Mixed Problems: Set 1

Transcript

0:00Introduction

Hello everyone! Okay, a small challenge to start. Here is some water with an unknown solute in it. I will not tell you which solute, or how much. Just one fact: it freezes at minus zero point nine three degrees Celsius. Now tell me its boiling point, and its vapour pressure. Impossible, right? Hold that thought. By the end of this video, you will do it in a few lines. This is set one of mixed problems on solutions, the short numericals that come up often in JEE Main. Two quick ideas, then two full problems. So, let's begin.

0:38Idea 1: Raoult's law and the vapour

Idea number one: Raoult's law, for an ideal solution of two volatile liquids, call them A and B, mixed together. Small p A, with A written below, is the partial vapour pressure of A in the solution. It equals x A, its mole fraction in the liquid, times p A nought, the vapour pressure of the pure liquid; the small circle on top, read as nought, means pure. Same for B, okay? Now the numbers. p A nought is one hundred and fifty, and p B nought is fifty millimetres of mercury, written m m H g. x A is zero point four, so x B is one minus that, zero point six. Right? Capital P, the total vapour pressure, is the sum of the two partial pressures. Zero point four times one hundred and fifty is sixty; zero point six times fifty is thirty. Sixty plus thirty: ninety millimetres of mercury. Clear? Now, x A is zero point four in the liquid. So in the vapour above it, the mole fraction of A is also zero point four, right? No, no, careful! The vapour has its own composition. By Dalton's law, y A, the mole fraction of A in the vapour, is p A over capital P, the total. Sixty over ninety: two thirds, about zero point six six seven; the wavy equals sign means approximately. With me? See the jump? Zero point four in the liquid, about zero point six seven in the vapour. Pure A has the higher vapour pressure, so it is the more volatile liquid, and for an ideal solution the vapour is always richer in it. Got it?

2:10Idea 2: two solutes? Add the moles

Idea number two. Zero point six gram of urea, plus one point eight grams of glucose, dissolved together in one hundred grams of water. Find the freezing point. Two solutes, one beaker: what do we do with them? Simple: add their moles! Colligative properties count particles, whichever solute they come from. Urea, molar mass sixty: zero point six over sixty is zero point zero one. Glucose, one hundred and eighty: one point eight over one hundred and eighty, zero point zero one again. Total: zero point zero two moles. Okay? Now the molality, small m: moles of solute over kilograms of water. Zero point zero two over zero point one kilogram: zero point two moles per kilogram, written mol k g to the power minus one. Right? Next, delta T f, the fall in freezing point: the triangle is the Greek capital delta, for change, and the small f is for freezing. It 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 two: zero point three seven two. Clear? Water freezes at zero, so T f is minus zero point three seven two degrees Celsius; the small circle is the degree sign. Both solutes are non-electrolytes, so there is no dissociation here. Got it?

3:34Example 1: freezing point of K3[Fe(CN)6], 90% dissociated

Alright, example one. K three, F e, C N six is ninety percent dissociated in water. The square brackets hold the F e C N six part, the round ones hold C N, and the small numbers below count the atoms. Find i, the van 't Hoff factor, and the freezing point of its zero point one molal solution; the small m after the number means molal. Pause the video, and try it first. Done? Okay, let's check. First, count the ions. One formula unit of this salt gives how many ions in water? Four, exactly! The arrow means, dissociates into. Three potassium ions, K plus, the plus up top being the charge, and one complex ion, F e C N six, with its three minus charge also up top. So n, the number of ions, is four. Okay? Ninety percent dissociated, four ions each. So i is four times zero point nine, three point six, right? Oh no, careful! The undissociated ten percent still counts, as whole formula units. The formula is i equals one, plus n minus one, times alpha, where alpha, the Greek letter, is the degree of dissociation, zero point nine here. One plus three times zero point nine: three point seven. Clear? Now delta T f equals i times K f times m. Three point seven times one point eight six times zero point one: about zero point six eight eight. With me? So the freezing point is about minus zero point six nine degrees Celsius. For an electrolyte, always find i first. Got it?

5:16Example 2: boiling point and vapour pressure from one Tf

Now the harder one: our challenge from the start. Here is a solution of a non-volatile solute in water. It freezes at minus zero point nine three degrees Celsius. Find its boiling point, and its vapour pressure at two hundred and ninety eight kelvin, where p nought, the vapour pressure of pure water, is twenty three point eight millimetres of mercury. Pause the video, and try it. Done? Okay, let's check. We do not know the solute or its mass. So which one quantity connects the freezing point, the boiling point and the vapour pressure? The molality, exactly! Water freezes at zero, so delta T f is zero point nine three. m equals delta T f over K f: zero point nine three over one point eight six, zero point five moles per kilogram. Okay? Boiling point next. Delta T b, small b for boiling, equals K b times m, and K b of water is zero point five two. Zero point five two times zero point five: zero point two six. So T b, the boiling point, is one hundred point two six degrees Celsius. Right? Now the vapour pressure. For that we need x B, the mole fraction of the solute. Molality zero point five means zero point five mole of solute in one kilogram of water. So how many moles of water is that? One thousand grams, over eighteen grams per mole: about fifty five point five six moles. Clear? So x B is zero point five, over zero point five plus fifty five point five six. That is zero point five over fifty six point zero six: about zero point zero zero eight nine. Got it? Now, the vapour pressure is p nought times x B, right? No, no, wait! p nought times x B is only the lowering, how much the vapour pressure falls. The solution keeps the rest: p equals p nought, times one minus x B, okay? Twenty three point eight times one minus zero point zero zero eight nine: about twenty three point five nine millimetres of mercury. See? One freezing point gave us both answers. Challenge solved!

7:28Quick revision: Raoult's law, adding moles, i and alpha

So, quick revision. An ideal solution of two volatile liquids: p A equals x A times p A nought, and the vapour is richer in the more volatile one. Several solutes: add the moles. And for an electrolyte, i equals one, plus n minus one, times alpha. Finally, one delta T f gives the molality. Then the molality gives delta T b, and p equals p nought, times one minus x B, for the vapour pressure. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!