Question: 79. Which of the following aqueous solution will exhibit highest boiling point?
(1) 0.01 M Urea
(2) 0.01 \mathrm{M} \mathrm{KNO_3}
(3) 0.01 \mathrm{M} \mathrm{Na_2SO_4}
(4) 0.015 \mathrm{M} \mathrm{C_6H_{12}O_6}
Answer: Option (3)
Explanation:
Elevation in boiling point is a colligative property and depends on the total number of solute particles present in solution.
The boiling point elevation is given by \Delta T_b = i K_b m,
where i is the van’t Hoff factor and m is molality.
Since the solvents are the same, comparison can be made using the product i \times m.
Urea is a non-electrolyte,
so i = 1. For 0.01 M urea: i \times m = 1 \times 0.01 = 0.01.
\mathrm{KNO_3} dissociates into two ions: \mathrm{K^+} and
\mathrm{NO_3^-}.
So, i = 2.
For 0.01 M \mathrm{KNO_3}: i \times m = 2 \times 0.01 = 0.02.
\mathrm{Na_2SO_4} dissociates into three ions: 2\mathrm{Na^+} and
\mathrm{SO_4^{2-}}.
So, i = 3.
For 0.01 M \mathrm{Na_2SO_4}: i \times m = 3 \times 0.01 = 0.03.
Glucose is a non-electrolyte, so i = 1.
For 0.015 M \mathrm{C_6H_{12}O_6}: i \times m = 1 \times 0.015 = 0.015.
Since \mathrm{Na_2SO_4} has the highest value of i \times m,
it produces the maximum elevation in boiling point.
Therefore, the aqueous solution of 0.01 \mathrm{M} \mathrm{Na_2SO_4} exhibits the highest boiling point.