Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Two beakers of sugar water sit side by side in a sealed box. Each has fifty grams of water, but the second has twice as much sugar. No pipe between them, and nobody touches them. Leave them for a long time. Will anything change? Believe it or not, water quietly moves from one beaker to the other! Which way, and how much? That is a JEE Advanced level problem from this chapter, the unit problem bank for Solutions. Today: how to use this bank, and this exact puzzle, solved. So, let's begin.

0:39How to use this chapter

So, this chapter has three parts. Part A has six solved problems that mix the whole unit, then four more for practice. Okay? The instruction: attempt Part A as a thirty minute test, after Chapters one to six. Sit with a clock, and open the solutions only at the end. Right? Then Part B: concept checks, the kind often asked as statements or assertions. With me? And Part C: longer problems, at JEE Advanced level. Our sugar water puzzle lives there. Clear?

1:14Part A: name the law before you calculate

Now, the real skill in a mixed test: before you calculate, name the law. Here is a Part A problem: C O two, carbon dioxide, dissolved in water at a partial pressure of two point five bar. How much dissolves? Which law is that? Henry's law, yes! The double arrow means this leads to. Small p equals K H times x: p is the partial pressure of the gas, x its mole fraction in the water, and K H, the K with a small H, is Henry's law constant. From Chapter one. Okay? Next: an ideal solution of two liquids, A and B, with its vapour. Find capital P, the total pressure, and y A, the mole fraction of A in the vapour, the y with a small A below it. Which law now? Raoult's law, correct! p A equals x A times p A nought: the mole fraction of A in the liquid, times the vapour pressure of pure A, okay? The small circle, read as nought, means pure. Chapter two, and this exact problem is our Example one. Got it? One more. The small m on screen means molal, moles per kilogram of solvent. One molal glucose, potassium chloride K C l, potassium sulphate K two S O four, or aluminium chloride A l C l three. Which has the highest boiling point? Same molality, so count the particles with the van 't Hoff factor i. Glucose stays as one particle: i is one. Taking each salt as fully dissociated, K C l gives two ions, K two S O four three, and A l C l three four. More particles, a bigger rise: so A l C l three wins. The van 't Hoff factor from Chapter five, used for a boiling point. That is what mixed means.

3:11Part B: read every word

Clear? Now for Part B, here is an assertion and reason, okay? Assertion: molality is preferred to molarity in colligative property measurements. With me? Reason: molality does not change with temperature. Two checks, always. Is each statement true? And does the reason explain the assertion? Both true, and yes, the reason explains it! Molality uses the mass of the solvent, which does not change with temperature; molarity uses volume, which does. Right? Next, a true or false. An ideal solution has delta mix S equal to zero. Delta, the little triangle, means a change; mix says it is the change on mixing; and S is entropy. Ideal, nothing special happens, so zero, right? True or false? If you said true, careful! It is false. For an ideal solution, delta mix H, the enthalpy change, is zero, and delta mix V, the volume change, is zero. But delta mix S is greater than zero; the pointed sign means greater than. One small symbol flips the answer, so read every word. Okay?

4:23Example 1: total pressure and mole fraction of A in vapour

Now, Example one, from Part A, okay? An ideal solution of one mole of A, with p nought two hundred and fifty, and four moles of B, with p nought fifty, in m m H g, millimetres of mercury, is in equilibrium with its vapour. Find capital P, the total pressure, and y A, its mole fraction in the vapour. Pause the video, and try it first. Done? Okay, let's check. One mole of A out of five in total: x A is zero point two. And x B is four over five, zero point eight. Right? Now Raoult's law for each liquid, and add: capital P equals x A times p A nought, plus x B times p B nought. That is zero point two times two hundred and fifty, plus zero point eight times fifty. With me? Fifty plus forty: ninety millimetres of mercury. And ninety lies between fifty and two hundred and fifty, as it must. Perfect! Now for y A, okay? The liquid is zero point two A, so the vapour is zero point two A as well, right? No, no, careful! That is the x and y trap. y A is the share of A in the pressure: small p A over capital P, that is fifty over ninety, zero point five five six. Much richer in A, the more volatile liquid. Got it?

5:54Example 2: two sugar solutions sharing water in a sealed box

And now, our puzzle, exactly as it appears in Part C, okay? Two beakers: fifty grams of water with zero point zero one mole of sugar, and fifty grams of water with zero point zero two mole of sugar, placed together in a sealed box. What are the masses of water in each, at equilibrium? Pause the video and try it. Hint: what must be equal at the end? Done? Okay, let's check. First, the molalities. Fifty grams is zero point zero five zero kilogram. Beaker one: zero point zero one over zero point zero five zero, that is zero point two molal. Beaker two: zero point four molal. Right? Sugar is non-volatile and does not dissociate, so more sugar per kilogram means a lower vapour pressure. So tell me: which beaker loses water? Beaker one, yes! It is more dilute, so its vapour pressure is higher. Water evaporates from beaker one and condenses into beaker two; the arrow shows the direction. Clear? Now, in another Part C problem, a beaker of pure water sits next to a solution, and all the pure water goes across. So here too, beaker one dries up completely, right? No, no, wait! Beaker one still has its sugar. As it loses water it gets stronger, and its vapour pressure falls; beaker two gets weaker, and its vapour pressure rises. So the transfer stops when the vapour pressures are equal, and for dilute solutions of the same solvent, that means equal molalities. With me? Now a topper's trick. Water is only moved, never lost: still one hundred grams in all, zero point one kilogram, with zero point zero three mole of sugar. So the common molality, small m, is zero point zero three over zero point one zero zero: zero point three molal. Okay? So beaker one holds zero point zero one over zero point three: zero point zero three three three kilogram, thirty three point three grams. Right? And beaker two: zero point zero two over zero point three: zero point zero six six seven kilogram, sixty six point seven grams. Together, one hundred. Perfect! Water crossed the box with no pipe at all. Puzzle solved! Got it?

8:22Quick revision: name the law, check the reason, sealed box

So, quick revision. Part A: a thirty minute test after Chapters one to six, and name the law before you calculate. Part B: check each statement, then check whether the reason explains the assertion. And in a sealed box, water goes from the dilute solution to the stronger one, until the molalities are equal. Now open Part A, start your clock, and begin. All the best!