Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a one-minute question, in the JEE Main style. Thiourea has nitrogen and sulphur in the same molecule. Fuse it with excess sodium, and test the extract. Nitrogen and sulphur together give a blood red colour, so we will see blood red, right? Hold that thought for a few minutes. This chapter gives you three sets of mixed problems, so let us see how to use them. Let's begin.

0:28How this chapter works

So, first, why this chapter matters. This unit usually gives one or two questions in a JEE Main shift. Okay? They come in three types: a question on purification or chromatography, one on Lassaigne's test, like our thiourea, and a quick numerical. Right? Now the three sets. Set one covers purification and chromatography, Set two, detection of the elements, and Set three, estimation and formulae. Got it? And the real JEE Main questions, from twenty twenty one to twenty twenty six, are at the end of Chapters one to five. These problems are original, in the same style, and each should take one to two minutes. With me?

1:09Tool for Set 1: match the method to the property

Now, the tool for Set one: match the method to the property. Camphor contains a little sand. Which method purifies it? Sublimation, yes, exactly! Camphor sublimes: on heating it passes straight from solid into vapour, and the sand stays behind. So a sublimable solid with a non-sublimable impurity means sublimation. Right? Next, liquids. On screen, b p is short for boiling point. When the boiling points are far apart, simple distillation works. When they are close, we need fractional distillation. Clear? Now chromatography. R f, capital R with a small f written low, is the distance of the spot over the distance of the solvent front; the long line is a fraction bar. Measure both from the base line, not from the edge of the plate. Okay? Now a statement from Set one. In column chromatography on silica, the most polar component is eluted first. True or false? It is false, and many students fall for it! Polar compounds are strongly adsorbed on silica, so they move slowly. The double arrows mean so: more polar, held more, lower R f, eluted later. Got it?

2:27Tool for Set 2: Lassaigne's test

Now Set two, detection of the elements. The chart on the right shows the tests on the sodium fusion extract. Fused with sodium, N a, the covalent nitrogen, sulphur and halogens become ionic sodium salts; X stands for any halogen, and the arrow means become. Okay? The first box says nitrogen gives Prussian blue. But nitrogen is caught only as cyanide or thiocyanate, and both need carbon. So hydroxylamine, N H two O H, with no carbon at all, gives no nitrogen test. Right? Now nitrogen and sulphur together, with only a little sodium. They stay together as sodium thiocyanate, and iron three chloride turns it blood red, as the third box shows. Clear? So, back to our thiourea, fused with excess sodium. Both nitrogen and sulphur are there, so blood red, right? No, no, careful, that is the trap! Excess sodium breaks the thiocyanate: sodium thiocyanate plus two sodium gives sodium cyanide plus sodium sulphide. So Prussian blue and the sulphur tests of the second box, black lead sulphide and violet, are positive, but not blood red. And that settles our opening question. Got it? Now the halogens. First boil with nitric acid, HNO three, to expel cyanide and sulphide, which would also precipitate with silver nitrate. Then add silver nitrate, A g N O three, and read the colour from the fourth box. The bottom box, phosphorus, gives a yellow precipitate with ammonium molybdate. With me?

4:04Tool for Set 3: the estimation formulae

Now Set three, the numericals. From combustion, per cent C equals twelve over forty four, times m C O two over w, times one hundred; that is the per cent sign, and the cross means times. Small m, with C O two written low, is the mass of carbon dioxide, and w is the mass of the compound. Hydrogen is the same, with two over eighteen for water. Okay? Carius works the same way. Chlorine is thirty five point five over one hundred and forty three point five of the silver chloride, A g C l. And sulphur is thirty two over two hundred and thirty three of the barium sulphate, B a S O four. Right? For nitrogen, Dumas' method collects nitrogen gas. Per cent N equals twenty eight over twenty two thousand four hundred, times V naught in millilitres over w, times one hundred. V naught, V with a small zero, is the volume at STP, standard temperature and pressure. Clear? Kjeldahl's method catches the nitrogen as ammonia, N H three. Per cent N equals one point four times the milliequivalents of acid used by the ammonia, over w in grams; m e q means milliequivalents. With me? Now, two warnings. For sulphuric acid, H two S O four, milliequivalents are two times millimoles, m m o l on screen, because it has two acidic hydrogens. And Kjeldahl's method fails for nitrogen in nitro and azo groups, or in a ring. Got it?

5:32Example 1: spot distance for Rf 0.65 and an 8 cm front

Now Example one, an easy one from Set one. On the right is a developed TLC plate, with the base line, the solvent front, and two spots; the arrows marked d show the distances, and spot B, the more polar one, has the lower R f. Our compound has an R f of zero point six five. How far does it move when the solvent front moves eight centimetres? Pause the video and try it first. Done? Okay, let's check. Some students take eight over zero point six five, which is about twelve point three centimetres; the wavy equals sign means about. Can that be right? No, no, wait, the spot can never travel farther than the solvent front, and the front moved only eight centimetres. So R f is always at most one; the sign on screen means less than or equal to. Clear? So we multiply instead. The distance of the spot equals R f times the distance of the solvent front. Okay? That is zero point six five times eight, which gives five point two centimetres. And that is less than eight, as it must be. Right?

6:45Example 2: Kjeldahl back titration with H2SO4, percent N

Now Example two, a medium problem from Set three. The ammonia from zero point two grams of a compound goes into fifty millilitres of zero point one molar sulphuric acid. The excess acid needs sixteen millilitres of zero point two five molar N a O H, sodium hydroxide. Find the per cent of nitrogen. Here g is grams, m L is millilitres, and capital M is molar. Pause the video and try it first. Done? Okay, let's check. Fifty times zero point one is five, so the acid taken is five milliequivalents, right? Oh no, careful, many students slip right here! Five is only the millimoles, and sulphuric acid has two acidic hydrogens. So the acid taken is ten milliequivalents. Okay? Sodium hydroxide used is sixteen times zero point two five, which gives four milliequivalents. So the ammonia used ten minus four, that is, six milliequivalents. Right? Now the formula: one point four times six, over zero point two, gives forty two per cent. So always double for sulphuric acid. Got it?

7:56Quick revision: method choice, thiourea trap, meq of H2SO4

So, quick revision. Set one: match the method to the property; on silica, more polar means lower R f. Set two: nitrogen needs carbon, nitrogen and sulphur with little sodium give blood red, and excess sodium separates them. Set three: the right fraction for each element, and for sulphuric acid, milliequivalents are two times millimoles. Now try the three practice problems at the end of each set. All the best!