Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Dissolve acetic acid in benzene: one mole of acid for every kilogram of benzene. Benzene's freezing-point constant is five point one two, so the freezing point should drop by five point one two kelvin, right? But the experiment shows a drop of only two point five six. Exactly half! So where did half the particles go? Hold on to that question. Today: Set two of our mixed problems on solutions, the ideas it tests, and two solved problems. The second one is this exact puzzle. Questions like these come up often in JEE Main. So, let's begin.

0:42Set 2: eight problems, four ideas

So, Set two has eight solved problems, built on four ideas. First, Raoult's law for two volatile liquids: two problems. Okay? Second, Henry's law, for a gas dissolved in water: one problem. Right? Third, the freezing point: two problems. Clear? And fourth, the van 't Hoff factor, the small italic i: three problems. Those last two cover five of the eight, so today we go deep into them. With me?

1:13Freezing point and molar mass

Now, the freezing point. Look at the graph: vapour pressure p going up, temperature T going across. The green curve, the solution, lies below the dark blue curve of the pure solvent, so it meets the grey curve of the solid at a lower temperature. So T f is below T f nought; the little circle means pure solvent. That gap is delta T f, the little triangle meaning change. The right side, with the dashed line at one atmosphere, is about boiling; today we need the left. And delta T f equals K f, the freezing-point constant of the solvent, times m, the molality. This holds for a dilute solution of a non-volatile solute, with no association or dissociation. Okay? And molality m is moles of solute over kilograms of solvent. Moles of solute: w sub B, its mass in grams, over M sub B, its molar mass. Kilograms of solvent: w sub A, its mass in grams, over one thousand. So B is the solute, and A is the solvent. Right? Now think. In the lab, you weigh the solute and the solvent, and you measure delta T f. You know K f for the solvent. So what can you find? The molar mass of the solute, yes! Put this m into delta T f equals K f m, and rearrange. M sub B equals one thousand times K f times w sub B, over delta T f times w sub A, with both masses in grams. The one thousand turns the solvent's grams into kilograms. Got it? Two values of K f to remember, in kelvin kilogram per mole; the minus one on mol means per mole. Water: one point eight six. Benzene: five point one two. Both of today's examples use benzene. Clear?

3:00The van 't Hoff factor i

Alright, the next idea: the van 't Hoff factor. Some solutes split into ions, and some molecules join together. Either way, the number of particles changes. So the van 't Hoff factor i is the observed colligative property, over the calculated one. Okay? And every formula just gets an i. Delta T f equals i K f m. And osmotic pressure pi, the Greek letter, equals i CRT: i times molar concentration, times the gas constant, times temperature. Right? Now, dissociation. If each formula unit splits into n ions, and the degree of dissociation is alpha, the Greek letter, then i equals one plus, n minus one in brackets, times alpha. More particles, so i goes above one. With me? Next, the opposite: association. Here n molecules join into one, with degree of association alpha. Then i equals one minus, in brackets one minus one over n, times alpha. Fewer particles, so i drops below one. Clear? Now your turn. Two molecules pair up into one dimer, so n is two. What is i? One minus half alpha, exactly! One minus one over two is one half, so i equals one minus half alpha. And if every molecule pairs up, alpha is one, and i is one half. Got it? Now, the angled signs mean greater than and less than. i greater than one means dissociation. i less than one means association. And non-electrolytes like glucose and urea have i equal to one.

4:41Example 1: molecular formula of sulphur from benzene's Tf

Okay? Now, Example one, a medium problem from Set two. Twenty five point six grams of sulphur, in one thousand grams of benzene, lowers the freezing point by zero point five one two kelvin. K f is five point one two. Find the molecular formula of sulphur. Pause the video, and try it first. Done? Okay, let's check. We know both masses, delta T f, and K f. So which formula do we need? The molar mass formula, correct! So w sub B is twenty five point six grams of sulphur, and w sub A is one thousand grams of benzene. Then M sub B equals one thousand times five point one two times twenty five point six, over zero point five one two times one thousand. The cross means times. Right? The thousands cancel. Five point one two over zero point five one two is ten. Ten times twenty five point six: two hundred and fifty six grams per mole. With me? So, the answer is two hundred and fifty six, right? No, no, wait! The question asks for the molecular formula, not the molar mass. One sulphur atom has an atomic mass of thirty two. So two hundred and fifty six over thirty two: eight. Clear? So each molecule of sulphur holds eight atoms. The molecular formula is S eight, with a small eight written below, a subscript. The freezing point told us how many atoms sit in one molecule. Nice, isn't it?

6:17Example 2: percentage of acetic acid as dimers in benzene

And now, Example two, a hard one, and it is our puzzle from the start. One point zero molal acetic acid in benzene, where the small m means molal, freezes two point five six kelvin below pure benzene. K f is five point one two. Find the percentage of acetic acid present as dimers. Pause the video, and try it first. Done? Okay, let's check. First, the drop we would expect if the molecules stayed single, with i equal to one. Delta T f equals K f m: five point one two times one point zero, five point one two kelvin. Right? But the observed drop is only two point five six. So i, observed over calculated, is two point five six over five point one two: zero point five. Less than one, so association. Okay? So i is one half, which means half the acid is in dimers: fifty percent, right? No, no, careful! This is the trap. i is not the fraction in dimers. Use the formula. For dimers, i equals one minus half alpha, and that equals zero point five. Clear? So half alpha is zero point five, and the double arrow means therefore: alpha equals one. Got it? Alpha is one: complete dimerisation. One hundred percent of the acetic acid is present as dimers. Puzzle solved! The particles were never lost; every two molecules paired up into one. Half the particles, so half the drop in freezing point. With me?

7:55Quick revision: molar mass formula, i and dimers

So, quick revision. Molar mass from the freezing point: M sub B equals one thousand K f w sub B, over delta T f w sub A, masses in grams. The van 't Hoff factor i is observed over calculated: greater than one means dissociation, less than one means association. And for dimers, i equals one minus half alpha, so i of zero point five means alpha is one: complete dimerisation, not fifty percent. Now go and try the four practice problems on this page. All the best!