2 Amount and Concentration: Making and Diluting Solutions
Jenny Koenig
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**Amount and Concentration: Making and Diluting Solutions**
Instructions:
The aim of these questions is to give you practice in calculating how to make solutions either in g/L or in M and how to make appropriate dilutions.
You can answer as many as you wish, in which ever order and as many times as you wish.
It is best to work out your answer with pen and paper before inputting it. Click on "Check Answer" to reveal the correct answer and appropriate feedback if you got it wrong.
pic001.jpg
credit: LTSN Bioscience © University of Leeds
**Making solutions in g/L, % w/v, % v/v**
Making solutions: % w/v
You have a very precious sample of purified enzyme and need to make a 0.5 mL solution at a concentration of 0.1 % w/v. How much do you need to weigh out?
Express your answer in mg to one significant figure.
0.5 mg
well done
A 0.1 % w/v solution is 0.1 g in 100 ml

which is 0.001 g in 1 ml

which is 1 mg in 1ml

which is 0.5 mg in 0.5 ml.
Making solutions
Concentrated solutions of ethanol in water are commonly used to sterilise surfaces. The concentration of ethanol that is used is 70 % v/v. You need to fill a 500 mL sprayer bottle. How much ethanol and water will you need?
volume of ethanol = 350 mL

volume of water = 150 mL
well done
70 % v/v is 70 ml ethanol in 100 ml total, so that means 70 ml ethanol plus 30 ml water.

Scale that up to 500 ml total to give 350 ml plus 150 ml.
Making solutions: g/L
You have weighed out 2.5 mg of an enzyme and need to make a 10 mg/ml solution. How much water do you need to dissolve it in?
0.25 ml
well done
volume = amount / concentration

= 2.5 mg / 10 mg/ml

= 0.25 ml
Convert % to g/L
The legal limit for alcohol in the blood whilst driving is 80 mg%. This is an obsolete unit (although you may occasionally see it in practice) which is equivalent to the number of milligrams per 100 g weight of solvent. Assume for this exercise that blood is the same density as water (i.e. 1 g per litre). Convert 80 mg% to g/L
0.08 g/L
0.8 g/L
8 g/L
80 g/L
800 g/L
well done
80mg/ 100 g water

= 80 mg / 100 ml

= 800 mg/L

= 0.8 g/L
Concentrations in g/L
pic002.jpg
*This marine dinoflagellate, Ceratium, is common member of the coastal phytoplankton. The central grove is where the flagellum is located. Dinoflagellate blooms cause phenomena known as red tides, some of which may be toxic to humans. credit: Dr Gordon Beakes © University of Newcastle upon Tyne*
The concentration of plankton in seawater at a depth of 4 m is roughly 5 mg/L.
What volume of seawater would contain 25 μg?
5 ml
well done
5 ml: volume = amount / concentration

= 25 x 10^{-6} g / 5 x 10^{-3} g/L

= 5 ml
**Convert amount from g to mol and calculate molar concentration.**
Calculating concentration (M)
What is the molarity (concentration in M) of the following solutions:
5 μmol dissolved in 25 mL
8 mmol dissolved in 40 L
3 μmol dissolved in 150 mL
4 μmol dissolved in 0.2 mL
2 μM
20 μM
200 μM
2 mM
20 mM
200 mM
well done
answers were 200 μM, 200 μM, 20 μM, 20 mM
Convert amount in g to mol
You need to measure out 0.001 mmol of purified thyroid stimulating hormone (molar mass = 28 000 Da). How much should you weigh?
The weighing balance is accurate to 4 decimal places. Therefore weigh out 0.0280 g
well done
0.001 mmol = 1 x 10^{-6} mol. Amount (in g) = amount (in mol) x molar mass (in g/mol) = 1 x 10^{-6} mol x 28 x 10^{3} g/mol

= 28 mg

= 0.0280 g to 4 decimal places.
Calculating concentrations in M
The concentration of a protein of molecular weight 21.5 kDa was determined to be 0.1 mg/L.

What is the concentration in nM?

How much protein would be present in a 10 μL sample?
Concentration = 4.7 nM (to two significant figures)

Amount in 10 μL = 47 fmol (to two significant figures)
well done
0.1 x 10^{-3} g/L / 21500 g/mol =0.47 x 10^{-8} mol/L = 4.7 nM. 10 x 10^{-6} L x 4.7 x 10^{-9} mol/L

= 47 x 10^{-15} mol = 47 fmol
Calculating concentration in M
pic003.jpg
*Blood cells in a blood vessel - picture taken by transmission electron microscope. credit: Dr Gordon Beakes © University of Newcastle upon Tyne*
A female patient's blood contains approximately 128 g of haemoglobin per litre. If the red blood cells were disrupted so that the haemoglobin was dispersed, what would the concentration be in M?
Molecular weight for haemoglobin = 64 kDa. Express your answer to two significant figures.
0.0020 M
well done
128 g/L / 64000 g/mol

= 0.0020 M

(for 2 significant figures you need the 0 after the 2)
Calculating concentration (M)
Calculate how much urea is present in the following (in μmol):
10 ml x 0.01 M
25 ml x 4 mM
20 μl x 0.5 M
5 μl x 2 mM
50 ml x 2 μM
0.01 μmol
0.1 μmol
1 μmol
10 μmol
100 μmol
1000 μmol
well done
first convert volume to L and concentration to mol/L, then multiply, then convert to μmol.

The answers were 100, 0.1, 10, 0.01, 0.1 μmol.

For example, the first one is 10 ml x 0.01 M

= 0.01 L x 0.01 mol/L

= 0.0001 mol = 0.1 mmol = 100 μmol.
Converting from mol to g
Following on from the previous question...
If one haemoglobin protein contains 4 iron atoms, how many moles of haemoglobin and iron and how many milligrams of iron would be present in a 1 ml sample?
Express your answers to two significant figures;
atomic mass for iron is 56.
Amount of haemoglobin = 2.0 μmol

Amount of iron = 8.0 μmol

Amount of iron = 0.45 mg
well done
amount of haemoglobin = 0.002 mol/L x 0.001L

= 2.0 μmol.

There are 4 iron molecules for each haemoglobin, so the amount of iron is 8.0 μmol.

Convert this to grams, 8 μmol x 56 g/mol

= 448 μg

= 0.45 mg to two significant figures.
**Convert from g/L to M and calculate concentration in M**
Convert % w/v to M
Isotonic saline is often expressed as 0.9 % w/v. What is the equivalent concentration in M? (Molecular weight of NaCl is 58.44 g/mol).
0.154 M (to 3 significant figures)
well done
0.9 g/100ml = 9 g/L = 9 g/l / 58.44 g/mol = 0.154 mol/L = 0.154 M
Calculating concentrations in M, converting from g to mol
Normal physiological solutions contain around 5 mM KCl but raised concentrations of KCl - usually 150 mM - are often used to depolarise neurons to cause an action potential. The molar mass of KCl is 74.5 g/mol.
To make 100 ml of a 5 mM solution you would need 0.037 g (2 significant figures)

To make 100 ml of a 150 mM solution you would need 1.1 g (2 significant figures)
well done
5 mM = 0.005 mol/1000ml = 0.0005 mol/100ml.

0.0005 mol x 74.55 g/mol = 0.037 g

to get 150 mM multiply 0.037 g by 30 to give 1.1 g (2 significant figures)
Calculating concentration or amount (M or mol)
*The picture below shows an electron micrograph of a striated muscle. Muscle tissue is dominated by the regular arrays of interleaving fibres, which slide together to bring about muscle contraction. Mitochondria are abundant between the layers of fibres and supply the ATP energy needed for contraction.*
*Dr Gordon Beakes © University of Newcastle upon Tyne*
pic004.jpg
In an experiment to measure the rate of an enzyme reaction, 15 μl of 0.04 M ATP solution was added to each tube which had a total volume of 2 ml. How much ATP was present in each tube?
0.6 μmol (to 1 significant figure)
well done
15 x 10^{-6} L x 0.04 mol/L = 0.6 x 10^{-6} mol = 0.6 μmol
Calculating concentration (M)
Recent research has identified several peptide hormones which are important in food intake. One such peptide, called orexin (which increases food intake) is present in very low concentrations. In an experiment with cultured neuronal cells, you need to add 0.1 fmol to a volume of 100 μl. What will the final concentration be?
1 aM
10 aM
1 fM
1 pM
10 pM
1 nM
well done
Answer is 1 pM.

Working: 0.1 fmol = 0.1 x 10^{-15} mol.

Concentration = 0.1 x 0^{-15} mol / 0.1 x 10^{-3} L

= 1 x 10^{-12} M = 1 pM
Calculating amount in moles
Following on from the previous question. How many molecules of orexin are there in 0.1 fmol?
6 x 10^{7}
6 x 10^{8}
6 x 10^{9}
6 x 10^{10}
6 x 10^{12}
6 x 10^{14}
well done
the answer was 6 x 10^{7}

since 0.1 fmol = 0.1 x 10^{-15} mol.

Multiply by Avogadro's number (which is 6 x 10^{23}) to get 0.6 x 10^{8} mol = 6 x 10^{7}
**Making Dilutions**
Making dilutions
The concentration of a laboratory stock solution of the antibiotic streptomycin is 25 mg/ml. How much of the stock solution and how much water would be needed to prepare 100 ml of a working solution with an streptomycin concentration of 2.5 mg/ml?
10 ml stock solution of streptomycin and 90 ml water.
well done
This is a 1:10 dilution since 25 / 2.5 = 10.

Therefore it is one part stock solution plus nine parts water.

To make 100 ml add 10 ml stock to 90 ml water.
Making dilutions
You are given a solution of 1M NaCl and are asked to take 1.5 ml and add that to 8.5 ml of a complex nutrient growth mixture. What is the resulting concentration of NaCl?
0.15 M
well done
1 M x 1.5 ml / 10 ml = 0.15 M
Making dilutions
*The mitochondria are the cellular power stations which carbohydrates to produce the chemical energy required to fuel all cellular processes. The mitochondria are typically capsule shaped bodies about half a micron in diameter. In animals the infolded inner membrane forms flat plate-like cristae. Cristae: these infolds of the inner mitochondria membrane contain molecules for the generation of ATP in the process known as oxidative phorphorylation*.
pic005.jpg
*Photo credit: Gordon Beakes © University of Newcastle upon Tyne image courtesy Centre for Bioscience, the Higher Education Academy, ImageBank http://www.bioscience.heacademy.ac.uk/imagebank/.*
Antimycin is an inhibitor of mitochondrial function and is fairly insoluble in water. It can be dissolved in ethanol at 1 mM and then diluted into water to make a 1 μM solution. How would you make 10 mL of a 1 μM solution of antimycin from the 1 mM stock solution? What would the final concentration of ethanol be?
Add 10 μL of 1mM antimycin solution to 9.99 mL water.

The resulting ethanol concentration would be 0.1 % v/v.
well done
The dilution factor is 1000 (10^{-6} M / 10^{-3} M).

To make 10 ml you need 0.01 ml (=10/1000) + 9.99 ml (= 10 - 0.01) water.

The original solution was 100% ethanol which was diluted 1:1000 to become 0.1 % (v/v)
Making dilutions
ATP is an important molecule required for energy-dependent processes and is often added to in vitro experiments for enzymes to function correctly. If you were to add 15 μl of a 0.04 M ATP solution to a final volume of 2 ml, the resulting concentration of ATP would be ...
0.3 mM
well done
15 μl = 0.015 ml. The dilution factor = 2/0.015 = 133.333. Therefore the resulting concentration is 0.04 M / 133.3333 = 0.0003 M = 0.3 mM
Making dilutions
You have completed an experiment in which you estimated the protein concentration of a urine sample to be 0.04 mg/ml. The sample had previously been diluted by adding 0.05 ml of urine to 0.95 ml water. What was the original concentration of protein in the urine?
0.8 mg/ml
well done
The dilution factor was 1/0.05 = 20. Therefore the concentration of the original solution was 20 x 0.04 mg/ml = 0.8 mg/ml.
Making dilutions
Sodium dodecyl sulfate (SDS) is a commonly used detergent found in the lab and in a number of household products. In the lab it is used in the separation of proteins by electrophoresis. Say you have a 3% (w/v) SDS stock solution and you need to prepare a sample of liver mitochondria for electrophoresis. You have 5 ml of the liver mitochondria sample and the final SDS concentration needs to be 0.2 % (w/v). What volume of 3% SDS should you add? (to three significant figures)
0.357 mL of 3% SDS plus 5 ml liver mitochondria suspension will give a final SDS concentration of 0.2%.
well done
C_{1}V_{1} = C_{2}V_{2};

in this case C_{1} = 3 %; C_{2} = 0.2 %; V_{2} = 5 ml + V_{1} and V_{1} is what we're looking for.

Substituting in gives ... 3V_{1} = 1 + 0.2V_{1} and simplifying gives V_{1}= 1/2.8 = 0.357 ml
Making dilutions
pic006.jpg
*Zebrafish: has poison glands contained in the Dorsal, Pelvic and Anal fins. They feed on crusteaceans and small fish. This photo was taken on the eastern coast of Australia. Photo credit: Andrew Holland © Andrew Holland image courtesy Centre for Bioscience, the Higher Education Academy, ImageBank http://www.bioscience.heacademy.ac.uk/imagebank/.*
You have been asked to prepare a small volume of a 20 nM solution of a poison of a zebrafish from a 1 mM stock solution. Which of the following might be the best way to do it?
1 ml stock solution + 499 ml water
1 ml stock solution + 49999 ml water
0.01 ml stock solution + 499.99 ml water
0.01 ml stock solution + 4.99 ml water then 0.01 ml of the resulting solution + 0.99 ml water
0.0001 ml stock solution + 4.9999 ml water
well done - you need to dilute it in two stages.
It is impossible to pipette less than 0.01 ml accurately. Therefore you have to do this in at least two stages. Go from 1 mM to 2 μM (which is 1:500 or 0.01 ml + 4.99 ml) then from 2 μM to 0.02 μM (= 20 nM) which is 1:100 or 0.01 ml + 0.99 ml.
Making dilutions
You have a stock solution of 1 mg/ml lysozyme (MW 14700) and need to prepare 5 ml of a solution of 500 pmol/ml. What volumes of the stock solution and water do you require (to the nearest μL)?
37 μL x 1 mg/ml lysozyme + 4.963 mL water
well done
1 mg/ml = 1 g/l = 1g/l / 14700 g/mol = 6.8 x 10^{-5} mol/l.

Need to dilute to 500 pmol/ml = 500 nmol/l = 5 x 10^{-7} mol/l.

Therefore dilution factor is 6.8 x 10^{-5} divided by 5 x 10^{-7} = 136.

You want 5 ml volume so need 5 ml /136 = 37 μL.

The volume of water required is 5 - 0.037 = 4.963 mL
Making dilutions
How much 0.2 M NaCl solution must be added to 10 ml of enzyme solution to bring the enzyme solution to a concentration of 0.1 mM NaCl?
0.005 ml
well done
C_{1}V_{1}=C_{2}V_{2};

C_{1} = 0.2 M = 200 mM; C_{2} = 0.1 mM; V_{1} is what we want to find and V_{2} = 10ml + V_{1}.

Substituting in we get 200V_{1}= 1 + 0.1 V_{1}

so V_{1} = 0.005 ml.