Okay everyone we will start with the discussion of the two questions that we had taken as

homework last time. Okay, so the first question was, calculate the volume of concentrated

sulphuric acid whose density is 1.84 gram per millilitre and contains 98.0% H2SO4 by weight,

that would contain 40.0 grams pure H2SO4. How many of you have calculated this? How

much answer are you getting? Approximately 22.2 millilitre. Okay, let’s see how this is

done. Let’s discuss that. What should we do here, we will assume here that they have

asked us the volume of concentrated acid. Say for question one, this is the first question

and this is second question. For the first one we will assume that, let the volume of

concentrated acid be equal to what V millilitre. What will be the advantage of taking this in

millilitre? The advantage will be, what have we taken the density of this, 1.84, see always

remember one thing, whenever you are given density, it relates to two things. Mass of

solution and….. You have been given 98.0% mass but mass of solution is not given. So

therefore what is the way to go about this question. You have to simply assume some

millilitre as you want volume. You cannot assume as 1 litre. You will have to assume it as V

millilitre. As you will take in V millilitre you will have to say that this implies the mass of

solution is equal to what, 1.84 gram per millilitre into what, V millilitre. And millilitre here,

millilitre-millilitre we cancel, so what is the mass of solution we get. 1.84 gram, okay, here

what is the weight of solute, weight of solute is going to be 98% mass of solution, so 98

divided by 100 into what, 1.84, okay here we forgot the V, so it will be V grams. Okay. So

here how much weight of solute is there, 98%, how much should it be, 40.0 gram, since this

is the solute I wanted. From the original solution what I should have done. From the original

solution I have to tell that volume in which what is present, 40 gm of H2SO4. Okay, so what

did I do, I took V as volume and to that I multiplied density and what did I find out, mass of

solution and what is 98.0% of this, mass pure of H2SO4. So what did I want of this 40.0

grams, so what did I do of this, I equalled it to this. So again you may solve and you will get

that V is equal to what, 22.2 millilitre approximately, you will get this.

Now for the second question, the density of a 3 molar solution of NaCl is 1.25 gram per

millilitre. Calculate the molality of the solution. How many of you have done this, what

answer did you get. 2.79, okay this is the question from the NCRT book, if you had opened

that book you would have got the solution, you would have seen the full answer there.

Anyway, now we will see how should we tackle this question, for question number two, we

have to obtain molality. See first of all take the formula of molality, what should we find to

get molality. Number of moles of solute divided by the mass of solution. You will get into a

problem, mass of not solution, it is solvent. Yes, molality, that is why I tell you to do with

concentration. Okay, so we require these two things. Okay, do you know the mass of

solvent, no, do you know the number of moles of solute, no, but you know morality and

what else do you know. Density. Okay, I know I have taken the solute of what, of NaCl, and

it is aqua solution, okay, if nothing is mentioned what you will assume it to be. Aqua

solution, okay. Even if you do not assume it to be aqua solution, then too it is okay. Now

you will see what I will do. From this, let volume of solution be equal to 1 litre. What will be

the advantage of this, morality, yes right, what will the number of moles be morality into

volume. See what will morality be, number of moles divided by the volume. So now what it

will become, morality into volume. So what was the morality – 3 moles per litre, what is

molar, mole per litre into what, please do it with unit, I am again reminding you again and

again, and throughout the course of our calculations I will always remind you, do the

calculations with units wherever possible. Right. Okay, here 3 moles have come, it is come

from where, from NaCl, so what is going to be the weight of solute, 3 moles times 58.5

grams per mole , again from here what will we cancel, the moles and what will we get the

mass of solute in grams, what did we get. Okay, what we will take this to be approximately

how much. 175 grams. So therefore you can find out the weight of solvent, how much will

it become. To remove the weight of solvent we will require the weight of solution. We

know the density, so to 1.25 gram per millilitre how much millilitre is there here, 1000, how

much will this become after multiplication, 1250, so you subtract 1250 how will you

subtract. How much did you get, 1075 grams, in kilogram how much will this become,

1.075, so now you do one thing, you got both the things, so now what can you remove,

molality, what will it be equal to 3 divided by what should be written, into 1000. See what

can you cancel with this 1000, cancel it with 25, so what will you get here 40 and here what

will you get 43. Okay, so what will this be, 160 divided by 43, you solve this you will get it as,

sorry, how much, 120, I am sorry, I mistook this to be 4, so 3 into 40 will be 120, 120 divided

by 43 will be approximately 2.79. That is correct. So all of you have done both these

questions. Excellent.

Now, we will do some more calculations today. Let’s take one more question based on the

calculation of molality and molarity, let’s see. Yes. The question is a 49 gram, a 49 % mass

by mass solution of H2SO4 in water has a density of 1.2 gram per millilitre. Calculate the

molarity and the molality of the solution. Okay, the formula that was derived to you, you

can do it with that also. You will get the right answer, okay. But please try to do just like the

first principle as we had done it the first time. It is as if assuming either the volume of

solution to be 1 litre or the mass of solution to be what, 100 grams. Please take any one

assumption from these two and solve it. Okay, do it.

How should we do this. I have already told you the technique of this and we have got the

formula of this. You can do it as per the formula, what was it 10 row x by m naught. Then

you will get the molarity immediately. Okay, you got 6, you can do it this way too or else do

it, okay, you assume, I will show both the approaches again to you all. Let volume of

solution to be equal to what, 1 litre, taking 1 litre what will you get the density to be 1.2

gram per millilitre. What will be the weight of solution – 1200 grams. What will be the 49%

of this? Weight of solute what will it become – 49 by 100 into 1200 grams. Strike off the

zeros, you will get 49 into 12 grams. Okay, a lot of you will calculate this as, I am again

telling that apply distributive law, or like in lay man language we call it as the Pathani

method. Okay, basically what is 12 into 49, so what you will write it as 12 into 50 minus 1,

12 into 50 – 600 minus 12 is what 588, so you can directly mark this as what, 588 grams.

Okay, now what will you do after this, what will you solve after this, for weight of solvent.

What will be the weight of solvent – 1200 minus 588 is 600 and…