# 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

#### Answer:

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")#