass equal to our density of our gas, multiplied by the gas constant R. Times temperature and then divided by pressure of our gas. Before we go ahead and plug in our known values. We need to recognize that we're going to have to convert our pressure value from Tour to A. T. M. Because that's the proper unit for pressure in this equation. And we want to go ahead and also convert our temperature from Celsius to kelvin. So what we're going to do is start out by converting our pressure. So we're given the pressure of 835 tour and we want to go ahead and convert from Tour into a. T. M's by recalling our conversion factor that for 1 80 M we have 60 Tour. So now we're able to cancel out our units of tours leaving us with a T. M. As our pressure unit. And this is going to give us a pressure equal to 1.10 ATMs. Now moving on to our temperature, we want to go ahead and convert our temperature given from 34°C. And we're going to not multiply but rather add to 73 . Kelvin to convert to Kelvin. And this is going to give us our proper temperature equal to 307.15 Kelvin. So now we're able to go ahead and solve for molar mass. This is going to equal our density value given in the problem as 1.457 g/l. We're going to then multiply this by r. Gas constant R. Which we should recall is equal to 0.8206. Leaders times a T. M's, divided by moles, times kelvin. And then this is also multiplied by our temperature, which we converted to 307.15 Kelvin In our denominator. We're going to plug in our pressure, which we converted to ATM as 1.10 ATMs. So now we're going to cancel out our units. We can get rid of ATMs. We can get rid of leaders. Sorry, Leaders is over here. We can still get rid of kelvin's and this leaves us with gramps Per mole as our final units, which is what we would agree with for molar mass. And so what we would get is that our molar mass of our gas is equal to a value of 33.4 g per mole. And this is going to be our final answer to complete this example. So I hope that everything I explained was clear. If you have any questions, leave them down below and I will see everyone in the next practice video.