20 g of fructose is being dissolved in 150g of H2O at 25°C.Calculate the boiling n freezing point of the solution.what is the osmotic pressure of the solution( density of water=100g/L)?

1 Answer
Sep 12, 2015

Answer:

Apply the equations for colligative properties and osmotic pressure.

Explanation:

The change in boiling point will be

#DeltaT_b = i * b * K_b#

There is a similar equation for freezing point depression

#DeltaT_f = i * b * K_f#

osmotic pressure is

#prod = i * MRT#

The trick to solving this, is looking up the value of molal boiling point elevation constant for water, once you have that you need to calculate the molality (#b#) of the solution.

It is useful to memorize the definition for molality

#"molality" = "moles of solute"/"kilograms of solvent"#

Look up the molar mass of fructose, calculate the number of moles, and then divide by 0.15 kg of #"H"_2"O"# (since that is the solvent).

Fructose does not dissociate so #i=1#. Once you have calculated #DeltaTb#, you can calculate the new boiling point to be

#T_b = 100 + DeltaTb#

You can go to a similar process to calculate the new freezing point, by using the freezing point equation (#b# will be the same, #i# is the same, but #K_f# is different than #K_b#).

Note the new freezing point will be

#T_f= 0 - DeltaT_f#

The formula for osmotic presure is slightly different, you need #M# which is molarity

#"molarity" = "moles of solute"/"liters of solution"#

Use the proper value of #R#, and #T# is expressed in Kelvin.