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Definition of Dilution Factor
You may come across something like, "prepare a 1:50 dilution of the solution".
What it means is, take a known volume of the stock solution (V_{initial}) and add enough solvent to it so that the solution has a new volume, V_{final}, of 50 x V_{initial}.
V_{final} = 50 × V_{initial} 
The "1:50" tells you the dilution factor, the ratio of volumes, to use to prepare the new solution.
In this case it tells us that V_{1} = 1 and V_{2} = 50
so the dilution factor, DF, = V_{2} ÷ V_{1} = 50 ÷ 1 = 50
That is, the new, diluted solution will have a volume 50 times greater than the volume of the original, undiluted, solution:
V_{final} = DF × V_{initial} 
A dilution factor does not tell you what the initial volume is, neither does it tell you what the final volume is, it only tells you what the ratio of the initial to final volume is.
You can use the equation V_{final} = DF × V_{initial} to find the final volume of solution after dilution if you know the initial volume of the solution.
The table below gives you a number of different options for preparing a 1:50 dilution:
V_{intial} (initial volume) 
1 mL 
1 L 
0.1 mL 
2 mL 
25 μL 
V_{final} (final volume) 
50 mL 
50 L 
5 mL 
100 mL 
1250 μL 
ratio of volumes used V_{initial} : V_{final} V_{1} : V_{2} 
1 : 50 1 : 50 
1 : 50 1 : 50 
0.1 : 5 1 : 50 
2 : 100 1 : 50 
25 : 1250 1 : 50 
dilution factor DF = V_{2} ÷ V_{1} 
50 ÷ 1 = 50 
50 ÷ 1 = 50 
50 ÷ 1 = 50 
50 ÷ 1 = 50 
50 ÷ 1 = 50 
Note that in each example above, the final volume = 50 times the initial volume:
that is, V_{final} = 50 × V_{initial}
or, V_{final} = dilution factor × V_{initial}
or, V_{final} = DF × V_{initial}
Now when the ratio of volumes is known, that is V_{1}:V_{2} is known,
then dilution factor, DF = V_{2} ÷ V_{1}
so V_{final} = DF × V_{initial}
or V_{final} = V_{2}/V_{1} × V_{initial}
Effect of Dilution Factor on Concentration
Remember, dilution factor tells us the ratio of the initial (undiluted) volume of solution to the final (diluted) volume of solution:
V(undiluted) : V(diluted) or V_{1} : V_{2}
and, concentration in moles per litre (c)
c = amount of solute in solution (n) ÷ total volume of solution (V)
c = n ÷ V
Consider a 250 mL volumetric flask containing ONLY a 5 mL aliquot of a solution with a concentration of c_{initial} (this will look like a thin layer of solution in the bottom of the otherwise empty flask).
V_{initial} = the initial volume of undiluted solution = 5 mL
c_{initial} = ? (we have not been told the concentration of this solution)
The amount of solute in the solution (n) = concentration (c_{initial}) × volume (V_{initial})
n(solute) = c_{initial} × 5 mL
When more solvent is added to the volumetric flask until the bottom of the meniscus just sits on the mark on the neck of the flask, the new volume of the diluted solution, V_{final}, is 250 mL
V_{final} = final volume of diluted solution = 250 mL
Note that adding more solvent to the flask does NOT add more solute, neither does it remove any solute.
The same amount of solute is still present in the volumetric flask, that is
amount of solute in diluted solution = amount of solute in solution before dilution
amount of solute in diluted solution = n(solute) = c_{initial} × 5 mL
The concentration of the final solution after dilution (c_{final}) = amount of solute in solution ÷ total volume of solution
c_{final} 
= 
n(solute) V_{final} 

from the definition of concentration 

= 
c_{initial} × 5 mL 250 mL 

substituting for n and V_{final} used to make the dilute solution 

= 
c_{initial} × 5 mL 250 mL 

simplifying the expression by dividing by 5, that is, 5/5 and 250/5 Notice that the units for volume also cancel out. 

= 
c_{initial} × 1 50 

the simplified expression 

= 
c_{initial} × 
1 50 

separating the concentration term from the ratio of volumes term 

= 
c_{initial} × 
V_{initial} V_{final} 

generalising the expression 
This equation shows us the relationship between concentration and the ratio of the inital volume, V_{initial}, and the final volume, V_{final}.
Now the ratio of the volumes actually used to make the dilute solution, V_{initial} : V_{final}, will be the same as the ratio of volumes used to determine the dilution factor,
V_{1} : V_{2}
So
V_{initial} V_{final} 
= 
V_{1} V_{2} 
So we can also write c_{final} = c_{initial} × V_{1}/V_{2}
We can rearrange this new equation to find the relationship between dilution factor and concentration:
c_{final} 
= 
c_{initial} x 
V_{1} V_{2} 

the expression 
c_{final} × V_{2} 
= 
c_{initial} × 
V_{1} × V_{2}
V_{2} 

multiply both sides of the expression by V_{2} 
c_{final} × V_{2} V_{1} 
= 
c_{initial} × 
V_{1}
V_{1} 

divide both sides of the expression by V_{1} 
c_{final} × V_{2} V_{1} 
= 
c_{initial} 


rearranged expression 
Since dilution factor = V_{2} ÷ V_{1} = DF
we see that:
c_{initial} = dilution factor × c_{final}
c_{initial} = DF × c_{final}
and by rearranging this equation,
c_{final} = c_{initial} ÷ dilution factor
c_{final} = c_{initial} ÷ DF
Examples
Calculating Volumes
Question 1. What volume of stock solution must be added to a 100 mL flask to produce a 1:25 dilution once the flask is filled to the mark with solvent?
 What is the question asking you to do?
Calculate the initial volume of solution, V_{initial}.
 What information (data) has been given in the question?
V_{final} = final volume of solution after dilution = 100 mL
dilution factor as a ratio of volumes, V_{1}:V_{2} is 1:25
therefore V_{1} = 1 and V_{2} = 25
 What is the relationship between what you know and what you need to find out?
(i) DF = dilution factor = V_{2} ÷ V_{1}
(ii) V_{final} = DF × V_{initial}
So, V_{initial} = V_{final} ÷ DF
 Calculate the dilution factor, DF
DF = V_{2} ÷ V_{1} = 25 ÷ 1 = 25
 Calculate the initial volume of the stock solution used
V_{initial} = V_{final} ÷ DF = 100 mL ÷ 25 = 4 mL
Question 2. A 5 mL pipette is used to transfer an aliquot of stock solution to a volumetric flask.
What is the volume of the volumetric flask used to produce a 1:200 dilution of the solution?
 What is the question asking you to do?
Calculate the final volume of the solution, V_{final}.
 What information (data) has been given in the question?
V_{initial} = intial volume of solution = 5 mL
dilution factor as a ratio of volumes, V_{1}:V_{2} = 1 : 200
therefore V_{1} = 1 and V_{2} = 200
 What is the relationship between what you know and what you need to find out?
(i) DF = dilution factor = V_{2} ÷ V_{1}
(ii) V_{final} = DF × V_{initial}
 Calculate the dilution factor, DF
DF = V_{2} ÷ V_{1} = 200 ÷ 1 = 200
 Calculate the final volume of solution
V_{final} = DF × V_{initial} = 200 × 5 mL = 1000 mL
Calculating Concentrations
Question 1. What is the concentration of the diluted solution after 0.020 mol L^{1} NaCl(aq) is diluted 1:150 ?
 What is the question asking you to do?
Calculate the final concentration of the solution, c_{final}.
 What information (data) has been given in the question?
c_{initial} = initial concentration = 0.020 mol L^{1}
dilution factor as a ratio of volumes, V_{1}:V_{2} is 1:150
So, V_{1} = 1 and V_{2} = 150
 What is the relationship between what you know and what you need to find out?
(i) DF = dilution factor = V_{2} ÷ V_{1}
(ii) c_{final} = c_{initial} ÷ DF
 Calculate dilution factor, DF
DF = V_{2} ÷ V_{1} = 150 ÷ 1 = 150
 Calculate final concentration
c_{final} = c_{initial} ÷ DF = 0.020 ÷ 150 = 0.00013 = 1.3 × 10^{4} mol L^{1}
Question 2. What was the concentration of the stock solution if, after 1:75 dilution, the final solution has a concentration of 4.85 × 10^{3} mol L^{1} ?
 What is the question asking you to do?
Calculate the initial concentration, c_{initial}.
 What information (data) has been given in the question?
dilution factor as a ratio of volumes, V_{1}:V_{2} is 1:75
so V_{1} = 1 and V_{2} = 75
c_{final} = final concentration of solution after dilution = 4.85 × 10^{3} mol L^{1}
 What is the relationship between what you know and what you need to find out?
(i) DF = dilution factor = V_{2} ÷ V_{1}
(ii) c_{initial} = DF × c_{final}
 Calculate dilution factor
DF = V_{2} ÷ V_{1} = 75 ÷ 1 = 75
 Calculate initial concentration
c_{initial} = DF × c_{final} = 75 × 4.85 × 10^{3} = 0.364 = 3.64 × 10^{1} mol L^{1}