What is the osmotic pressure for #"1.95 g"# of sucrose dissolved in #"150 mL"# solution at #25^@ "C"#? #"FW"# #=# #"342.2965 g/mol"#

1 Answer
Truong-Son N.
May 31, 2017

#Pi = "0.9292 atm"#.

You can read more about osmotic pressure (and other colligative properties) here.

The osmotic pressure #Pi#, the pressure needed to stop the solvent flow from low to high concentration across a semi-permeable membrane, is given by:

#Pi = icRT#,

where:

  • #i# is the van't Hoff factor, as seen in freezing point depression and boiling point elevation. It can be approximated by the number of particles per formula unit in solution.
  • #c# is the concentration in whatever units are appropriate. In this case, if #Pi# is in #"atm"#, then #c# is in #"mol/L"# when...
  • ...the universal gas constant #R# is #"0.082057 L"cdot"atm/mol"cdot"K"#.
  • #T# is the temperature in #"K"#.

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We are indeed given the correct info to calculate the molarity of the sucrose solution.

#1.95 cancel"g suc" xx "1 mol suc"/(342.2965 cancel"g suc")#

#=# #"0.005697 mols suc"#

So, the molarity is:

#c = "0.005697 mols suc"/(150 xx 10^(-3) "L soln")#

#=# #"0.03798 M"#

Therefore, the osmotic pressure at #25^@ "C" = "298.15 K"# for the NONELECTROLYTE sucrose is:

#color(blue)(Pi) = (1)(0.03798 cancel"mol/L")(0.082057 cancel"L"cdot"atm/"cancel"mol"cdotcancel"K")(298.15 cancel"K")#

#=# #color(blue)("0.9292 atm")#,

since nonelectrolytes have #i = 1#, having hardly any further dissociation at all.

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