Question #cbbe3

2 Answers
Jan 19, 2016

Answer:

#color(blue)(1.25 %)#
I have given both the shortcut method and also explained in detail what makes the shortcut work

Explanation:

#color(blue)("Assumption 1")# The concentration is specified by percentage

#color(blue)("Assumption 2")# The original solution is 100%

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The wording; "diluted to" is stating the #0.25# forms part of the #20#

Thus, expressed as a fraction we have: #0.25/20#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Changing this to a percentage")#

#color(brown)("Shortcut method")#

#color(green)((0.25 -: 20) xx100-> 0.25/20xx100=1.25%)#

#:::::::::::::::::::::::::::::#

#color(brown)("Introduction to first principles to demonstrate what percentage is")#

What #0.25/20# actually stating is that for every 20 of the whole we have 0.25 of what we are looking at.

To change this to percent we need to have #("some unknown value")/100# which is the standardisation we call percent. That is how many have we got of what we are looking at for every 100 of the whole.

#color(brown)("Applying first principles to solve the problem")#

For a moment lets think of the #0.25/20# as a ratio

If we can retain this ratio but change the bottom number (denominator) to 100 we have our answer

Known: #20xx5=100#

So if we multiply #0.25/20# by 1 but in the form of #5/5# we change the way our number look without actually changing the intrinsic value of the whole.

#color(blue)("As a percentage")color(brown)(-> 0.25/20xx5/5= 1.25/100=1.25%)#

The #%# is just another way of writing: how many out of a count of 100

#color(brown)("The shortcut just by passes the bit where I multiplied by 1 in the ")##color(brown)("form of "5/5)#

Jan 21, 2016

Answer:

#1:80#

Explanation:

Let me offer a more chemical-oriented approach to this problem.

As you know, a solution's molarity is defined as the number of moles of solute divided by the volume of the solution - expressed in liters.

#color(blue)("molarity" = c = "moles of solute"/"liters of solution" = n/V)#

Now, the idea behind diluting a solution is that you can decrease its concentration by

  • keeping the number of moles of solute constant
  • increasing the volume of the solution by increasing the volume of the solvent

The underlying principle of a dilution

http://pharmrx.yolasite.com/concentration-and-dilutions.php

So, your initial sample has a total volume of #"0.25 mL"# and an unknown concentration #c_1#.

The target solution has a total volume of #"20 mL"# and an unknown concentration #c_2#.

The number of moles of solute present in these solutions will be

#n_1 = c_1 * V_1" "# and #" "n_2 = c_2 * V_2#

Since a dilution implies keeping the number of moles of solute constant

#n_1 = n_2#

you will get

#color(blue)(c_1V_1 = c_2V_2) -># the equation for dilution calculations

Your goal is to determine the dilution factor, so you can rearrange the above equation to get

#c_1/c_2 = V_2/V_1#

Plug in your values to get

#c_1/c_2 = (20 color(red)(cancel(color(black)("mL"))))/(0.25color(red)(cancel(color(black)("mL")))) = 80#

Therefore, the initial solution was eighty times as concentrated as the target solution

#c_1 = 80 xx c_2#

And that's how you find the dilution factor. Simply divide the final volume of the solution by the initial volume of the sample

#color(blue)("D"."F" = V_"final"/V_"initial") -># the equation for finding the dilution factor

A dilution factor of #80# will thus represent a #1:80# dilution.