375mL of a 0.455M sodium chloride (#NaCl#) solution is diluted with 1.88L of water. What is the new concentration in molarity?

1 Answer
Apr 24, 2016


#"0.0767 mol L"^(-1)#


The thing to keep in mind when you're dealing with a dilution is that the number of moles of solute must remain constant.

That means that the stock solution and the diluted solution will contain the same number of moles of solute.

So, your strategy here will be to calculate how many moles of solute, which in your case is sodium chloride, #"NaCl"#, are present in the stock solution.

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

You will have

#n_(NaCl) = "0.455 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(375 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(green)("volume expressed in liters"))#

#n_(NaCl) = "0.1706 moles NaCl"#

This is exactly how many moles of sodium chloride must be present in the diluted solution. Calculate the volume of the diluted solution first

#V_"diluted" = V_"stock" + V_"water"#

#V_"diluted" = 375 * 10^(-3)"L" + "1.88 L" = "2.255 L"#

This means that the concentration of the diluted solution will be

#c_"diluted" = "0.1706 moles"/"2.255 L" = color(green)(|bar(ul(color(white)(a/a)"0.0767 mol L"^(-1)color(white)(a/a)|)))#

The answer is rounded to three sig figs.