The vapor pressure of ethanol (C2H5OH) at 19 °C is 40.0 torr. A 1...

Question

The vapor pressure of ethanol (C2H5OH) at 19 °C is 40.0 torr. A 1.00-g sample of ethanol is placed in a 2.00 L con- tainer at 19 °C. If the container is closed and the ethanol is allowed to reach equilibrium with its vapor, how many grams of liquid ethanol remain?

Accepted Answer

Hello everyone today. We have the following problem. A 1.5 g of hep pain. With a vapor pressure of 35.5 tour at 20 degrees Celsius is placed in a 2.5 liter vessel at 20 degrees Celsius assume that the vessel is closed and the volatile substance has reached equilibrium with its vapor. How much liquid hep pain in grams is left. So at this equilibrium point that was mentioned, so equilibrium, our heP 10 will in fact continue to evaporate so it will continue to evaporate until of course the flask is filled with That 35.5 tour. So until 35.5 tour is reached. So what we need to do is we need to determine the mass of the liquid obtain left. So the mass and grams of heavy rain left is going to be equal to the initial mass. I've had 18 minus the mass Of the head, obtain vapor. And that's how we can find the mass of liquid obtained left. So before we get into that we want to quickly just transform this 28°C into Kelvin. We do that by adding to 73.15 to get to 93.15 Kelvin. And of course going based on the periodic table, our molar mass of hep dane can be found to be 100.2 g per mole. So now we have these two values. So Let's look at our pressure, we have 35.5 tour and we need to convert that to atmospheres. We can say that in one atmosphere There are Tour our units of tour will cancel out giving us 0.047 atmospheres as our pressure. And then we can use our ideal gas law actually with our values that we have. So gas laws, the pressure times the volume is equal to the number of moles times our gas constant, times our temperature. And of course if we were to rewrite this, this would be the same thing as saying that the pressure times the volume is equal to our mass over our molar mass, times our gas constant times our temperature. That's what the N. Represents, rearranging us once more to solve for mass. We get that mass is equal to the molar mass times are pressure times volume over r gas constant times our temperature. And so now we can find our mass of Neptune vapor. So we have our mass of obtain vapor and that is going to be equal to basically the values that we have plugged into fashion that we have. So to start we have our 100. grams per mole. We're gonna multiply that by our pressure, 0.047 atmospheres and then times are volume which was 2.5 liters. And then we're going to place all of that over our gas constant which is 0.08 leaders times atmospheres over moles, times kelvin. And then we're gonna multiply finally by our temperature which was 2 93. kelvin. And so once all of our units cancel out, we will be left with 0.49 g four hour mass of hefting vapor. So going back to our initial equation of the mass of hefting left, we said that it was the initial mass. Our initial mass was 1.50 g and then the mass of the vapor was 0.49 g. This is essentially going to equal 1.1 g of our liquid hep 10 left. And so with that we have answered the question overall. I hope this helped. And until next time.

The vapor pressure of ethanol (C2H5OH) at 19 °C is 40.0 torr. A 1.00-g sample of ethanol is placed in a 2.00 L con- tainer at 19 °C. If the container is closed and the ethanol is allowed to reach equilibrium with its vapor, how many grams of liquid ethanol remain?

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