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"`?

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`.

`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")`