Question

What is the boiling point of an ionic solution containing 29.7 grams $N{{a}_{2}}S{{O}_{4}}$ and 84.4 grams water, assuming 100% ionization?

Answer 1

The solute in this question is sodium sulfate ($N{{a}_{2}}S{{O}_{4}}$) and the solvent is water. The formula that can be used to solve the question will be:
$\Delta {{T}_{b}}=i\text{ x }{{\text{K}}_{b}}\text{ x m}$
Where $\Delta {{T}_{b}}$ is the elevation in the boiling point of a solution when the solute is added to a solvent, i is the van’t Hoff factor, ${{K}_{b}}$ is the elevation in boiling point constant, and m is the molality of the solution.

Complete answer:
We have to the boiling point of the solution in which sodium sulfate and water are present. The solute is sodium sulfate ($N{{a}_{2}}S{{O}_{4}}$) and the solvent is water and the question says that there is 100% ionization of the solute molecules. The reaction will be:
$N{{a}_{2}}S{{O}_{4}}\to 2N{{a}^{+}}+SO_{4}^{2-}$
So, we can see that the solute is dissociating into 3 ions.
The formula that can be used to solve the question will be:
$\Delta {{T}_{b}}=i\text{ x }{{\text{K}}_{b}}\text{ x m}$
Where $\Delta {{T}_{b}}$ is the elevation in the boiling point of a solution when the solute is added to a solvent, i is the van’t Hoff factor, ${{K}_{b}}$ is the elevation in boiling point constant or ebullioscopic constant, and m is the molality of the solution.
Here it will be 3 because 3 ions are formed.
Number of moles for 29.7 grams of $N{{a}_{2}}S{{O}_{4}}$ will be:
$Moles=\dfrac{29.7}{142.04}=0.2091$
So, the molality of the solution will be:
$Molality=\dfrac{0.2091}{0.0844}=2.477$
As we know that the ebullioscopic constant of water is ${{0.512}^{\circ }}C\text{ }kg\text{ }mo{{l}^{-1}}$. Now putting all these values in the formula, we get:
$\Delta {{T}_{b}}=3\text{ x 0}\text{.512 x 2}\text{.477 = 3}\text{.80}{{\text{5}}^{\circ }}\text{C}$
So, the rise in the temperature is ${{3.805}^{\circ }}C$, this is equal to the difference of the boiling point of the pure solvent and solution. This is written as:
$\Delta {{T}_{b}}={{T}_{b}}-T_{b}^{\circ }$
$3.805={{T}_{b}}-100$
${{T}_{b}}=100+3.805={{103.85}^{\circ }}C$
So, the boiling point of the solution is ${{103.805}^{\circ }}C$.

Note:
In the question, the number of moles is calculated by taking the ratio of the given mass of sodium sulfate and the molecular mass of sodium sulfate, and the molality of the solution is calculated as the ratio of moles of sodium sulfate and the mass of the water in kg.