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

Consider #AlCl_3# to be a strong electrolyte.

1 Answer
Jul 23, 2016

Answer:

#"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.

chemwiki.ucdavis.edu/Wikitexts/University_of_California_Davis/UCD_Chem_002B/UCD_Chem_2B%3A_Gulaca

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.