# What is the osmotic pressure of a 0.050 M solution of AlCl_3 in water that is at 0.00°C?

## Consider $A l C {l}_{3}$ to be a strong electrolyte.

Jul 23, 2016

$\text{4.5 atm}$

#### Explanation:

Osmotic pressure is all about the number of particles of solute present in solution.

As you know, a solution's osmotic pressure represents the minimum pressure that must be applied in order to prevent water from flowing inward through a semipermeable membrane.

Osmotic pressure is directly related to osmosis, which represents the movement of water through a semipermeable membrane from a region of lower solute concentrations to a region of higher solute concentration.

Simply put, the higher the concentration of particles of solute in a solution, the greater the intake of water through the semipermeable membrane, and thus the higher the osmotic pressure.

Osmotic pressure, $\Pi$, is calculated using the equation

color(blue)(|bar(ul(color(white)(a/a)Pi = i * c * RTcolor(white)(a/a)|))

Here

$i$ - the van't Hoff factor, equal to $1$ for non-electrolytes
$c$ - the concentration of the solution
$R$ - the universal gas constant, useful here as $0.0821 \left(\text{atm" * "L")/("mol" * "K}\right)$
$T$ - the absolute temperature of the solution

Now, you're dealing with a $\text{0.050 M}$ solution of aluminium chloride, ${\text{AlCl}}_{3}$, which is said to be a strong electrolyte.

Aluminium chloride dissociates completely in aqueous solution to produce aluminium cations and chloride anions

${\text{AlCl"_ (3(aq)) -> "Al"_ ((aq))^(3+) + 3"Cl}}_{\left(a q\right)}^{-}$

Notice that every mole of aluminium chloride that dissociates produces $1$ mole of aluminium cations and $3$ moles of chloride anions.

The van't Hoff factor essentially tells you how many moles of particles of solute you get per mole of solute dissolved in solution. In this case, $1$ mole of solute dissolved in solution produces $4$ moles of ions, which means that the van't Hoff factor is equal to

$i = 4$

Plug in your values into the above equation and solve for $\Pi$ -- do not forget to convert the temperature from degrees Celsius to Kelvin

$\Pi = 4 \cdot 0.050 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mol")))/color(red)(cancel(color(black)("L"))) * 0.0821("atm" * color(red)(cancel(color(black)("L"))))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 0.00)color(red)(cancel(color(black)("K}}}}$

$\Pi = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{4.5 atm}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs.