Question

Calculate the freezing point of the solution.After mixing these 2 bottles together, set Kf of water = 1.86 ° C / m.
Bottle 1 contained 0.3 grams of glucose in 1000 grams of water.
The 2nd bottle contains 0.5 mol fructose in 1000 grams of water.

Answer 1

Answer : The freezing point of solution is 273.467 K

Explanation : Given,

Mass of glucose (solute) = 0.3 g

Mass of water (solvent) = 1000 g = 1 kg

Moles of fructose (solute) = 0.5 mol

Mass of water (solvent) = 1000 g = 1 kg

Molar mass of glucose = 180 g/mole

First we have to calculate the moles of glucose.

$\text{Moles of glucose}=\frac{\text{Mass of glucose}}{\text{Molar mass of glucose}}=\frac{0.3g}{180g/mole}=0.00167mole$

Now we have to calculate the total moles after mixing.

$\text{Total moles}=\text{Moles of glucose}+\text{Moles of fructose}$

$\text{Total moles}=0.00167+0.5=0.502moles$

Now we have to calculate the molality.

$\text{Molality}=\frac{\text{Total moles}}{\text{Mass of water(solvent in kg)}}$

$\text{Molality}=\frac{0.502mole}{(1+1)kg}=0.251mole/kg$

Now we have to calculate the freezing point of solution.

As we know that the depression in freezing point is a colligative property that means it depends on the amount of solute.

Formula used :  

$\Delta T_f=K_f\times m$

$T^o_f-T_f=K_f\times m$

where,

$\Delta T_f$ = change in freezing point

$T_f^o$ = temperature of pure water = $0^oC$

$T_f$ = temperature of solution = ?

$K_f$ = freezing point constant of water = $1.86^oC/m$

m = molality = 0.251 mole/kg

Now put all the given values in this formula, we get

$0^oC-T_f=1.86^oC/m\times 0.251mole/kg$

$T_f=-0.467^oC=273.467K$

conversion used : $K=273+^oC$

Therefore, the freezing point of solution is 273.467 K