If an aqueous solution starts at #"36 g/L"# and #"4.98 bar"#, what is the new concentration in #"g/L"# needed to accomplish an osmotic pressure of #"1.52 bar"#?

2 Answers
May 21, 2017

Answer:

This is essentially a proportional reasoning question: the concentration will be #(1.52)/(4.98)xx 36# #gL^-1# = #11# #gL^-1#.

Explanation:

The osmotic pressure is directly proportional to the concentration, when the temperature is kept constant (as it is in this case).

If a concentration of #36# #gL^-1# yields an osmotic pressure of #4.98# bar, then a concentration of #x# #gL^-1# will yield an osmotic pressure of #1.52# bar.

After that it's just a matter of solving for #x#.

May 21, 2017

#D_2 = "11 g/L"#


Osmotic pressure #Pi# is given by:

#bb(Pi = icRT)#

where:

  • #i# is the van't Hoff factor. For non-electrolytes, #i = 1#, as they hardly dissociate.
  • #c# is the concentration in the appropriate units.
  • #R# is the universal gas constant.
  • #T# is the temperature in #"K"#.

Given an osmotic pressure in #"bar"#, the units of #c# must be:

#cancel"bar"/(("L"cdotcancel"bar""/mol"cdotcancel"K") xx cancel"K")#

#=# #"mol/L"#

Converting to a concentration in #"g/L"# (i.e. mass concentration) would mean that:

#PiM = icMRT -= iDRT#,

where #M# is the molar mass in #"g/mol"#, and #D# is the mass concentration in #"g/L"#.

(This can be seen as analogous to the ideal gas law:

#P = n/VRT#

#=> PM = (nM)/VRT = DRT#.)

Given two states with the same temperature and van't Hoff factor (due to the same solute):

#Pi_1M = iD_1RT#
#Pi_2M = iD_2RT#

Thus, the new concentration is gotten as follows:

#Pi_1/D_1 = Pi_2/D_2 = (iRT)/M#

#=> color(blue)(D_2) = D_1 (Pi_2/Pi_1)#

#= "36 g"/"L" xx ("1.52 bar"/"4.98 bar")#

#=# #color(blue)("11 g/L")#