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
This is essentially a proportional reasoning question: the concentration will be
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
After that it's just a matter of solving for
#D_2 = "11 g/L"#
Osmotic pressure
#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
#cancel"bar"/(("L"cdotcancel"bar""/mol"cdotcancel"K") xx cancel"K")#
#=# #"mol/L"#
Converting to a concentration in
#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")#