The relative molecular weight of acetic acid is 60.05 and its density is 1.049g/ml. Describe how you would prepare 1L of 2M solution of acetic acid?

#"119.1 g"# of the starting solution can be measured out and poured into some water, and the rest of the volume is to be occupied by water up until the #"1-L"# mark on your volumetric flask.

Alternatively you can somehow measure out #"114.5 mL"# of the starting solution, but in practice, getting the mass is more accurate and convenient.

Well, you would use some stock acetic acid solution like this one for starters. It's basically #100%# acetic acid. The question is, what concentration do you get from that?

#(1.049 cancel"g HOAc")/cancel"mL" xx (1000 cancel"mL")/"1 L" xx "1 mol"/(60.05 cancel"g HOAc")#

#=# #"17.47 M"#

This would be your starting concentration. Since you want #"1 L"# of #"2 M"# solution (and let's get it at least 3 decimal places), you'll want to have #"1.000 L"# of final solution volume in order to get a

#17.47/2.000 = 8.735#-fold dilution.

That means your starting volume should be

#2.000/17.47 = 1.000/8.735 = 0.1145# times the final volume of #"1.000 L"#

(or #ul"0.1145 L"#).

And to confirm this beyond simply rationalizing it:

#M_1V_1 = M_2V_2#

#M# and #V# are molar concentration in #"M"# and volume in #"L"#, respectively.

#V_1 = M_2/M_1 V_2#

#= ("2.000 M")/("17.47 M") xx "1.000 L"#

#=# #ulcolor(blue)"0.1145 L"# as we got conceptually.

Since it's so concentrated acid,

#"1) "# I would first put in roughly #"50 mL"# of water, let's say, then record the starting mass (or tare the scale that weighs this).

#"2) "# Use the density of the acetic acid solution to measure out a mass of it:

#114.5 cancel"mL" xx "1.049 g HOAc"/cancel"1 mL" = "119.1 g HOAc soln"#

Add the acetic acid stock solution to the water until you have added this much mass to the water. If you've tared the scale, it should be easy to tell.

#"3) "# Then simply add water to fill the flask up to the #"1-L"# mark.

That ensures that you make a proper #"2 M"# solution to approx. the third decimal place, and avoids issues with the change in solution volume due to the presence of solute.