Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take five hundred grams of water with some glucose dissolved in it, and cool it to a little over two degrees below zero. Some ice forms. But not all the water freezes; part of it just stays liquid. Why does the freezing stop partway? Hold on to that question. Today: how this chapter of mixed problems works, the five kinds of question on solutions that come up often in JEE Main, the constants to use, and this exact puzzle, solved. So, let's begin.

0:35How this chapter works

So, let's open this chapter. Recent JEE Main questions on solutions are mostly short numericals. Read, spot the idea, calculate, done. Okay? And where are the real past questions, from twenty twenty one to twenty twenty six? They sit at the end of Chapters one to five. Right? This chapter is different. Its problems are original, written in the same style, so you can practise the pattern. With me? There are three sets of eight solved problems, and each set ends with four practice problems. Set one opens with the easy ones. Clear?

1:14Five kinds of question

Now, the five kinds of question. First: interconverting concentrations. Molality, small m, is n sub B, the moles of solute, over w sub A, the mass of the solvent in kilograms. Molarity, capital M, is n sub B over V, the volume of the solution in litres. So B is the solute and A is the solvent, okay? Second: Henry's law, for gases. p equals K sub H times x: the partial pressure of the gas equals Henry's law constant, K sub H, times x, the mole fraction of the gas in the solution. Now tell me. If a gas has a larger K sub H, is it more soluble, or less? Less soluble! Rearrange it: x equals p over K sub H, so at the same pressure, a larger K sub H gives a smaller mole fraction, and less gas dissolves. Students often flip this one, so careful. Right? Third: Raoult's law for two volatile liquids, forming an ideal solution. p sub A equals x sub A times p sub A nought. The little circle on top means pure, so p sub A nought is the vapour pressure of pure liquid A, and x sub A is its mole fraction in the liquid. Got it? Fourth: freezing points and osmotic pressures. Delta T f, with the little triangle for change, is the drop in freezing point, and it equals K f times m, the molality. And osmotic pressure pi, the Greek letter, equals CRT: molar concentration, times the gas constant, times temperature. These hold for a dilute solution of a non-volatile solute, with no association or dissociation. Clear? And fifth: the van 't Hoff factor i. When the solute splits into ions, or its molecules join together, the number of particles changes. So whenever the solute dissociates or associates, multiply by i: delta T f equals i times K f times m, and pi equals i CRT, okay? Forget the i, and the answer is wrong. With me?

3:19Constants to use, unless a question gives its own

Alright, the constants. The problems in this chapter use these values, unless a question says otherwise. K f of water, the freezing point constant: one point eight six kelvin kilogram per mole. The minus one on mol means per mole. Okay? K b of water, the boiling point constant: zero point five two, in the same units, kelvin kilogram per mole. Right? Now a quick one for you. Same solute, same molality, in water. Which moves more: the freezing point, or the boiling point? The freezing point, yes! K f is bigger than K b; the angled sign means greater than. So at the same molality m, delta T f is bigger than delta T b. So never swap the two constants. Got it? Then R, the gas constant: zero point zero eight three litre bar per kelvin per mole. This is the value for pi equals CRT, with pi in bar and C in moles per litre. Clear? And the trap that costs marks here: T must be in kelvin, never in Celsius, and the units of R must match those of pi and V, so check them before you multiply. With me?

4:34Example 1: freezing point of ethylene glycol in water

Now, Example one, an easy problem from Set one, okay? Find the molality of a solution of six point two grams of ethylene glycol in two hundred and fifty grams of water, and its freezing point. Pause the video, and try it first. Done? Okay, let's check. Ethylene glycol has a molar mass of sixty two grams per mole. So n, the moles of solute, is six point two over sixty two: zero point one mole. Right? Now molality. Divide zero point one mole by which mass? The whole solution, two hundred and fifty six point two grams? Or just the water, two hundred and fifty grams? Just the water, correct! Molality uses the mass of the solvent, not of the solution. That is a classic trap. So m equals zero point one mole over zero point two five zero kilogram: zero point four moles per kilogram. Clear? Next, the freezing point. Ethylene glycol does not split into ions, so i is one. Delta T f equals K f times m: one point eight six times zero point four, zero point seven four four kelvin. With me? Water freezes at zero degrees Celsius; the small circle is the degree sign. So the solution freezes at zero minus zero point seven four four: minus zero point seven four four degrees Celsius. Below zero, and that minus sign matters. Got it?

6:03Example 2: how much ice separates from glucose solution

Now, our puzzle from the start, as a medium problem from Set two. Five hundred grams of water containing zero point four mole of glucose is cooled to minus two point two three two degrees Celsius. How much ice separates? Pause the video, and try it first. Done? Okay, let's check. First, tell me this. When ice forms, what happens to the glucose? Does it freeze out along with the water? No! The ice is pure water. All zero point four mole of glucose stays in the liquid. So as ice forms, the same glucose sits in less and less water, the liquid gets more concentrated, and its freezing point drops further. That is why the freezing stops partway. Puzzle solved! Right? Freezing stops when the liquid's own freezing point reaches minus two point two three two, so delta T f is two point two three two. Glucose does not split into ions, so i is one. For the liquid left, m equals delta T f over K f: two point two three two over one point eight six, one point two moles per kilogram. Okay? Now, zero point four mole of glucose, at a molality of one point two. How much water holds it? Zero point four over one point two. The wavy equals sign means approximately: about zero point three three three kilogram, so about three hundred and thirty three grams. With me? So the ice is three hundred and thirty three grams, right? No, no, careful! Three hundred and thirty three grams is the water that is still liquid. The ice is the rest: five hundred minus three hundred and thirty three point three, about one hundred and sixty seven grams.

7:50Quick revision: 5 question types and default constants

Clear? So, quick revision. Five kinds of question: concentrations, Henry's law, Raoult's law, delta T f and pi, and the van 't Hoff factor i. Unless a question says otherwise: K f is one point eight six, K b is zero point five two, and R is zero point zero eight three. Molality uses kilograms of solvent, and on freezing, only the solvent separates. Now open Set one, try each problem before reading its solution, then do the practice problems. All the best!