Test Papers – ICSE – Class – X
Chemistry 08,Stoichiometric Calculation Using Strength of Solutions
Okay, let’s begin today’s discussion. First of all, we have to discuss these questions which
yesterday I had given as homework. I will show you how to tackle these two questions. In
the last lecture we had discussed neutralization of HCl and NaOH. So what we will do, for
first question, we will mark the reaction HCl plus NaOH gives you what, NaCl plus water. Is
this reaction balanced? Yes, okay, now we will do one thing, you see here, 2.5ml of solution
of HCl, we will mark it here, it is neutralizing what – 4.5 ml of NaOH solution, additionally
given is the molarity of NaCl is how much – 3 molars, so we have to find the molarity of this.
So, we will take it as x molars. Okay, now see how we will solve it. We will say, we will find
their milli moles, molarity into volume with millilitre is what? Millimoles. We will say how
many milli moles it has. 2.5x milli moles and milli moles of NaOH will be 3 into 4.5 millimoles
and we know that 1 mole HCl reacts with 1 mole NaOH, so 1 millimole will react with, so
these many millimoles will react with these many millimoles for complete neutralisation. So,
you have a mark here 2.5 into x is equal to what, 3 into 4.5, now you can cancel this, now
forget the decimal, 5 into 5 and 5 into 9, so this is what x into 27 by 5 which is equal to what
5.4 m, everybody understood how to do this.
Second question is very easy. You had to produce 34 grams H2SO4. What is the mole of
H2SO4 that you have to produce? See, why are you taking moles, you will say in data there is
gram why have we got moles? Because the molarity is co-related to what, volume and
moles. If we see number of moles, molar mass of H2SO4 is 34 grams per mole. So, number
of H2SO4 mole is how much? So, 1 mole of H2SO4 is produced by how many moles of
H2SO4, so therefore I can say that, I want 5 mole H2SO4, n number of moles of H2SO4
required will be equal to 5, which should be equal to the molarity of H2SO4 into the volume.
So what is the molarity given? 0.20 into V, what is 0.2, 1 by 5 into V, this implies V is equal to
what? So, we can write down 5 into 5, which is how much, 25Litres. Now everybody
understood, is there any problem. There is no problem, right. Can we do one more question
like this? All of us will do based on stoichiometry. If you want to take down this solution, you
can take it down and we will discuss the next question. Everyone has written down, okay.
See, the next one.
This question is from your NCERT book. The question is calcium carbonate plus 2HCl which
give CaCl2 plus H2O plus CO2. What mass of calcium carbonate is required to react with
25ml of 0.75 molar HCl. That means to react fully, how much calcium carbonate mass you
will need? Okay, everyone, I will demonstrate and then I will give you another question to
try. See this once more, I hope that you should be able to handle this question. 0.9375
grams is the answer, you are not getting it. We will write the reaction, calcium carbonate
plus 2HCl gives you calcium chloride plus H2O plus CO2, given things are 25ml HCl and 0.75
molar and we have to find the mass. Right, this is what is required. So, what will you do?
First of all find out molar mass of calcium carbonate. 100, you must have memorised it. So,
this is how much, W by 100 moles, okay, and how many moles this has and whose moles
should be double? HCl, so here how many moles should be there, 2W by 100. And this data
should be equal to the molarity in volume we have. Shall I make it equal? We will say 25, we
have to convert it into litre because we want moles. 25 divided by 1000 into what, 0.75. So
from here cancel two zeros. Now, if you solve, what will be W? W is equal to 25 into 0.75
divided by 20. This will be 5, 4 you may solve this will be 3.75 divided by 4 and the answer is
0.9375 g, okay. Everyone understood how we did it. You wrote this correctly, this should be
right and after this calculation we had. Everyone understood, completed this.
Let’s see one more question. Solve this question. See, if we talk about the question given
here, according to that we will write it. Na2SO4 reacts with what, what, very good, I am sure
if you have found out BaSO4 molar mass, it must be 233 grams. You can write down 2.33
grams divided by 233 gram per mole. From here you strike off.
Chemistry 07, Variation Of Strength Dilution And Mixing
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…
Chemistry 06, Problems Based On Strength Of Solution
So, let’s resume with the discussion of strength terms, in the previous lecture we had
discussed parts (00:13), now we will start with the next term and it is one of the very
important terms. The first term that we are going to do is, Molality. Molality is defined with
the symbol small m, okay, so first all of you write the definition, it represents the moles of
solute, it represents the moles of solute, it represents the moles of solute present in 1
kilogram of solvent, present in 1 kilogram of solvent. Please note, and where you have
written the definition make a box around it and write solvent. The word ‘solvent’ you again
make a mistake, molality is not moles of solute upon mass of solution, it is the mass of
solvent. Highlight it, it is not solution it is solvent. What will be the unit? Moles per kg, we
have already written it down, moles per kg and this is also called as molal represented by
small m. Okay, so the symbol also is this and what do we call it, molal represented by small
m. okay, now, if I take an example here, let’s say I have got a 2.5 molal solution of a solute,
so what is the meaning. What is the meaning? So can I say that 1 kg of solvent contains
what, if that solvent happens to be water, in case we assume it to be, so would we get to
know the moles of water? Tell me. Children, please participate. If we take the solvent to be
water, we know the mass of water. 18 gram per mole. Can we find out the moles of water?
If we know the moles of water, and if we know the mole of solute then we can get the mole
of fraction. Okay, assume I have the moles of solute and I know the molecular mass of
solute, then can I calculate here. What can I calculate here? Mass of solute. And if I know
the mass of solute, mass of solvent then I will also know the mass of solution. Can I remove
the mass percentage? Can you understand? You will get questions of inter-conversions.
You will have to understand from which information you can do the inter-conversions. Is
the matter clear? Okay, so molality is clear to all of you.
So let’s take the next term which is, this is molarity, represented by capital M. And write
down the definition, it represents the moles of solute, it represents the moles of solute, it
represents the moles of solute present in 1 litre of solution. It represents the moles of solute
present in 1 litre of solution. So as a formula we can represent it as, molarity is equal to
moles of solute, moles will be the unit of this, and volume of solution, litres. Okay, and if we
write it like this then we will get, what is the unit in this case. Unit is mole per litre, it is also
called as molar and it is symbolised as capital M again. Okay, even in the unit it will be
written as capital M. So just suppose that we have got for example 1.7 molar NaCl aqueous
solution, what does this imply. So in 1 litre of solution, what do you have? Correct, okay.
So what can you convert this into? Tell me, what data you will require to convert this into.
If we have the density then we can calculate the mass of solution, because we know the
volume of solution. If we know the mass of solution then if I may tell you we know the
molecules of NaCl mass. We will also get the NaCl mass. So what can you calculate? Mass
percent. Okay, if we remove the mass of solution and from there we remove the mass of
solute then what will be get, mass of solvent. Then we can find out molality also. And if
required we can convert this into mole fraction, for both of them, we know the moles for
the two of them. We have found out the mass of solvent and also got the moles. Are you
understanding how the inter-conversion is done. What should be kept in mind and which is
the term which will be the central term for us for inter-conversion – density. What does
density inter-relate? Mass of solution with the, you have to relate this and then the rest of
the inter-conversion will be done. We will shortly see the question, first write a comment
on molarity. Write down, molarity is generally taken to be synonymous to concentration. If
I do not have anything mentioned, that this is a concentrated solution, or it is a highly
concentrated, and there is a less concentrated solution, the indication is by the unit of
molarity, if it is highly concentrated solution then the molarity will be more, and if it is less
concentrated solution then the molarity will be less. So generally when we say a highly
concentrated solution and a low concentrated solution or a less concentrated solution we
are referring to which term. That is whichever terms we have studied till now, all of them
are concentration terms. But if no specific concentration is defined then you will assume it
to be Molarity. Did all of you understand this? So that is Molality and Molarity.
Now I am going to put up three questions in front of you, which you will have to do. Please
no one give the answer in between. Once you are done raise your hand, I will come and
check your answer. Okay. Do it yourself and then we will discuss. Okay. Solve these three
questions.
Chemistry 05, Strength Of Solution
We had earlier defined what is the strength of a solution. We said that, the strength of a
solution is the amount of solute, amount is not necessary it is just a mass, it can be moles, it
can be volume also. So it is the amount of solute dissolved per unit solution or solvent, is
called as the strength of solution. Typically in the lab we are used to see such setups, there
are bottles kept on the racks in the lab, like concentrated HCl and dilute HCl. What is the
difference between both of them? In one bottle, that is in one bottle the amount of HCl will
be more in concentrated HCl and in dilute the amount of HCl will be less. What are we
defining here, strength, how strong it is. The stronger it is, accordingly to our knowledge
the stronger it is, it is more dangerous, and it is difficult to handle. Right, dilute is easier to
handle and it is less dangerous. It is dangerous even then, okay, but still we can handle it,
okay. And, if we talk about, here, we discussed about mole fraction.
How did we define mole fraction. It is the moles of solute present in 1 mole of solution.
Right. And, today we will see more questions related to this, related to mole fractions. And
how had we defined mass percentage. It is the mass of solute present in …. So it is clear till
here, any confusion, anything you wish to ask. Let’s continue, a 2.0 grams sample
containing SiO2 and Fe2O3 on very strong heating leaves a residue weighing 1.96 grams.
The reaction responsible for loss of weight is: Fe2O3 giving you Fe3O4 and Oxygen, you are
asked what is the percentage of mass of SiO2 in the original mixture. Okay. Now this is not a
typical solution kind of a problem. It is more of a mixture kind of a problem. In which the
mass percentage is being asked and we can define it also. How do you do that? See to do
this first of all we will require to do two approaches, so let’s try approach one. Approach
one would be, we would say, let us say the mass of Fe2O3 in the mixture is w gram. And the
mass of Fe3O4 in the mixture, sorry not Fe3O4 but SiO2 is how much, 2 minus w gram. Now
based on the reaction given, is the reaction balanced as of now? How can we balance this
reaction? See, if we have to balance this, we have to do a simple thing, if we have to balance
the Iron, what do we put here, a 3 and what do we put here, a 2. Correspondingly how
much oxygen is here, 9 and here, how much is there? 8, 2 into 4, 8, plus another one, so it
has become more, right. So what do you do then, make this one less by making it 1/2. So
now what has it become, 3, 9, 2, 4, 8 and here what is it still, 9, oxygen, okay, now it is done.
So what do we do now, we should multiple this full equation by 2. Either multiple by 2 or
keep it as it is, it is upto you. Okay, we are fine with both. Or in case we keep it like this,
then we will say, 3Fe2O3 reacts to give you 2Fe3O4 plus 1/2 O2. What can you write for the
moles, tell me, correct? Okay, now corresponding to this we will write further. What is the
molar mass of Fe2O3? Yes, okay, what will be the molar mass of Fe3O4? So therefore what
can we say from here, this is how much, 3 into 180, how much is it. How much, sorry 3 into
160 how much will it be? 480 grams. Okay, of Fe3O4 we have made 2 moles, 2 into 232 how
much will that become, 464 grams, correct. We will not write for this, we have to just write
about Fe3O4. So for 480 grams it becomes 464 grams then we can say how much it will
become for 1 gram. 464 divided by 480 gram. Okay, therefore how much will it become for
w gram. 464 by 480 gram, is this correct. Now what we have written in the data we will go
corresponding to that. The data given to us is the reaction responsible for the loss of weight
is this in which Oxygen gas is going but Fe3O4 is left. And they have said how much the
residue mass would be. So what should we write now, what is the residue, Fe3O4 plus the
left SiO2 that is 2 minus w so what will be the equation you would write. 2 minus w plus
464 by 480 w is equal to what – 1.96, so from here what will you write, what will 2 minus
1.96 be? 0.04, correct, and after this you do one thing, you take it to the right hand side,
now what is left behind, w minus 464 by 480 w. Now do one thing, take w as common, you
take 480 on top, now 480 minus 464 is how much, divided by, now you will see that this will
get cancelled by 30, so this implies that w is how much. 0.04 into 30 which is, so what is the
mass of silica, 0.8, 0.8 divided by 2 into 100, how much will you get. Now you see the
approach is quite lengthy. You saw, right. Okay how many of you have done it like this. So
how did the others do? Okay, that is a good idea, that is a brilliant idea. Now what will we
do here, what is the reason for whatever weight loss that is there in the reaction. It is
because of Oxygen. What is the mass of Oxygen loss, 0.04, so we will say 1/2 mole that is
there, okay, from here, first we will remove the moles of Oxygen. So moles of Oxygen is
how much, 0.04 divided by how much, 32, okay. Now these moles that we have got, now we
can say that if we want ½ mole or we will multiple this by 2, then we will tell corresponding
to 1 mole what is the mole of 3Fe2O3, 6, so we can tell 0.04 by 32 mole corresponding to
oxygen how much mole will there be in Fe2O3, 6 times 0.04 by 32 mole of what, Fe2O3, yes,
so what can you remove from this. The mass of Fe2O3. What did we determine the molar
mass of Fe2O3 to be? So what should we do now, we will write mass of Fe2O3 will be
equal to 6 into 0.04 divided by 32 into what, 160, 32 will be cancelled into this by 5, 5 into 6
is 30, now see what 30 into 0.04 did you get again. Now see from here to here, three lines,
very quickly. Right, now 1.2 gram if we have got, now to remove the percentage it is very
easy. This you have understood? How many had applied the second approach. Very nice.
How have the others done it? You have not done it, yeah, homework you get for time pass,
right, it is like that, right? Do not do that. Okay, so this is done. Now let us look at the next
strength term. The next strength term is, mass by volume percent. It is denoted by
percentage w by v. Okay, and for this term, first of all you write the definition for the mass
by volume percent. Write down, a mass by volume percent mark, it represents the mass of
solute in grams present in 100 ml of solution. It represents the mass of solute in grams
present in 100 ml of solution. So this time the units are very important. If you remember
the last time where we had done the mass by mass percent, in that mass was on top as well
as below, so you could put it as gram by gram, kilogram by kilogram, milligram by milligram,
anything you want. But this time units are important, because the dimensions on top and
below are different, so you will have to put the mass in grams and you will have to put the
volume of solution in, what? Right, so you will have to be careful regarding this. Now, here
if we take 100 for the volume of solution, what will 100 on 100 be. So you all understood
the definitions, it is mass of solute in grams present in, now you will get the percentage.
Like, if you take an example, let’s say we take 10 percent weight by volume NaCl aqua
solution, what does this represent, tell me. 10, so let us mark it properly as 100 ml solution
will contain 10 grams of NaCl. Okay, you remember the last time.
