Elevation of Boiling Point

Transcript

0:00Introduction

Hello everyone! Okay, a small question first. Pure water boils at one hundred degrees Celsius, that is three hundred and seventy three point one five kelvin. Now I dissolve some glucose in it. Will the solution boil at the same temperature, higher, or lower? If you said higher, correct! But why? And by how much: ten degrees, or just a tenth of a degree? Today we answer both, with one short formula: the elevation of boiling point. It comes up often in JEE, so let's begin!

0:35Why a solution boils at a higher temperature

So, first, when does a liquid boil? When its vapour pressure, small p, reaches one atmosphere, written one a t m. On the graph, p goes up, temperature, capital T, goes along, and the horizontal dashed line is one atmosphere. Okay? Remember the last video? Dissolve a non-volatile solute, one that does not evaporate, and the vapour pressure drops. So p of the solution is less than p of the solvent; the pointed sign means less than, and s o l n is short for solution. On the graph, the dark blue curve is the pure solvent, and the green curve, the solution, sits below it. With me? Now we heat both. Both curves climb towards the dashed line. Which one touches one atmosphere first? The pure solvent, yes! The solution's curve is lower, so it needs more heating to reach one atmosphere. So T b, with a small b below, the boiling point of the solution, is greater than T b nought, the boiling point of the pure solvent; the sign now points the other way, greater than. The small circle on top, read as nought, means pure. Right? That difference, T b minus T b nought, is delta T b, the elevation of boiling point; the capital Greek delta means change. On the graph, it is the orange double arrow on the right. The grey curve and delta T f on the left are about freezing, our next video. Clear?

2:01How big is the rise?

So, how big is the rise? Delta T b equals K b times m. K b is a constant of the solvent; more on it in a minute. Okay? And small m is the molality: moles of solute, n with subscript B, over the mass of the solvent in kilograms. So its unit is moles per kilogram. Right? But this formula needs three conditions. The solution is dilute. The solute is non-volatile. And the solute does not dissociate into ions, or associate into bigger particles. Glucose and urea are fine. Electrolytes need the van 't Hoff factor i, which comes in chapter five. Clear? Now, a trap. Ten percent sucrose by mass. In one hundred grams of this solution, how many grams of water? Ninety, correct! Ten grams are sucrose, so ninety grams are water; w by w means weight by weight. Not one hundred, the whole solution! Molality always uses the mass of the solvent. Got it?

3:04Kb belongs to the solvent

Now, K b. Its name is the ebullioscopic constant. Its unit is kelvin kilogram per mole, written K, k g, mol to the power minus one. Okay? Values to remember: water, zero point five two; benzene, two point five three; chloroform, three point six three. It depends only on the solvent, not on the solute. Right? Quick question. Zero point one molal glucose, and zero point one molal urea, both in water; the small m means molal. Which one boils at a higher temperature? Neither! The same temperature. Both are non-electrolytes in the same solvent, with the same molality, so delta T b is the same. The nature of the solute does not matter, only the number of particles: that is what colligative means. Got it? And where does K b come from? K b equals R times T b squared times M A, over one thousand times delta vap H, okay? R is the gas constant, T b is the boiling point of the pure solvent, M A is its molar mass in grams per mole, and delta vap H is its enthalpy of vaporisation, in joules per mole, not kilojoules. With me? Now, benzene's K b is much larger than water's. Look at the formula. Why? Two reasons. Benzene's molar mass is seventy eight, against eighteen for water, so the top is bigger. And benzene has no hydrogen bonds, written H bonds, so its delta vap H is smaller, and that sits at the bottom. Bigger top, smaller bottom: bigger K b. Clear?

4:40Molar mass from the rise

Now the form problems use most. Dissolve w B grams of a solute of molar mass M B, in w A grams of solvent. Moles of solute: w B over M B, the slash means divided by. Kilograms of solvent: w A over one thousand. Molality is one over the other. Okay? Put this into delta T b equals K b times m. We get one thousand times K b times w B, over M B times w A, right? Now we want M B, so just swap M B and delta T b. M B equals one thousand times K b times w B, over delta T b times w A, clear? Measure the rise, and you get the molar mass of an unknown solute. That's it!

5:25Example 1: boiling point of glucose in water

Now, our hook, with real numbers. Eighteen grams of glucose is dissolved in five hundred grams of water, and K b for water is zero point five two. Find the boiling point. Pause the video, and try it first. Done? Okay, let's check. Glucose has a molar mass of one hundred and eighty, so n B is eighteen over one hundred and eighty: zero point one mole. Now the molality. Do we divide by five hundred, or by zero point five? Zero point five, yes! Molality needs kilograms of solvent. Zero point one over zero point five kilogram: zero point two moles per kilogram. Clear? So delta T b is zero point five two times zero point two: zero point one zero four kelvin. Right? And T b is three hundred and seventy three point one five, plus zero point one zero four: about three hundred and seventy three point two five kelvin; the wavy equals sign means approximately. So, our hook: higher, yes, but only by about a tenth of a degree! Got it?

6:28Example 2: molar mass from boiling point rise in benzene

Now a harder one. One point zero gram of a non-electrolyte in fifty grams of benzene raises its boiling point by zero point four zero kelvin. K b of benzene is two point five three. Find the molar mass. Pause the video, and try it. Done? Alright. The data: w B is one point zero gram, w A is fifty grams, and delta T b is zero point four zero kelvin. Now, in Example one we changed grams to kilograms. So here, w A becomes zero point zero five, right? No, no, careful! In this formula, the one thousand already changes grams to kilograms, so w A stays fifty. So M B equals one thousand times two point five three times one point zero, over zero point four zero times fifty. Okay? The top is two thousand five hundred and thirty, the bottom is twenty. So M B is one hundred and twenty six point five: about one hundred and twenty seven grams per mole. And notice why the question says non-electrolyte: no dissociation, so our formula holds. That's it!

7:37Quick revision: delta Tb = Kb m and the molar mass

So, quick revision. Delta T b equals K b times m, for a dilute solution of a non-volatile solute that does not dissociate. Molality uses kilograms of solvent; the molar mass formula, with its one thousand, uses grams. And K b depends only on the solvent: zero point five two for water, two point five three for benzene. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!