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# Solutions and Concentration Calculations

This is a common sort of chemistry exam or homework question to come across,
and you are usually required to calculate something like how many grams
or moles of a chemical are in a given volume of solution or what volume
of solution will you need to have a certain amount of some chemical. They
are an important type of chemical calculation because when carrying out
reactions, you need to know exactly how much of each chemical is being used
if you are working with solutions. Many other types of calculations in chemistry
involving solutions depend on getting these ones right.

I hope you find this chemistry tutorial very useful, and try some of those
in the menu as well.

### How to do these chemical calculations

The method to solve these chemistry problems is sometimes presented in a magical maths formula. But some students have problems using this magical formula as they are not sure what each bit of it means. So here's a straightforward way to do the sums. There are several versions of these questions, but once you learn how each type is done, it gets easier.

Remember that concentration is the number of moles of a chemical dissolved in 1 dm^{3},

and as you are not always given that, it is moles divided by volume measured in dm^{3}

**Type One** - given a number of grams of
a chemical present in a certain volume of solution, what volume of solution
will contain a certain number of moles of that chemical?

**Ex 1** If a solution contains 4.9 grams of H_{2}SO_{4}
in 100 ml (or 100cc or 0.1 litre or 0.1 dm^{3}), what volume of solution
will contain 2 moles of H_{2}SO_{4}?

**Step 1** - convert grams of H_{2}SO_{4} into
moles

Calculate the relative molecular mass of molar mass of H_{2}SO_{4}
by adding together all the atomic masses for each atom present - H = 1,
S = 32, O = 16, so the RMM of H_{2}SO_{4} is 98 (if you
need help with calculating rmm, use our tutorial and practice examples pages in
the menu).

We have 4.9 g of H_{2}SO_{4}, so that gives us 4.9/98 =
0.05 moles, which are present in the 0.1 dm^{3}

**Step 2** - calculate the moles per dm^{3} (also written
as mol dm^{-3}) i.e. the concentration of the solution.

We have 0.05 moles present in 0.1 dm^{3}, so dividing moles by
volume will give us the concentration. We MUST use dm^{3} (also
called litres) as the volume in this calculation. (a common trick is to
give the volume in ml or cc or cm^{3}, in which case you divide by 1000 to get the dm^{3}).

So it's 0.05/0.1 = 0.5 mol dm^{-3}

**Step 3** - calculate the volume that contains the desired
number of moles.

To do this we divide the required number of moles by the concentration

In our example it becomes 2/0.5 = 4dm^{3} (4 litres), the final
answer.

It's important to get this basic calculation correct, as the other variations on solution calculations are just this set of steps with some extra steps added on.

Side note: dm^{3} means the volume of the solution, while dm^{-3}
means "per dm^{3}", for example in mol dm^{-3} =
moles per dm^{3}

**Type Two** - given the number of moles
of some chemical in a given volume of a solution, what volume will contain
a certain number of grams.

These solution calculations start at Step 2 above (the basic solution calculation backbone) and just need extra steps to complete them. Let's see how it's done.

**Ex 2** If a solution contains 0.025 moles of H_{2}SO_{4}
in 0.5dm^{3}, what volume of this solution will contain 9.8 g of H_{2}SO_{4}?

**Step 1** - calculate the concentration of the solution, the
moles per dm^{3}

Divide the number of moles by the volume in dm^{3}, in this case
0.025/0.5 = 0.05 mol dm^{-3}

**Step 2** - convert the required 9.8 g H_{2}SO_{4}
into moles, by dividing by the relative molecular mass of H_{2}SO_{4}

Well, earlier we calculated the rmm of H_{2}SO_{4} as 98,
so I will not repeat that bit here, the last part of this step is the 9.8/98
= 0.1 moles.

**Step 3** - So we simply have to find what volume of solution
of concentration 0.05 mol dm^{-3} is needed to get a total
of 0.1 moles of the acid.

We do that by dividing the required moles by the
concentration.

In this case it's 0.1/0.05 = 2 dm^{3} and note that here the answer
is always a volume measured in dm^{3}.

Type Three - give the concentration of a solution, which volume would contain a certain number of grams of the chemical given

This is the reverse of the above Type One, isn't it, and quite similar to Type Two (you'll start to notice these similarities and that makes things easier as we go on).

**Ex 3** What volume of a solution of 0.2 mol dm^{-3} NaOH would contain 1 g of NaOH?

**Step 1** - calculate the rmm (relative molecular mass) of NaOH

We do this by adding up the ram (relative atomic mass) of each atom in NaOH - Na = 23, O = 16 and H = 1, which gives us 40

**Step 2** - convert the required number of grams into moles

We do this by diving the required number of grams by the rmm of the NaOH

so it's 1/40 = 0.025 moles required

**Step 3** - find the volume of the solution that contains the required number of moles

To get the answer we divide the required number of moles by the solution's concentration

so it's 0.025/0.2 = 0.125 dm^{3} - the answer is always a volume in dm^{3}

Type 4 - given the concentration of a solution, how much would be needed to react completely with a given volume of another solution whose concentration is also given.

**Ex 4** What volume of NaOH solution of concentration 0.025 mol dm^{-3} will be required to react completely with
200 cm^{3} of H_{2}SO_{4} of concentration 0.010 mol dm^{-3}

For solution calculations like these, there is a mathematical equation which you can juggle about to get th eanswer, but converting everything into moles makes things very much clearer.

**Step 1** - use the balanced equation for the reaction to find the ratio of the moles reacting

In this case it is 2NaOH + H_{2}SO_{4} ---> Na_{2}SO_{4} + H_{2}0

i.e. two moles of NaOH reacts with every one mole of H_{2}SO_{4}

**Step 2** - find how many moles of H_{2}SO_{4} are being used in the reaction.

Multiply the volume in dm^{3} by the concentration in mol dm^{-3}

In this case that's 200/1000 x 0.010 = 0.2 x 0.010 = 0.0020 moles of H_{2}SO_{4}

**Step 3** - find the number of moles of NaOH needed to react with this amount of H_{2}SO_{4}

So that's 2 x 0.0020 = 0.0040 moles of NaOH.

**Step 4** - find the volume of NaOH that contains that number of moles

Here we divide the required number of moles of NaOH by the NaOH solution's concentration

So it's 0.0040/0.025 = 0.16 dm^{3} (or 160cm^{3})

The above question could have been made more complicated by giving you the number of grams of NaOH dissolved in the given volume.

That would mean you'd start by doing the first two steps of question Type 1, to get the concentration.

Or you could have be given the number of grams of H_{2}SO_{4} dissolved in a given volume, instead of its concentration.

So again you'd have to use those first two steps of Type 1 to get the concentration.

Notes:

Given grams, convert it to moles by dividing by the relative molecular mass.

Ensure any volume in the calculation is in dm^{3}

Compare any question to the above types and check whether all you need to do is
combine part of one type with another type.

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