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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.

2016-06-13T14:04:50+00:00 Categories: Archive - 2015-16|Tags: , |0 Comments
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