# 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)?

Sep 12, 2015

Apply the equations for colligative properties and osmotic pressure.

#### Explanation:

The change in boiling point will be

$\Delta {T}_{b} = i \cdot b \cdot {K}_{b}$

$\Delta {T}_{f} = i \cdot b \cdot {K}_{f}$

osmotic pressure is

$\prod = i \cdot M R T$

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

$\text{molality" = "moles of solute"/"kilograms of solvent}$

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

Fructose does not dissociate so $i = 1$. Once you have calculated $\Delta T b$, you can calculate the new boiling point to be

${T}_{b} = 100 + \Delta T b$

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

${T}_{f} = 0 - \Delta {T}_{f}$

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

$\text{molarity" = "moles of solute"/"liters of solution}$

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