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

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.

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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("atm" * "L")/("mol" * "K")`
`T` - the absolute temperature of the solution

Now, you're dealing with a `"0.050 M"` solution of aluminium chloride, `"AlCl"_3`, which is said to be a strong electrolyte.

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

`"AlCl"_ (3(aq)) -> "Al"_ ((aq))^(3+) + 3"Cl"_ ((aq))^(-)`

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 * 0.050 color(red)(cancel(color(black)("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 = color(green)(|bar(ul(color(white)(a/a)color(black)("4.5 atm")color(white)(a/a)|)))`

The answer is rounded to two sig figs.