What is the molality of the resultant solution when equal masses of a #"2 molal"# solution of #"NaOH"# and a #"4 molal"# solution of #"NaOH"# are combined?

About #"2.966 molal NaOH"#. This should make sense since it is close to a simple average.

You can always add up the mols of #"NaOH"# at the end. They are extensive. The solution volumes are extensive as well since the solutes are identical and the solvents are identical, and they can also be added separately at the end.

We begin with:

#"2 mols NaOH"/"1 kg water"#, soln 1

#"4 mols NaOH"/"1 kg water"#, soln 2

Using a molar mass of #"38.988 g NaOH/mol"#, we can rewrite this as:

• #"77.976 g NaOH"/"1000 g water"# for #"2 molal"# (#"1077.976 g"#), soln 1
• #"155.952 g NaOH"/"1000 g water"# for #"4 molal"# (#"1155.952 g"#), soln 2

Molality, as with all concentrations, is intensive, so it can be scaled however we want.

We want to scale down the second solution ratio to equal #"1077.976 g"#, since these solutions have apparently the same mass, so we have two equations and two unknowns:

#m_(NaOH)/(1000m_(H_2O)) = 0.155952# #" "" "" "" "" "bb((1))#

#m_(NaOH) + 1000m_(H_2O) = 1077.976# #" "" "bb((2))#

with the mass units of the solute in grams and of the solvent in kilograms.

Solving for #m_(NaOH)# in #(2)# gives:

#m_(NaOH) = 1077.976 - 1000m_(H_2O)#

So, plugging back into #(1)#:

#(1077.976 - 1000m_(H_2O))/(1000m_(H_2O)) = 0.155952#

#1.077976/(m_(H_2O)) - 1 = 0.155952#

#1.077976/(m_(H_2O)) = 1.155952#

So, solving for #m_(H_2O)#, the #"kg"# mass of water in the solution needs to be:

#m_(H_2O) = 1.077976/1.155952 = "0.9325 kg"#

And from #(2)#, the mass of #"NaOH"# is therefore:

#m_(NaOH) = 1077.976 - 932.5 = "145.476 g NaOH"#

We can check that this is correct in two ways:

1. This still gives a molality for the first solution as

#(77.976 cancel"g NaOH" xx "1 mol NaOH"/(38.988 cancel"g NaOH"))/"1.0000 kg water"#

#=# #"2 molal NaOH"#

as specified in the question, as well as a molality of

#(145.476 cancel"g NaOH" xx "1 mol NaOH"/(38.988 cancel"g NaOH"))/"0.9325 kg water"#

#~~# #"4 molal NaOH"#

for the second solution.

2. The total masses of each solution before combination are indeed equal as specified in the question:

Solution 1: #"77.976 g NaOH" + "1000 g water" = "1077.976 g"#
Solution 2: #"145.476 g NaOH" + "932.5 g water" = "1077.976 g"#

So, now that we are absolutely sure, we can add the mols and volume to get the resultant solution molality:

#("2 mols NaOH" + "3.731 mols NaOH")/("1 kg water" + "0.9325 kg water")#

#=# #color(blue)("2.966 molal NaOH")#