Overview

Transcript

0:00Introduction

Hello everyone! Okay, one small situation before we start. You are sitting in the JEE Main exam. There is a chemistry question with no options, only a box where you type a number. You work it out, and you get twenty five point eight. The question says, give the answer to the nearest integer. So what do you type: twenty five, or twenty six? Hmm, sounds easy, right? But marks get lost on exactly such small things. And look, this unit gives one or two questions in almost every JEE Main shift. So today I will show you how to use this chapter of mixed problems like a real test, and then we will solve two problems together. So, let's begin.

0:45Why this unit matters in JEE Main

So, first, why does this unit matter so much? Some Basic Concepts in Chemistry gives one or two questions in almost every JEE Main shift, okay? Almost every shift! For just one unit, that is a lot of marks. Now tell me, what kind of question is it, usually? Options to choose from, or something else? That's right, very often it is a numerical-value question! No options to pick from. You solve it completely, and you type the number yourself. So your method must be clear, and your calculation must be clean. Clear?

1:21Six question types to master

So what do these questions ask? There are six types, okay? Let's go one by one. First, moles and particles: moles, molecules, atoms, even protons and electrons. Second, percentage composition and formulae. Third, limiting reagent and yield: which reactant runs out first, and how much product you actually get. Fourth, gas volumes. Four done, two more to go. Let's go! Fifth, molarity and molality conversions. And sixth, simple titrations. Six types, that's it. And you have already learnt every one of them in Chapters One to Eight. Here they simply come mixed together.

2:06How this chapter is built

Got it? Now, where are the real exam questions? The real JEE Main questions from twenty twenty one to twenty twenty six are already at the end of Chapters One to Eight, okay? This chapter is different. It has three sets of fifteen problems, all original, written in the same style as those real questions. So you get fresh practice, on problems you have never seen before. Set one is on the mole concept, composition and formulae. Look, both our examples today come from this set, so you will see its style in a few minutes. Set two is on stoichiometry and gases: limiting reagent, yield, and gas volumes, all of it. And set three is on concentration, equivalents and titrations. So between the three sets, the whole unit is covered, from start to finish. Fair enough?

2:55Treat each set as a real test

Now the most important part: speed. Each problem in this chapter should take one to two minutes. Just one to two minutes, okay? So tell me, fifteen problems, one to two minutes each. How long should one whole set take? Fifteen into one is fifteen, and fifteen into two is thirty. So fifteen to thirty minutes for a set, exactly! Sit down with a clock, and treat every set as a small test. Alright, now the trap. You are on problem seven, you feel a little stuck, and you think, let me just check the solution once, right? You read it, you nod, and you feel, yes, I knew this! No, no, careful! That feeling is the trap. Reading a solution is not the same as solving the problem. So the rule of this chapter is clear. Attempt the whole set before reading any solution. Your own answers, on paper, right or wrong. Then compare, and learn from every mistake. And one more rule, straight from the exam. Where the question asks for the nearest integer, give the nearest integer. Practise it here exactly the way you will do it in the exam hall.

4:05Topper's trick: read the badges

Got it? Now let me share a small trick that toppers use. Every problem on the page carries a badge: easy, medium or hard. Take set one: four easy problems, eight medium ones, and three hard ones, okay? So finish the easy ones fast and clean, close to one minute each. The arrow on screen means, so: the time you save there becomes extra time for the hard ones. That is how you finish the whole set in time. Simple and practical.

4:35Example 1: percentage of sulphur in hypo

Now, let's practise exactly the way this chapter says. This one is from set one, and it is an easy one. Find the percentage of sulphur in hypo. Its formula is N a two, S two, O three, dot five H two O, and the dot five H two O on screen means five molecules of water of crystallisation. The molar mass is given: two hundred and forty eight. Give the answer to the nearest integer. Pause the video and try it. Your two minutes start now. Done? Okay, let's check. First, how much sulphur is in one mole of hypo? Look at the formula: S two, so two sulphur atoms, okay? Each sulphur atom is thirty two, so sixty four grams of sulphur in one mole. Many students take only thirty two here. So always count the atoms first, remember that. Now divide by the molar mass, and multiply by one hundred. Sixty four by two hundred and forty eight, into one hundred. Twenty five point eight percent. The box wants an integer, so drop the point eight and type twenty five, right? No, no, careful! This is our puzzle from the start. Nearest integer means you round off; you do not just cut the decimal. Twenty five point eight is closer to twenty six than to twenty five. So the answer is twenty six. Puzzle solved! Type twenty six, and those marks are yours.

6:06Example 2: mass of CO2 in a CO and CO2 mixture

Now a hard one, also from set one, okay? One point eight grams of a mixture of C O and C O two contains three point zero one one into ten to the power twenty two molecules. Find the mass of C O two in it, as x into ten to the power minus one gram. Pause the video and try it. Hint: start from the molecules. Done? Okay, let's check. Molecules to moles: divide by the Avogadro constant, six point zero two two into ten to the power twenty three. That gives zero point zero five mole, of both gases together. With me? Now, how do we split this zero point zero five mole between the two gases? Simple: give the unknown a name! Let C O two be b mole. Then C O is whatever is left: zero point zero five minus b mole, okay? Now use the mass. C O is twenty eight grams per mole, and C O two is forty four. So twenty eight into zero point zero five minus b, plus forty four b, equals one point eight. Twenty eight into zero point zero five is one point four, and forty four b minus twenty eight b is sixteen b. So sixteen b is zero point four, and b is zero point zero two five mole. So the mass of C O two is zero point zero two five into forty four. One point one gram. Now the box is waiting. One point one gram, so you type one point one, right? No, no, wait! Read the question once more. The question wants the mass as x into ten to the power minus one gram. One point one gram is eleven into ten to the power minus one gram. So x is eleven, and you type eleven. Look, in numerical-value questions, always read the answer format. That one line decides your marks. Got it?

8:07Quick revision: question types, timing and rounding

So, quick revision. Six question types, and one to two minutes a problem. Attempt a whole set before reading any solution. And round to the nearest integer, and always read the answer format. Now open set one, start your clock, and begin. You can do this, I promise. All the best!