Colligative Properties; Lowering of Vapour Pressure

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Two beakers, same amount of water. In the first, I dissolve glucose, a fairly big sugar molecule. In the second, urea, a much smaller one. Same number of solute molecules in each. Which one lowers the vapour pressure of the water more? Most students say the bigger molecule, right? Hold that thought. Today we meet properties that do not care what the particles are, only how many there are. They are called colligative properties, and they come up often in JEE, so let's begin.

0:37What is a colligative property?

So, here is the big idea. Some properties of a solution depend only on the number of solute particles. We call them colligative properties. Okay? Compared with what? With the number of solvent particles. And not on their nature at all: big molecule or small, it does not matter. One condition: a dilute solution. Size, shape, what the molecule is made of: none of it matters, only the count. Right? Now, there are four colligative properties. One: relative lowering of vapour pressure. Two: elevation of boiling point, the boiling point goes up. Okay? Three: depression of freezing point, so the freezing point goes down. And four: osmotic pressure. This chapter takes all four, and today we start with the first. Clear?

1:29Trap: which ones are colligative?

Now, a classic trap. Is the vapour pressure of a solution itself a colligative property? No, no, careful! The vapour pressure of a solution depends on the solvent: water and benzene give completely different values. Same for the boiling point itself. So these two are not colligative. Got it? What is colligative is the change: the relative lowering of vapour pressure, and the elevation of boiling point. Osmotic pressure is colligative too. So for vapour pressure and boiling point, look for the change, not the value itself. Right?

2:04Relative lowering of vapour pressure

Now, the first one in detail. p nought, with the small circle on top, is the vapour pressure of the pure solvent. And small p is the vapour pressure of the solution, when the solute is non-volatile, meaning it does not evaporate at all. Okay? So think. Is p more than p nought, or less? Less, exactly! That is the less-than sign on screen, p less than p nought. The solute does not evaporate, and it lowers the share of the solvent, so the vapour pressure falls. The drop, p nought minus p, over p nought, is the relative lowering of vapour pressure. It equals x B, the mole fraction of the solute; the small B below is a subscript, B for solute, A for solvent. And x B is n B over, n A plus n B, with me? In a dilute solution, n B is tiny next to n A, so we drop it from the bottom. The wavy equals sign means approximately: about n B over n A, okay? And moles are mass over molar mass: small w is the mass in grams, capital M the molar mass. So n B over n A becomes w B times M A, over M B times w A, clear?

3:15The same result in terms of molality

One more form, very handy. Small m in italics is the molality: moles of solute per kilogram of solvent, the k g on screen. Okay? Then, for a dilute solution, the relative lowering is about m times M A, over one thousand, where M A is the molar mass of the solvent. Why one thousand? Because molality counts the solute per one thousand grams of solvent. Right? Quick check, both in water. Zero point one molal glucose, and zero point two molal urea; the small m after a number means molal. What is the ratio of their relative lowering of vapour pressure? One is to two, correct! Same solvent, same M A, so the relative lowering is proportional to m; the curly sign on screen means proportional to. Glucose is the bigger molecule, but urea has twice the particles: twice the effect. And our opening puzzle? Same number of particles in the same water, so the same lowering. The bigger glucose molecule gets no extra credit! Got it? And the conditions: a dilute solution, and a non-volatile solute that does not dissociate or associate, that is, it does not break into ions, and its molecules do not join together. Clear?

4:32Example 1: relative lowering for 12 g urea in water

Alright, example one. Find the relative lowering of vapour pressure of a solution of twelve grams of urea in one hundred and eighty grams of water. The g on screen means grams. Pause the video, and try it first. Done? Okay, let's check. Urea has molar mass sixty, so n B is twelve over sixty: zero point two mole. The m o l on screen means moles. Right? Water has molar mass eighteen, so n A is one hundred and eighty over eighteen: ten moles. Okay? Now, the relative lowering is x B, remember? So is it zero point two over ten, or zero point two over ten point two? Zero point two over ten point two, yes! The total moles go at the bottom: ten plus zero point two. That gives zero point zero one nine six. With me? So the relative lowering is zero point zero one nine six, with no unit: a pressure over a pressure. Zero point two over ten, zero point zero two, is only the dilute shortcut. Close, but use the exact form. Clear?

5:39Example 2: molar mass from vapour pressure at 373 K

Now a harder one, example two. The vapour pressure of a five percent, w by w, aqueous solution of a non-volatile non-electrolyte, at three hundred and seventy three kelvin, is seven hundred and forty five millimetres of mercury. w by w means by mass, K is kelvin, and m m H g is millimetres of mercury. Find the molar mass of the solute. Pause and try. Hint: what is p nought here? Done? Okay, let's check. We need p nought, but the question does not give it! So think: what is the vapour pressure of pure water at three hundred and seventy three kelvin? Seven hundred and sixty, exactly! Three hundred and seventy three kelvin is the normal boiling point of water, where the vapour pressure equals one atmosphere, the a t m on screen. That is seven hundred and sixty millimetres of mercury. Clear? So the relative lowering is seven hundred and sixty minus seven hundred and forty five, over seven hundred and sixty. Fifteen over seven hundred and sixty: zero point zero one nine seven. And that is x B, okay? Now the moles of water. Five percent by mass, so five grams of solute in one hundred grams of water, right? Oh no, careful! Five percent by mass means five grams of solute in one hundred grams of solution. So the water is ninety five grams, and n A is ninety five over eighteen: five point two eight moles. With me? Now, solve it exactly. x B equals n B over, n A plus n B, okay? Cross multiply and collect the n B terms: n B equals x B times n A, over one minus x B; the double arrow means implies. Put in the numbers: zero point one zero six mole. Right? So the molar mass is five grams over zero point one zero six mole: about forty seven grams per mole, which is the g mol to the power minus one on screen. Got it? And the dilute shortcut? n B is about x B times n A, zero point one zero four mole, giving about forty eight. Close, but the exact answer is forty seven. Okay?

7:53Quick revision: colligative properties, relative lowering

So, quick revision. Colligative properties depend only on the number of solute particles, not on their nature. The relative lowering of vapour pressure equals x B, about n B over n A, about m times M A over one thousand, for a dilute solution of a non-volatile solute. And vapour pressure itself is not colligative. Now go and try the practice problems on this page. All the best!