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}}=(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}=(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* m`

`T^o_f-T_f=K_f* 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* 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