Vapour Density and Average Molar Mass

Transcript

0:00Introduction

Hello everyone! So, one small puzzle before we start. Take phosphorus pentachloride, PCL five. Its molar mass is two hundred eight point five. Its vapour density, as you will see in a minute, should be half of that: one hundred four point two five, okay? Now we heat it, and measure the vapour density of the gas. It comes out sixty two point five five. Wait, the gas became lighter? We did not take anything out, so where did the mass go? By the end of this video, you will solve this mystery yourself. And look, these are quick, sure marks in JEE once you see the one idea behind them. So, let's begin.

0:44Vapour density: half the molar mass

So, first, what is vapour density? It tells you how heavy a gas is compared with hydrogen gas. We write it as V D, okay? And the rule is simple: V D equals M by two, where M is the molar mass of the gas. Why two? Because hydrogen gas, H two, has molar mass two. Clear? Now tell me. Carbon dioxide, C O two, has molar mass forty four. What is its vapour density? Yes, twenty two! Just half of forty four. As simple as that. And the reverse is just as easy. Whenever a question gives you the vapour density, first double it to get the molar mass: M equals two times V D, okay? See, this looks too simple, but in JEE many students put the vapour density straight into a formula that needs the molar mass. We will catch this trap in today's last example. Got it?

1:39A mixture of gases: average molar mass

Alright, now a harder question. Suppose a gas is a mixture: some heavy molecules, some light molecules. Now what is the molar mass of this mixture? Which one do we use? Neither one alone, right? We use an average molar mass, written M with a bar on top, M bar, okay? But not a simple average. M bar equals the sum of x i times M i. Here x i is the mole fraction of each gas, M i is its molar mass, and the big sigma sign on screen just means: add them all up. So each gas counts as much as its share of the moles. That is why we call it the mole fraction weighted mean. Clear? And here is the same idea in plain words. M bar equals total mass divided by total moles. That's it. Whenever you are confused, come back to this line: how many grams, how many moles, divide. One more thing that saves time in gas problems. For gases, the volume fraction is the mole fraction, okay? So if a question says seventy nine percent by volume, you can directly take the mole fraction as zero point seven nine. Just like that.

2:49When a gas dissociates

Now, the most interesting part. Some gases break up on heating. Each molecule of gas A breaks into n molecules of gas B, okay? And alpha, this Greek letter on screen, is the degree of dissociation: the fraction of A that actually breaks up. Now think. Start with one mole of gas A, okay? After heating, how many moles of gas do we have in total? Look, alpha moles of A broke up, so one minus alpha moles of A are left. And each mole that broke up gave n moles of B, so we get n alpha moles of B, okay? Add them: one minus alpha, plus n alpha, which is one plus n minus one, times alpha. With me? Now the key question. When the gas breaks up, does any mass go anywhere? And then, what happens to M bar? No mass is lost! The same atoms are still there, so the total mass stays the same. But the number of moles went up. And M bar is total mass divided by total moles, right? Same mass, more moles, so M bar falls, and the vapour density falls with it. So, put it together. M nought, the M with the little zero on screen, is the molar mass before dissociation. The mass is fixed, so M nought by M bar is just moles after, by moles before. So, M nought by M bar equals one plus n minus one, times alpha, okay? And here is the topper's trick: don't mug this formula up blindly. Just count the moles you get from one mole of gas, and you can write it in ten seconds, in any question.

4:28Example 1: average molar mass of air

Let's do our first example. Find the average molar mass of air, taken as seventy nine percent nitrogen and twenty one percent oxygen, by volume. Pause the video, and try it first. Done? Okay, let's check. Nitrogen, N two, is twenty eight, and oxygen, O two, is thirty two. So the answer is twenty eight plus thirty two, by two, that is thirty, right? No, no, careful! That is the simple average, and that is the trap. Air has much more nitrogen than oxygen, doesn't it? So, first, volume fractions are mole fractions: by volume means by moles. The mole fraction of nitrogen, x for N two, is zero point seven nine, and of oxygen, zero point two one. Now weight each molar mass by its mole fraction. M bar equals zero point seven nine times twenty eight, plus zero point two one times thirty two. Zero point seven nine times twenty eight is twenty two point one two. Zero point two one times thirty two is six point seven two. Add them: twenty eight point eight four grams per mole. And see, the answer sits much closer to twenty eight than to thirty two, because most of the air is nitrogen. That is your quick check.

5:48Example 2: degree of dissociation of PCl5 from VD

Now the harder one, and look, it is our puzzle from the start. Phosphorus pentachloride, molar mass two hundred eight point five, partly dissociates on heating, into PCL three and chlorine, C L two. The vapour density of the equilibrium mixture is sixty two point five five. Find the degree of dissociation. Pause the video and try it. Done? Okay, let's check. Many students write two hundred eight point five by sixty two point five five. That is about three point three three, so alpha is two point three three? Hold on! Alpha is a fraction, it can never be more than one! What went wrong? We divided a molar mass by a vapour density. That is exactly the trap I warned you about. So, first double the vapour density. M bar equals two times sixty two point five five, that is one hundred twenty five point one. Now count the moles. One PCL five gives one PCL three plus one C L two, so n is two, okay? One mole gives one plus alpha moles. So, M nought by M bar: two hundred eight point five by one hundred twenty five point one equals one plus alpha. The division gives one point six six seven. So alpha is zero point six six seven, about zero point six seven. Roughly two thirds of the PCL five has broken up. And now our puzzle is solved! The gas did not lose any mass. Each broken molecule became two, so there are more moles, and the same mass shared over more moles gives a smaller M bar, and a smaller vapour density.

7:32Quick revision: VD, mixture molar mass, dissociation

Got it? So, quick revision. Vapour density is half the molar mass, so double it before using any formula. For a mixture, M bar is the mole fraction weighted mean, or simply, total mass by total moles. And when a gas dissociates, M nought by M bar equals one plus n minus one, times alpha. Now go and try the three practice problems on this page. You will find them easy, I promise. All the best!