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When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. If 2.5 mol of aluminum nitrate is added to the same amount of water, by how much will the boiling point be changed? Show all calculations leading to your answer OR use 3 – 4 sentences to explain your answer.

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MissPhiladelphia
When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. If 2.5 mol of aluminum nitrate is added to the same amount of water, the boiling point will raise to 4 degrees Celsius. This is because aluminum nitrate will dissociate into 4 ions in the solution. The ions will act as a blanket for the solution thus increasing the boiling point even more.

Answer:The boiling point of the water by adding 2.5 moles of aluminium nitrate will be changed by [tex]4^o C[/tex].

Explanation:

When sugar is dissolved is x kilograms of water

[tex]\Delta T_1=1^o C=274 K(0^oC=273 K)[/tex]

Number of moles of sugar dissolved = 2.5 mol

i = Van'T Hoff factor of sugar = 1

[tex]\Delta T_1=iK_f\times molality=iK_f\times \frac{\text{moles of sugar}}{\text{weight of solvents in kg}}[/tex]

[tex]Delta T_1=274 K=1\times K_f\times \frac{2.5 mol}{x kg}[/tex]..(1)

When aluminium nitrate is dissolved x kilogram of water:

Number of moles of aluminium nitrate dissolved = 2.5 mol

[tex]Al(NO_3)_3(aq)\rightarrow Al^{3+}(aq)+3NO_3^-(aq)[/tex]

i' = Van'T Hoff factor of aluminium nitrate = 4

Aluminium nitrate being ionic compound will get dissociated into its constituting ions. The value of ' i' ' of the ionic compound is equal to number of discrete ions in a formula unit of a substance.

[tex]\Delta T_2=i'K_f\times \frac{\text{moles of }Al(NO_3)_3}{\text{weight of solvents in kg}}[/tex]

[tex]\Delta T_2=4\times K_f\times \frac{2.5 mol}{x kg}[/tex]...(2)

On dividing (1)and (2), we get

[tex]\Delta T_2=4\times 274 K=4\times(1^oC)=4^oC[/tex]  

The boiling point of the water by adding 2.5 moles of aluminium nitrate will be changed by [tex]4^o C[/tex].

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