Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle to start. I take a mixture of two volatile liquids, call them A and B, and only one quarter of its molecules are A molecules. I let it form vapour, I catch that vapour, cool it back into a liquid, and let that liquid form vapour once more. And now the vapour is three quarters A molecules! One quarter to three quarters, in just two rounds. How? Hold that thought; our last example cracks it. This is set three of mixed problems on solutions, short numericals of the kind that come up often in JEE Main. Two quick ideas, then two full problems. So, let's begin.

0:44Idea 1: colligative properties count particles

Idea number one. Zero point one molal aluminium sulphate, and zero point one molal potassium sulphate. The small m after each number means molal; the small numbers below count the atoms, and the small three after the brackets counts the sulphate groups. We want the ratio of delta T b, the elevation of boiling point: the triangle is the Greek capital delta, for change, and the small b is for boiling. Both have the same molality, in the same water. So the same elevation, right? No, no, careful! Delta T b equals i, times K b, times m. Here i is the van 't Hoff factor, K b is the ebullioscopic constant of water, and m is the molality. Same K b, same m, so the double arrow tells us: the ratio of the elevations is the ratio of the i values. Clear? So, now we count the particles. One formula unit of aluminium sulphate gives how many ions in water? Five, exactly! The single arrow means, dissociates into. Two aluminium ions, with a three plus charge written up top, and three sulphate ions, each with a two minus charge. Two plus three is five, so for complete dissociation, i equals five. Okay? Potassium sulphate: two potassium ions, K plus, and one sulphate ion. Three ions, so i equals three. Right? So the ratio is five to three. That needs both salts completely dissociated, as given, okay? Colligative properties count the particles, not their nature. Got it?

2:23Idea 2: osmotic pressure to freezing point

Idea number two. An aqueous solution has osmotic pressure pi, the Greek letter, of two point four nine bar at three hundred kelvin. The solute is a non-electrolyte, so there is no dissociation. Estimate the freezing point. These look like two different worlds. So what links them? The concentration, yes! The van 't Hoff equation: pi equals CRT, where C is the molar concentration, R the gas constant, and T the temperature in kelvin. So C equals pi over R T, with R as zero point zero eight three, since pi is in bar. Zero point zero eight three times three hundred is twenty four point nine, and two point four nine over that is zero point one; the capital M means molar, moles per litre. With me? But the freezing point formula needs the molality, moles per kilogram of water, not the molarity. For a dilute aqueous solution, one litre of solution is roughly one kilogram of water, so the molality is about the molarity; the wavy equals sign means approximately. So m is about zero point one moles per kilogram. Okay? Now, delta T f, small f for freezing, equals K f times m, and in this chapter K f of water is one point eight six. One point eight six times zero point one: zero point one eight six. Water freezes at zero, so T f is minus zero point one eight six degrees Celsius; the small circle is the degree sign. Clear?

3:54Example 1: degree of dissociation of MX2 from boiling point

Alright, example one. The boiling point of a zero point one molal aqueous solution of M X two is one hundred point one three degrees Celsius. The small two below means two X for each M, okay? Find its degree of dissociation. Pause the video, and try it first. Done? Okay, let's check. Water boils at one hundred degrees, so delta T b is one hundred point one three minus one hundred: zero point one three. Okay? Now find i. Delta T b equals i times K b times m, so i is delta T b over K b times m. K b is zero point five two, and times zero point one: zero point zero five two. Zero point one three over zero point zero five two: two point five. Right? Next, the ions. M X two dissociates into one M two plus ion and two X minus ions, with the charges written up top. So n, the number of ions, is three. With me? So, the degree of dissociation, alpha, is just i over n: two point five divided by three, about zero point eight three; the wavy equals sign again means approximately. Right? Oh no, careful! The part that does not dissociate still counts, one particle each. The correct formula: i equals one, plus n minus one, times alpha, where alpha, the Greek letter, is the degree of dissociation. With n equal to three, that is one plus two alpha, equal to two point five. So two alpha is one point five, and alpha is zero point seven five: seventy five percent. Got it?

5:30Example 2: vapour composition after condensing once

Now the harder one: our puzzle from the start. Two volatile liquids, A and B: pure A has a vapour pressure, p nought, of one hundred and eighty, and pure B, sixty millimetres of mercury, written m m H g; the small circle on top, read as nought, means pure. The liquid has x A, the mole fraction of A, equal to zero point two five. It comes to equilibrium with its vapour; that vapour is condensed, and comes to equilibrium once more. Find the composition of the final vapour. Pause the video, and try it. Done? Okay, let's check. We use Raoult's law, treating the solution as ideal: p A, the partial pressure of A, equals x A times p A nought. Zero point two five times one hundred and eighty: forty five. Then x B is zero point seven five, and times sixty, that is forty five too. Okay? Now the vapour. By Dalton's law, y A, the mole fraction of A in the vapour, is p A over the total pressure. Forty five over forty five plus forty five: zero point five. Already, one quarter has become one half. Clear? So the answer is zero point five, right? No, no, wait! Read the question again: the vapour is condensed and re-equilibrated once. So that vapour becomes a new liquid. What is x A in that liquid? Zero point five, exactly! Condensing does not change the composition, so the new liquid is half A and half B, okay? Raoult's law once more. Zero point five times one hundred and eighty is ninety; zero point five times sixty is thirty. With me? So the final y A is ninety over ninety plus thirty: zero point seven five. Three quarters! Puzzle solved. For an ideal solution like this, the vapour is always richer in the more volatile liquid, so each round pushes it further: the idea behind fractional distillation. Got it?

7:30Quick revision: i counts particles, osmotic pressure, vapour

So, quick revision. Colligative properties count particles: delta T b equals i times K b times m, and i equals n when the salt is completely dissociated. For partial dissociation, i equals one, plus n minus one, times alpha. Pi equals CRT gives the concentration, and for a dilute aqueous solution, the molality is about the same. And y A equals p A over capital P, the total pressure; condense and repeat, and the vapour gets richer in the more volatile liquid. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!