Part C: Advanced-Level Problems

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Two beakers, each with fifty grams of water. In the first, zero point zero one mole of sugar; in the second, zero point zero two mole. I seal both inside one box. Nobody touches them, but slowly, water moves from one beaker to the other! Which way does it go? And how much water ends up in each? Hold that thought; our last example cracks it. This is Part C of the solutions problem bank: advanced problems, JEE pattern. So, let's begin.

0:38Idea 1: solution and pure water in a closed container

Idea one: a simpler box first. Zero point one mole of a non-volatile solute in one kilogram of water, in a closed container, beside a beaker of pure water at the same temperature. Non-volatile means the solute itself never goes into the vapour. Okay? Now tell me: which one has the lower vapour pressure? The solution, yes, exactly! That is Raoult's law: a non-volatile solute lowers the solvent's vapour pressure. Right? So water evaporates from the pure water and condenses into the solution; the arrow means goes to. With me? Now, when does it stop? At some point the two must balance, right? No, no, wait! As long as any pure water is left, it has a higher vapour pressure than the solution. So the transfer goes on until the pure water has all gone. Clear?

1:33Idea 2: find i from the freezing point

Idea two: i, the van 't Hoff factor, from a freezing point. Take a zero point zero five molal solution of a non-electrolyte in water; the small m means molal. It freezes at minus zero point zero nine three degrees Celsius; the little circle means degrees. Non-electrolytes do not split into ions, so i is one. Okay? Delta T f, the fall in freezing point, equals i times K f times m; delta, the little triangle, means change, and the small f stands for freezing. So K f, the freezing point constant of water, is zero point zero nine three over one times zero point zero five: one point eight six. Right? Now a complex. Inside the square brackets, the coordination sphere: cobalt, four N H three groups, grouped in round brackets, and two chlorides; the small numbers are subscripts. One more chloride sits outside. Its zero point zero five molal solution freezes at minus zero point one eight six degrees Celsius. With me? So, three chlorines in the formula. If all come off, that makes four ions, so i is four, right? No, no, careful! Let the freezing point decide. i equals delta T f over K f times m: zero point one eight six over one point eight six times zero point zero five. Exactly two. Got it? Two ions: the complex ion, with a plus charge on top, and one chloride ion, C l minus. So only one chloride is outside the coordination sphere. One freezing point told us the structure!

3:13Idea 3: two solutes in the same water

Clear? Idea three: two solutes in the same water. Zero point one molal K three P O four, potassium phosphate, eighty percent dissociated; the small three and four are subscripts. And zero point two molal of another solute that dimerises, eighty percent; dimerise means two molecules join into one. Find the freezing point. Okay? Potassium phosphate gives three K plus ions and one P O four three minus ion; the arrow means gives, and the signs on top are charges. Three extra particles, each counted eighty percent: i is one plus three times zero point eight, three point four. With me? Now the dimer. Eighty percent of it goes into dimers, so i is one minus zero point eight, zero point two, right? Oh no, careful! Two molecules make one dimer, so you lose only half as many particles. i equals one minus zero point eight over two: zero point six. Less than one: association cuts the particles. Got it? Now, colligative properties count all the particles. So add i times m: zero point one times three point four, plus zero point two times zero point six. Zero point three four plus zero point one two: zero point four six. Okay? Delta T f is one point eight six, the K f we just found, times zero point four six: zero point eight five six. So the water freezes at about minus zero point eight six degrees Celsius; the wavy equals sign means approximately. Clear?

4:53Example 1: pure vapour pressures from vapour composition

Alright, example one. Liquids A and B form an ideal solution. At three hundred kelvin, a solution with x A equal to zero point two five, the mole fraction of A in the liquid, has a total pressure of zero point six atmosphere. And y A, the mole fraction of A in the vapour, is zero point five. Find p A nought and p B nought, the vapour pressures of the pure liquids; the little circle on top means pure. Pause and try it first. Done? Okay, let's check. Raoult's law says p equals x times p nought. So p A nought is zero point six over zero point two five, two point four atmosphere, right? No, no, careful! Zero point six is capital P, the total pressure of both liquids. Raoult's law needs p A, the partial pressure of A alone. The vapour gives it: p A equals y A times P, zero point five times zero point six, zero point three atmosphere. Okay? And p B is the rest: zero point six minus zero point three, also zero point three atmosphere. Right? Now Raoult's law, valid because the solution is ideal: p A equals x A times p A nought. Zero point three equals zero point two five times p A nought; the double arrow means so: p A nought is one point two atmosphere. With me? For B, x B is one minus zero point two five, zero point seven five. Zero point three equals zero point seven five times p B nought, so p B nought is zero point four atmosphere. Perfect!

6:41Example 2: two sugar solutions sharing water, sealed box

Now, back to our sealed box. Fifty grams of water with zero point zero one mole of sugar, and fifty grams with zero point zero two mole, sealed in together. What is the mass of water in each beaker at equilibrium? Pause and try. Hint: what is equal at the end? Done? Okay, let's check. Sugar is non-volatile, and beaker two has more of it in the same water, so its vapour pressure is lower. So water moves from beaker one into beaker two, like idea one. Right? So, does beaker one dry out, like the pure water did? No! This time both beakers hold solutions. As beaker one loses water, it gets more concentrated, and its vapour pressure falls. So the transfer stops when the vapour pressures are equal, that is, when the molalities are equal. Got it? Now the topper's trick: pool everything. Total sugar, zero point zero three mole. Total water, one hundred grams, zero point one kilogram. So the common molality is zero point zero three over zero point one: zero point three molal. With me? Beaker one has zero point zero one mole, so its water is zero point zero one over zero point three: zero point zero three three three kilogram, thirty three point three grams. Okay? And beaker two: zero point zero two over zero point three: zero point zero six six seven kilogram, sixty six point seven grams. Puzzle solved! Twice the sugar, twice the water. Very good!

8:18Quick revision: sealed box, i from Tf, vapour data first

So, quick revision. In a sealed box, water goes to the lower vapour pressure. Pure water empties completely; two solutions stop at equal molality. i is delta T f over K f times m; dissociation raises i, dimerisation lowers it; for mixed solutes, add i times m. With vapour data, p A is y A times P first, then Raoult's law. Now try the three practice problems on this page. All the best!