Question

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)?

Answer 1

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.