Solutions 5

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Name	Status	Author	Last Modified
Solutions 5	Ready to use	Jamie Antoun	01/03/2018 09:46
Dilutions	draft	David Goulding	27/04/2018 12:14
Vicky's copy of Solutions 5	draft	Vicky Hall	05/12/2022 13:12

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Question statement

Answer the following question. Please give your answers as decimals, not as fractions. Give your answers to 3 decimal places but only round your calculations at the final step, use the exact values in your calculator for intermediate steps.

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Give any introductory information the student needs.

			Name	Type	Generated Value

			Name	Type	Generated Value

stock_volume_used	integer	100
Cl	number	35.453
Na	number	22.990
NaCl	number	58.443
stock_mass_concentration	number	80
final_mass_concentration	number	32
stock_volume	number	2.5
stock_molarity	number	1.3688551238
final_volume	integer	250
final_molarity	number	0.5475420495
stock_mass	integer	200

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
100

JME function reference

Description

The volume of the "stock" solution used for dilution in ml.

Describe what this variable represents, and list any assumptions made about its value.

Can an exam override the value of this variable?

Generated value: `integer`

100

→ Used by:

final_mass_concentration
final_molarity

This variable doesn't seem to be used anywhere.

Parts

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1. Gap Gap 0. Number entry
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2. Gap Gap 1. Number entry
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A "stock" solution contained $\var{stock_mass}$ g of sodium chloride (NaCl) in $\var{stock_volume}$ L of solution. $\var{stock_volume_used}$ ml of the "stock" solution was diluted to $\var{final_volume}$ ml. What is the concentration of the final solution in g/L and M? [Relative atomic masses (Da): Na $\var{Na}$ ; Cl $\var{Cl}$ ].

Gap 0. g/L

Gap 1. M

Advice

We have been asked to do quite a lot of calculations in this example so let's break it up into several smaller parts. To work out the concentrations in g/L and M after dilution, we need to first work out the concentrations in g/L and M of the "stock" solution. Let's start by working out the concentration of the "stock" solution in g/L.

The "stock" solution contains $\var{stock_mass}$ g of sodium chloride dissolved in $\var{stock_volume}$ L of solution. To work out the concentration in g/L we use the formula

$\dfrac{\text{mass of substance (in grams)}}{\text{volume of liquid (in litres)}} = \text{concentration (in g/L)}.$

Putting in our numbers we find that the concentration of the "stock" solution in g/L is

$\dfrac{\var{stock_mass}}{\var{stock_volume}} = \var{stock_mass / stock_volume} \text{ g/L}.$

Next, we can work out the concentration of the stock solution in M (mol/L). The formula for this is

$\dfrac{\text{number of moles of substance}}{\text{volume of liquid (in litres)}} = \text{concentration (in mol/L)}.$

We can see that we need to work out how many moles of sodium chloride (NaCl) there are in $\var{stock_mass}$ g. The formula for this is

$\dfrac{\text{mass of substance (in grams)}}{\text{molecular weight}} = \text{number of moles}.$

We can see that to do this, we are going to need to work out the molecular weight of sodium chloride (NaCl). We have been told the atomic weights of each of the atoms in sodium chloride (NaCl) so we can do this by adding together the atomic weights of all the atoms to get the molecular weight which is

$1 \times \var{Na} + 1 \times \var{Cl} = \var{NaCl} \text{ Da}.$

We can now use this to work out how many moles of sodium chloride there are in $\var{stock_mass}$ g using the formula above. Putting in the numbers we find that there are

$\dfrac{\var{stock_mass}}{\var{NaCl}} = \var{stock_mass / NaCl} \text{ moles}$

of sodium chloride in $\var{stock_mass}$ g. We can now work out the concentration of the "stock" solution in M (mol/L). The formula is

$\dfrac{\text{number of moles of substance}}{\text{volume of liquid (in litres)}} = \text{concentration (in mol/L)}.$

We have $\var{stock_mass}$ g of sodium chloride dissolved in $\var{stock_volume}$ L of liquid and we have calculated that there are $\var{stock_mass / NaCl}$ moles of sodium chloride in $\var{stock_mass}$ g. Putting these numbers into the formula we find that the concentration of the "stock" solution in M is

$\dfrac{\var{stock_mass / NaCl}}{\var{stock_volume}} = \var{stock_molarity} \text{ M}.$

Let's recap what we have calculated so far. We have worked out that the concentration of the "stock" solution in g/L is

$\var{stock_mass_concentration} \text{ g/L}$

and in M is

$\var{stock_molarity} \text{ M}.$

Since we are diluting $\var{stock_volume_used}$ ml of the "stock" solution, we need to work out how much sodium chloride there is in $\var{stock_volume_used}$ ml of the solution using the concentrations we have calculated. We need to work this amount out in both grams and moles, let's start with grams. If we rearrange the formula

$\dfrac{\text{mass of substance (in grams)}}{\text{volume of liquid (in litres)}} = \text{concentration (in g/L)}.$

we find that

${\text{volume of liquid (in litres)}} \times \text{concentration (in g/L)} = \text{mass of substance (in grams)}.$

Now, $\var{stock_volume_used}$ ml is equal to

$\dfrac{\var{stock_volume_used}}{1000} = \var{stock_volume_used / 1000} \text{ L},$

so in $\var{stock_volume_used}$ ml of "stock" solution, there are

$\var{stock_volume_used / 1000} \times \var{stock_mass_concentration} = \var{(stock_volume_used / 1000) * stock_mass_concentration} \text{ g}$

of sodium chloride. Similarly, we can work out the number of moles of sodium chloride using the formula

${\text{volume of liquid (in litres)}} \times \text{concentration (in mol/L)} = \text{number of moles of substance}.$

Putting in our numbers, we find that in $\var{stock_volume_used}$ ml of "stock" solution, there are

$\var{stock_volume_used / 1000} \times \var{stock_molarity} = \var{(stock_volume_used / 1000) * stock_molarity} \text{ moles}$

of sodium chloride.

Finally, we can work out the concentration of the diluted solution. We know that $\var{stock_volume_used}$ ml of "stock" solution is diluted to $\var{final_volume}$ ml so the final volume of liquid is $\var{final_volume}$ ml. We have worked out that in $\var{stock_volume_used}$ ml of stock solution there are $\var{(stock_volume_used / 1000) * stock_mass_concentration}$ g ( $\var{(stock_volume_used / 1000) * stock_molarity}$ moles) of sodium chloride and we have not added any more sodium chloride so in our diluted solution we have $\var{(stock_volume_used / 1000) * stock_mass_concentration}$ g ( $\var{(stock_volume_used / 1000) * stock_molarity}$ moles) of sodium chloride in $\var{final_volume}$ l of liquid. The formula for the concentration in g/L is

$\dfrac{\text{mass of substance (in grams)}}{\text{volume of liquid (in litres)}} = \text{concentration (in g/L)}.$

$\var{final_volume}$ ml is equal to $\var{final_volume / 1000}$ L and so the concentration of the diluted solution in g/L is

$\begin{align}\dfrac{\var{(stock_volume_used / 1000) * stock_mass_concentration}}{\var{final_volume / 1000}} & = \var{final_mass_concentration} \\ & = \var{precround(final_mass_concentration, 3)} \text{ g/L to 3 d.p.}\end{align}$

The formula for concentration in M (mol/L) is

$\dfrac{\text{number of moles of substance}}{\text{volume of liquid (in litres)}} = \text{concentration (in mol/L)}$

so the concentration of the diluted solution in M is

$\begin{align}\dfrac{\var{(stock_volume_used / 1000) * stock_molarity}}{\var{final_volume / 1000}} & = \var{final_molarity} \\ & = \var{precround(final_molarity, 3)} \text{ M to 3 d.p.}\end{align}$

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Christian Lawson-Perfect 7 years, 2 months ago

Christian Lawson-Perfect 10 years, 1 month ago

Jamie Antoun 10 years, 10 months ago

Error in variable testing condition

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Generated value: `integer`

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Solutions 5

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Christian Lawson-Perfect 7 years, 2 months ago

Christian Lawson-Perfect 10 years, 1 month ago

Jamie Antoun 10 years, 10 months ago

Error in variable testing condition

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Warning

Generated value: integer

Parts

Gap-fill

Generated value: `integer`