What information is needed to convert from molality to molarity?

A molality is defined as:

#"mols solute"/"kg solvent",#

whereas a molarity is defined as:

#"mols solute"/"L solution"#.

Note that a solvent is not a solution, so you need to account for the volume contributed by the solute. Therefore, you need the known density of the pure solute.

Let's say we had a #"0.5000 m"# solution made using stock #38%# #"HCl"# in water at #4^@ "C"#, at which #rho_w = "1.00 g/mL"# and the density of the the #38%# #"HCl"# we used is #rho_"38% HCl" ~~ 1".189 g/mL"#.

We know that the molar mass of the solute is #M_r^"HCl" = 1.0079 + 35.453 = "36.4609 g/mol"#. What we need to do is convert #"kg solvent"# to #"L solution"#.

The solvent was water, but the solution must incorporate #"HCl"# and water, so we need to:

• assume volumes are additive
• use the density of #38% "HCl"# and add the resultant volume.

Based on the molality, we used #V_"solvent" = "1 L water"# and #"0.5000 mols"# #"HCl"#. So, let's get #"0.5000 mols"# #"HCl"# into volume #"HCl"#.

#V_"solute"#

#= 0.5000 cancel("mols HCl") xx (36.4609 cancel("g HCl"))/cancel("mol HCl") xx "1 mL"/(1.189 cancel("g HCl")) xx "1 L"/(1000 cancel("mL")#

#=# #"0.01533 L HCl"#

Finally, we have:

#V_"soln" ~~ V_"solute" + V_"solvent"#

#~~# #"1.01533 L"#

Therefore, the resultant molarity would be:

#color(blue)(["HCl"])#

#= "mols HCl"/("L HCl + H"_2"O")#

#~~# #color(blue)("0.4925 M")#