The vapor pressure of water at 25 °C is 23.76 torr. If 1.25 g of
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Question

The vapor pressure of water at 25 °C is 23.76 torr. If 1.25 g of
water is enclosed in a 1.5-L container, will any liquid be present?
If so, what mass of liquid?

Accepted Answer

Hey everyone, we're told that at negative 45.4 degrees Celsius, the vapor pressure of ammonia is 400 millimeters of mercury. Is there a liquid present? If 1.5 g of ammonia is enclosed in a 2.0 liter vessel. If so determined the mass of liquid ammonia present. To answer this question, we first need to calculate the number of moles of ammonia in the gas phase and we can do so by using our ideal gas law equation, which is our pressure times volume equals the number of moles times r gas constant times our temperature. But since we want to solve for the number of moles of ammonia, we can go ahead and rearrange this equation into the number of moles is equal to our pressure times volume divided by our gas constant times our temperature. So let's go ahead and write out our values first. We were told that our pressure is going to be of mercury. Now we want to convert this into atmospheric pressure And we can do so by using dimensional analysis and we know that we have 760 of mercury Per one atmospheric pressure. So when we calculate this out, we end up with 0.5263 atmospheric pressure For our volume. We were told that we had 2.00 leaders And our temperature was said to be -45.4°C, which we want to change into Kelvin. So we can go ahead and add 273.15, which gives us a temperature of 227.75 Kelvin. So let's go ahead and solve for the number of moles of our ammonia in the gaseous state. So plugging in our values, we have 0.5263 atmospheric pressure for our pressure. And this is going to be multiplied by 2.00 leaders. Next, we're going to divide this all by our gas constant, which is 0. leaders times atmospheric pressure divided by mole times kelvin. And we're going to multiply this value by our temperature of .75 Kelvin. Now, when we calculate this out and cancel out all of our units, We end up with 0.0563 mol. Next we're going to calculate the mass of ammonia in our gas phase. To do so we're going to take the number of moles which we calculated to be 0.056, three more. And we're going to use dimensional analysis here to convert it instagrams. Using the molar mass of ammonia, we have 17.31 g per one mole of ammonia. Now this gets us to a value of 0.9588 g. Now we're going to determine the mass of ammonia remaining as a liquid. So to calculate the mass of ammonia as a liquid, we're going to take our total mass and subtract the mass of ammonia in its gas phase. So, plugging in these values, we have 1.05 g which was said to us in our question stem and we're going to subtract 0. g. Now this gets us to a total of 0.0912 g and this is going to be our final answer. So yes, there is zero .0912g of liquid ammonia present. Now, I hope this made sense and let us know if you have any questions.

The vapor pressure of water at 25 °C is 23.76 torr. If 1.25 g of water is enclosed in a 1.5-L container, will any liquid be present? If so, what mass of liquid?

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