A 20.0-L flask contains 0.776 g of He and 3.61 g of CO2 at
300 K....

Question

A 20.0-L flask contains 0.776 g of He and 3.61 g of CO2 at
300 K.
(a) What is the partial pressure of He in mm Hg?

Accepted Answer

Hi everyone. This problem reads 0.672 g of sulfur dioxide and 1.28 g of argon at 289 kelvin are both present and a 25 liter flask in millimeters of mercury. What is the partial pressure of argon? So the question that we want to answer is the partial pressure of argon in millimeters of mercury. Okay, so for this problem, we're going to need to recall the gas law. The ideal gas law and that is PV equals N. R. T. Okay, pressure times, volume is equal to the number of moles times are the gas constant and t is temperature. So we want to solve for partial pressure. Okay, so we need to rearrange this ideal gas law so that we're solving for that. So we'll go ahead and divide both sides by volume. And then we're going to get pressure. Partial pressure is equal to N R T over V. So this is what we're going to need to solve for the partial pressure. So let's go ahead and plug in what we know based off what's given in the problem. So N stands for moles. Okay, so we know that there is and we're solving for argon. Okay, so there's 1.28 g of argon. We need to know the number of moles of argon. So we need to convert this grams of argon two moles of argon and we're going to do that using its molar mass. Okay, so in one mole of argon, looking at the periodic table, the mass is 39. g. Okay, So our unit of grams of argon cancel. And our moles is going to equal 0.3 to zero moles of argon. So let's go ahead and plug that in for moles. Okay, so 0.3 to zero moles of argon. Okay, r is the gas constant and this is a value we should have memorized and that is 0.8 to 06 Leaders times, atmospheres over mole times kelvin and t is temperature in kelvin, which they give us the temperature in the unit kelvin. So this is great. So Calvin And this is all over volume were given the volume is 25 leaders. Okay, so the unit for pressure is going to be atmosphere. So let's just make sure all of the units cancel here. So we have mole cancels. We have Kelvin cancels leaders canceled. And we have a unit of atmosphere left, which is what we want. So let's go ahead and solve Down here. When we do the calculation, we get the partial pressure is going to equal 0. atmospheres. But the question asks us for the partial pressure and mm of mercury. So we need to go from atmospheres two of mercury. Okay, so we're gonna go ahead and do that by using the conversion in one atmosphere there is 760 millimeters of mercury. Okay, So we'll make sure that our units of atmosphere cancel which they do. And we're left with units of millimeters of mercury. So once we do the calculation, we get 23.1 millimeters of mercury. So our final answer is going to be the partial pressure of argon Is equal to 23.1 mm of mercury. And this is going to be our final answer. And that is it for this problem. I hope this was helpful.

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