# 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"?

May 21, 2017

This is essentially a proportional reasoning question: the concentration will be $\frac{1.52}{4.98} \times 36$ $g {L}^{-} 1$ = $11$ $g {L}^{-} 1$.

#### 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 $36$ $g {L}^{-} 1$ yields an osmotic pressure of $4.98$ bar, then a concentration of $x$ $g {L}^{-} 1$ will yield an osmotic pressure of $1.52$ bar.

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

May 21, 2017

${D}_{2} = \text{11 g/L}$

Osmotic pressure $\Pi$ is given by:

$\boldsymbol{\Pi = i c R T}$

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 $\text{K}$.

Given an osmotic pressure in $\text{bar}$, the units of $c$ must be:

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

$=$ $\text{mol/L}$

Converting to a concentration in $\text{g/L}$ (i.e. mass concentration) would mean that:

$\Pi M = i c M R T \equiv i D R T$,

where $M$ is the molar mass in $\text{g/mol}$, and $D$ is the mass concentration in $\text{g/L}$.

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

$P = \frac{n}{V} R T$

$\implies P M = \frac{n M}{V} R T = D R T$.)

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

${\Pi}_{1} M = i {D}_{1} R T$
${\Pi}_{2} M = i {D}_{2} R T$

Thus, the new concentration is gotten as follows:

${\Pi}_{1} / {D}_{1} = {\Pi}_{2} / {D}_{2} = \frac{i R T}{M}$

$\implies \textcolor{b l u e}{{D}_{2}} = {D}_{1} \left({\Pi}_{2} / {\Pi}_{1}\right)$

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

$=$ $\textcolor{b l u e}{\text{11 g/L}}$