Question

Which of the following aqueous solutions should have the highest boiling point?

• A. $1.0MKN{O}_3$
  
• B. $1.0MN{H}_{4}N{O}_3$
  
• C. $1.0MNaOH$
  
• D. $1.0MN{a}_{2}S{O}_4$
  

Answer 1

The correct option is D $1.0MN{a}_{2}S{O}_4$

• Van’t Hoff factor is the ratio of number of particles after association/dissociation to the number of particles before association/dissociation.
• As we know greater the value of van't Hoff factor, larger will be the elevation in boiling point and hence higher will be the boiling point of the solution.

(A) $1.0MNaOH\to N{a}^{+}+O{H}^{-}$

Total number of ions = $2$

van't Hoff factor = 2

(B) $1.0MN{a}_{2}S{O}_4\to 2N{a}^{+}+S{O}_4^{2-}$
Total number of ions = $3$

van't Hoff factor = 3

(C) $1.0MN{H}_{4}N{O}_3\to N{H}_4^{+}+N{O}_3^{-}$

Total number of ions = $2$

van't Hoff factor = $2$

(D) $1.0MKN{O}_3\to K^{+}+N{O}_3^{-}$
Total number of ions = $2$

van't Hoff factor = $2$

So, the van’t Hoff factor of $1.0MN{a}_{2}S{O}_4$ is highest.

Hence, correct option is (B).