You want to produce 579 mL of a 0.440 M solution of NaCl. You have a 1.40 M solution. How many mL of it should you use and dilute with water?
1 Answer
Explanation:
Let's start by assuming that you're not familiar with the formula for dilution calculations, so that you can't just plug in your values and do a quick calculation.
So, the idea with dilution calculations is that the number of moles of solute remains constant when diluting a given sample.
In essence, to dilute a solution you need to decrease the concentration of the solute by keeping the number of moles of solute constant and Increasing the volume.
As you know, molarity is defined as moles of solute, which in your case will be sodium chloride,
#color(blue)(c = n/V)#
Use the molarity and volume of the target solution to determine how many moles of solute it must contain
#color(blue)(c = n/V implies n = c * V)#
#n_(NaCl) = "0.440 M" * 579 * 10^(-3)"L" = "0.25476 moles NaCl"#
This is exactly how many moles of solute must be present in the sample of stock solution used to prepare the target solution.
This means that you can use the definition of molarity to determine what volume of the stock solution would contain this many moles of sodium chloride
#color(blue)(c = n/V implies V = n/c)#
#V = (0.25476 color(red)(cancel(color(black)("moles"))))/(1.40 color(red)(cancel(color(black)("moles")))/"L") = "0.182 L"#
Expressed in milliliters, the answer will be
#V_"stock" = color(green)("182 mL")#
Alternatively, you can use the formula for dilution calculations, which expresses the exact same concept
#overbrace(c_1V_1)^(color(purple)(stackrel("moles of solute")("in stock solution"))) = overbrace(c_2V_2)^(color(red)(stackrel("moles of solute")("in target solution")))" "# , where
In this case, you would get
#V_1 = c_2/c_1 * V_2#
#V_1 = (0.440 color(red)(cancel(color(black)("M"))))/(1.40color(red)(cancel(color(black)("M")))) * "579 mL" = color(green)("182 mL")#
