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An alkali metal in the form of a cube of edge length $0.171 \mathrm{cm}$ is vaporized in a $0.843-\mathrm{L}$ container at $1235 K$. The vapor pressure is $19.2 mmHg$. Identify the metal by calculating the atomic radius in picometers and the density. (Hint: You need to consult Figures $8.5,11.22,11.29,$ and a chemistry handbook.)

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$\mathrm{Na}\left(186 \mathrm{pm} \text { and } 0.965 \mathrm{g} / \mathrm{cm}^{3}\right)$.

Chemistry 102

Chapter 11

Intermolecular Forces and Liquids and Solids

Liquids

Solids

Rice University

Drexel University

University of Maryland - University College

Lectures

04:08

In physics, a solid is a state of matter characterized by rigidity and resistance to changes of shape or volume. Solid objects have a definite volume, they resist forces (such as pressure, tension and shear) in all directions, and they have a shape that does not change smoothly with time. The branch of physics that studies solids is called solid-state physics. The physical properties of solids are highly related to their chemical composition and structure. For example, the melting point of ice is significantly lowered if its crystal structure is disrupted.

03:07

A liquid is a nearly incompressible fluid that conforms to the shape of its container but retains a (nearly) constant volume independent of pressure. As such, a liquid is one of the four fundamental states of matter (the others being solid, gas and plasma). A liquid is made up of tiny vibrating particles of matter, such as atoms, held together by intermolecular bonds. Water is, by far, the most common liquid on Earth. Like a gas, a liquid is able to flow and take the shape of a container. Most liquids resist compression, although others can be compressed. Unlike a gas, a liquid does not disperse to fill every space of a container, and maintains a fairly constant density. A distinctive property of the liquid state is surface tension, leading to wetting phenomena.

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02:47

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05:20

01:02

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01:25

Refer to the table. $$\beg…

first thing we're gonna do when we approach this problem is we're gonna convert that pressure that were given which is given as being equal thio 19.2 millimeters of mercury. We're gonna convert that over to kill a pascal's to make our lives a little easier. And if we do that, they see that it equals about 2.56 killer Pascal's. Now that we've done that, we're gonna go ahead and use, um, the ideal gas law PV equals NRT. If we rearrange it, we can say that end equals P V over rt. Now, from here we substitute in the values we know, which is 2.56 killer Pascal's times our volume which is given to us us being 0.843 leaders and will divide that by our, which is 8.314 Jules and Permal College. And multiply that by our temperature off 1235. Calvin, run that calculation and we see that we have about 2.10 times 10 to the minus for molt that we're working with. Okay, now that we've figured out how many moles were working with. Let's figure out how many atoms that is so we can do that by just multiplying this. Buy avocados number number of atoms equals moules times, atoms per mole, which is 6.22 times 10 to the 23. And that is going to give us one 0.266 times 10 to the 20 Adams. Okay, And now that we know how many atoms were working with, we're gonna be able to figure out the volume of the cell eventually. And that's really what we want to do here. Because if we can figure out the volume of the cell, then we can figure out the edge length of the cell. And from there we can figure out the radius of an atom. Okay, now that we've gotten that, let's go ahead and figure out how many atoms we have in each cell. Well, conveniently figure 11 point to nine shows us the breakdown of all the different cell types that we could possibly have. And from this, we know that all alkali are B. C. C. That makes our life really easy because we Onley ever need to consider the case where there's going to be to Adams. Purcell. So now that we've not, that we can figure out the number of atoms, Purcell. So we can do that simply by saying the number of atoms Purcell is going to equal the number of atoms divided by to Oh, sorry. That's that's going to give us the number of cells that we have, so I'll make that more clear here. We'll move our Adams back up here, go ahead and delete that for a second there. So this is gonna give us the number of cells that were working with. And the reason for that is that we've considered the number of atoms and we know that every Adam is going there is gonna be exactly two atoms in each cell. So if we just divide the number of atoms by two, that's gonna give us the number of cells that way must be working with. And if we can then figure the volume of each cell, we can just take the total volume divided by ourselves him, which will eventually do so. We run this calculation and we see that we have 6.3 to 9 times tend toothy 19 shells Okay, so now that we've got the number of cells, let's go ahead and fear at the total volume that we're working with. And this is a simple calculation. We're just gonna take 0.171 centimeters, which we know is the edge of the initial Cuba working with no cube that value. Um, the way the original question is worded is a little poorly in that you think that each of the individual cells has an edge link the 0.171 centimeters. This is just a cube of the al Caylee metal that we are working with here. So we go ahead and we'll take that and we find that we're working with 0.0 five centimetres cubed as our total volume here. So now that we know the total volume and the total number cells, we can figure out the volume of each cell, which is just going to be a total volume over the number of cells. Okay, so we go ahead and we divide this value by all the different cells that were working with. And we get 7.9 times 10 to the negative 23 centimeters Cube Purcell, which, if we convert into the more convenient units of Pekka meters by multiplying by 10 of the 30 will get 7.9 times 10 to the seven Pekka meters cubed personal. Okay, now that we've got that, that's rule on down and we can determine the edge length, which is just gonna be equal to the cube root of the volume of the cell. Okay, now that we've got that will say that this is gonna be equal to or we can calculate, this will be equal to about 429.1 p commuters. And if we follow the hint in the question that tells us to use figure 11 0.22 this gives us a convenient little equation that says a is equal to four are over the cube root of three. Where is our edge length? So we can go ahead and use that relation and just swap it so that we're solving for the radius. So will say that our is going to equal swear it of three times are edge all divided by four. And if we calculate that, we'll get a radius of 186 p. Commuters. Okay, now that we've got our radius, let's take a look at one of these figures here. Specifically, figure 8.5. Revere 8.5 gives us the breakdown of various different radio. And specifically we notice that there is one that has 186 p commuters, and that is sodium. So congratulations. We have identified what we're working with, but that's not the end of the story for today. Next up, we need to figure out what the density is, and this is in grams per centimetre. Cute. So let's take the molar mass of sodium, which is 22.99 grams Permal. And let's fear out what this is gonna be in grams per centimeter cubed. So first thing we need to do is get rid of the mole and we need to add in a centimeter. Cute. So what values do we have? It could possibly do that. Well, we know from way up initially that we were working with a volume of 2.1 times 10 to the minus four moles which existed in V total which was 0.5 centimeters cubed so that an over V total, which is going to give us what we need to know. So again, that end was two point 10 times 10 to the minus four moles in the initial volume we were working with of the total 0.5 centimeter acute. Okay, we run that calculation and we see that we have a density off about 0.9 six far grams per centimetre. Cute. So there we have our density. We have our radius, and we know exactly what we are working with, sodium. Simply because we knew the initial conditions that we started with, we were able to figure out all of this. So this is a very convenient way that you can actually determine what you're gonna be working with by systematically working through what the possible conditions are and what your constraints might be. It really helped us out in this case that we knew we were working with BCC because alternatively, we would have to go ahead and calculate the density and the radius and compare both of these against our data values that we already knew to figure this out. But in our case, we didn't have to because we already had this provided to us.

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