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"#?

Question

Olivia Anton · Accepted Answer

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

When the temperature is held constant, as it is in this instance, the osmotic pressure is directly proportional to the concentration. 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#.

Wyatt Adkins · Answer

#D_2 = "11 g/L"# Osmotic pressure #Pi# is given by: #bb(Pi = icRT)# where: 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")#

Jeremiah Adams · Answer

undefined To find the new concentration $ C_2 $ needed to achieve an osmotic pressure of $ P_2 = 1.52 $ bar:

$$ C_2 = \frac{{P_1 \cdot C_1}}{{P_2}} $$

Given:
Initial concentration $ C_1 = 36 $ g/L,
Initial osmotic pressure $ P_1 = 4.98 $ bar,
Target osmotic pressure $ P_2 = 1.52 $ bar.

$$ C_2 = \frac{{4.98 \, \text{bar} \times 36 \, \text{g/L}}}{{1.52 \, \text{bar}}} $$
$$ C_2 = \frac{{179.28}}{{1.52}} $$
$$ C_2 \approx 118 \, \text{g/L} $$

So, the new concentration needed to achieve an osmotic pressure of 1.52 bar is approximately 118 g/L.