Question #64184

1 Answer
Jul 17, 2015

Answer:

You'd need 62 mL of concentrate for the second solution.

Explanation:

So, you know that 5 L of concentrate are used to make a 166-L solution.

Assuming that you're dealing with a volume by volume percent concentration, you know that

#"%v/v" = V_"solute"/V_"solution" * 100#

The volume by volume percent concentration of a solution is defined as the volume of solute divided by the total volume of the solution and multiplied by 100.

In your case, the solution will have a v/v percent concentration of

#"%v/v" = (5cancel("L"))/(166cancel("L")) * 100 = "3.01%"#

Now, for your second solution, the total volume of the solution will be

#V_"solution" = V_"solute" + V_"water"#

This means that, in order to get a 3.01% v/v solution that uses 2 L of water, you need to add

#"%v/v" = V_"solute"/V_"solution" * 100#

#"3.01%" = V_"solute"/(V_"solute" + "2 L") * 100#

#3.01 * V_"solute" + 6.02 = 100 * V_"solute"#

#V_"solute" = 6.02/96.99 = "0.0621 L" = "62 mL"#

So, to prepare a 3.01% v/v solution with 2 L of water, you need to use 62 mL of concentrate.

ALTERNATIVE METHOD

Alternatively, you can use something called the dilution factor, which is defined as

#"DF" = V_f/V_i#, where

#V_i# - the volume of the aliquot;
#V_f# - the volume of the solution.

The first solution will have a dilution factor of

#"DF" = (166cancel("L"))/(5cancel("L")) = "33.2"#

For the second solution, you know that the volume of the solution is

#V_f = V_"solute" + V_"water" = V_"solute" + "2 L"#

The dilution factor must be the same, so you have

#"DF" = (V_"solute" + "2 L")/V_"solute" = 33.2#

#V_"solute" = "2 L"/32.2 = "0.0621 L" = "62 mL"#