Solutions of Non-volatile Solutes

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Glucose does not evaporate, not even a little, right? Yet dissolve some glucose in water, and the vapour pressure of the water goes down. How can something that never enters the vapour change the vapour pressure? Strange, isn't it? And a second puzzle. Urea is a small molecule; sucrose is a big, heavy one. Same mole fraction of each in water: which one lowers the vapour pressure more? Hold that thought. This idea comes up often in JEE, so let's begin.

0:38Only the solvent makes the vapour

So, the solvent we call capital A, here water, and the solute capital B, here glucose. Non-volatile means the solute does not evaporate at all, okay? Now think. Who puts molecules into the vapour above this solution: the solvent, the solute, or both? Only the solvent, exactly! The solute adds nothing to the vapour. Right? And the solvent obeys Raoult's law, from the last lesson: p equals x A times p A nought. Small p is the vapour pressure of the solution, and x A is the mole fraction of the solvent; the small A below is a subscript, just a label. And p A nought, with the small circle on top, is the vapour pressure of the pure solvent at the same temperature. Clear? Now tell me. Once some glucose is dissolved, is x A still equal to one, or less than one? Less than one, correct! The solute takes a share of the moles. On screen, the double arrow means implies, and the sign pointing left means less than. So p is less than p A nought: the vapour pressure is lowered. First puzzle solved! Glucose never enters the vapour; it just lowers the mole fraction of the water. With me?

1:52Relative lowering of vapour pressure

Now, by how much is it lowered? Compare the drop with the starting value: p A nought minus p, over p A nought. This is called the relative lowering of vapour pressure. Put in p equals x A times p A nought, cancel, and you get one minus x A, okay? So, in a solution of just two components, what is one minus x A equal to? x B, yes! The mole fraction of the solute, because x A plus x B is one. So the relative lowering of vapour pressure equals the mole fraction of the solute. That is today's key result. Got it? And x B is n B over, n A plus n B, with n B and n A the moles of solute and of solvent. This is the exact form. Right? In a dilute solution, n B is tiny compared with n A, so we drop it from the bottom, and x B becomes approximately n B over n A, with the wavy equals sign meaning approximately equal. But careful! If a question asks for the exact answer, keep n B in the denominator. Okay? And the conditions: an ideal or dilute solution, and a solute that does not dissociate or associate, like glucose or urea. Clear?

3:06Does the type of solute matter?

Now, back to our second puzzle. Urea or sucrose, the same mole fraction of each in water. Many students say sucrose lowers the vapour pressure more, because it is heavier. Is that right? No, no, careful! The relative lowering equals x B, so it depends only on the mole fraction of solute particles. Right? It does not depend on the nature of the solute. So urea and sucrose at the same mole fraction give equal vapour pressures. Second puzzle solved! With me?

3:39Raoult's law: a special case of Henry's law

One more link, worth knowing for theory questions. Henry's law, from chapter one: p equals K H times x. Here x is the mole fraction, and K H, with a small H below, is Henry's law constant. Okay? Now, which value of K H turns Henry's law into Raoult's law? p nought, exactly! Put K H equal to p nought, and you get p equals x times p nought: Raoult's law. So Raoult's law is a special case of Henry's law. Got it? And in an ideal solution, Raoult's law holds for every component, at every composition. Clear?

4:17Example 1: vapour pressure of a glucose solution

Alright, example one. Eighteen grams of glucose is dissolved in one hundred and seventy eight point two grams of water, at a temperature where p nought is twenty four millimetres of mercury. On screen, g is grams, and m m H g is millimetres of mercury. Find the vapour pressure of the solution. Pause the video, and try it first. Done? Okay, let's check. Glucose has molar mass one hundred and eighty, so n B is eighteen over one hundred and eighty: zero point one mole. The m o l on screen means moles. Water has molar mass eighteen, so n A is one hundred and seventy eight point two over eighteen: nine point nine moles. Right? Now, Raoult's law needs a mole fraction. Whose do we use here: the glucose's, or the water's? The water's, correct! Only the water makes the vapour. So x A is nine point nine over, nine point nine plus zero point one: zero point nine nine. Okay? So p is zero point nine nine times twenty four; the cross on screen means times. That gives twenty three point seven six millimetres of mercury. Clear? Quick check with today's result. Twenty four minus twenty three point seven six, over twenty four, is zero point zero one. Exactly x B, the mole fraction of glucose. With me?

5:44Example 2: molar mass of a solute in benzene

Now a harder one, example two. Five grams of a non-volatile solute is dissolved in one hundred grams of benzene. The vapour pressure of this solution is ninety eight point five percent of that of pure benzene. Find the molar mass of the solute: first with the dilute solution approximation, then exactly. Pause and try. Hint: find the relative lowering first. Done? Okay, let's check. Ninety eight point five percent, so the relative lowering is zero point nine eight five, right? No, no, wait! Zero point nine eight five is p over p nought, the part that is left. The relative lowering is the part that is lost: one minus zero point nine eight five, that is zero point zero one five. Clear? Next, benzene, C six H six, has molar mass seventy eight; the capital M on screen is the molar mass. So n A is one hundred over seventy eight: one point two eight two moles. Okay? The dilute way first: n B over n A equals zero point zero one five, so n B is zero point zero one nine two mole. Then M is five over zero point zero one nine two: about two hundred and sixty. The g mol to the power minus one means grams per mole. Right? Now the exact way: n B over, n A plus n B, equals zero point zero one five. Multiply across and collect the n B terms: zero point nine eight five times n B equals zero point zero one five times n A, with me? So n B is zero point zero one nine five mole, and M is five over zero point zero one nine five: about two hundred and fifty six grams per mole. Compare with two hundred and sixty: close, because the solution is dilute. But when a question says exact, use the exact form.

7:40Quick revision: relative lowering = solute mole fraction

Got it? So, quick revision. With a non-volatile solute, p equals x A times p A nought, so the vapour pressure falls. The relative lowering equals x B, about n B over n A when dilute. And only the number of solute particles matters, not their nature. Now go and try the practice problems on this page. All the best!