Question

Which of the following aqueous solutions should have the highest boiling point?
A. 1.0M NaOH
B. 1.0M $N{{a}_{2}}S{{O}_{4}}$
C. 1.0M $N{{H}_{4}}N{{O}_{3}}$
D. 1.0M $KN{{O}_{3}}$

Answer 1

Hint:. The molarity of all these solutions is given to be the same. So the boiling point has to be decided by the number of ions that can be formed when the given compounds are dissociated into their respective ions.

Complete step by step answer:
-Many types of compound can be used as electrolytes but the most preferred compounds are acids, bases and salts. Electrolytes dissolve in water to form a solution and that solution conducts electricity by dissociating into ions.
Eg. $NaCl\left( s \right)\to N{{a}^{+}}\left( aq \right)+C{{l}^{-}}\left( aq \right)$
Water is added as solvent so that the ions can become mobile. Only then they can conduct electricity.

-Now coming to the strength of the electrolytes. There are two types of electrolytes, strong and weak. Strong electrolytes are those which can dissociate to a very large extent to give the respective cations and anions. Weak electrolytes are those which dissociate very less and so the ions formed are not much which reduces the conductivity of the solution.

-Weak electrolytes do not dissociate completely into their ions and the difference in their dissociation can be shown in the form of a quantity called the dissociation degree $\left( \alpha \right)$

-Van't hoff factor is the measure of the degree of dissociation for weak electrolytes. It is directly proportional to the degree of dissociation. It means that if the value of the van't hoff factor is more, more electrolyte will dissociate itself to form ions. If the value is less, then lesser amount of electrolyte will dissociate.

-There is a relationship between the van't hoff factor i and degree of dissociation $\left( \alpha \right)$
\[\alpha =\dfrac{i-1}{n-1}\]
Here n is the number of ions that can be produced by 1 mole of the electrolyte on dissociation.
${{A}_{x}}{{B}_{y}}\rightleftharpoons x{{A}^{y+}}+y{{B}^{x-}}$
If the degree of dissociation is $\left( \alpha \right)$, then equal moles will be formed as ions and the moles of the electrolyte left will be given as $\left( 1-\alpha \right)$ .

-The boiling point elevation is directly proportional to the van't hoff factor. So this factor decides the boiling point in the given question.

-In the option A, C and D, the van't hoff factor is 2 as there are 2 ions into which the compound dissociates. In option B, there are 3 such ions and so the van't hoff factor is 3. So it has more boiling point than other options.
So, the correct answer is “Option B”.

Note: The expression by which the boiling point elevation is connected to the van't hoff factor is $\Delta T=i\times {{K}_{b}}\times m$ where K is the boiling point elevation constant and m is the molality of the solution.